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Relation between sides and angles, Solutions of triangles Questions in English
Class 11 Mathematics · Trigonometrical Equations · Relation between sides and angles, Solutions of triangles
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MediumMCQ
If $p_1, p_2, p_3$ are altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively,and $\Delta$ is the area of the triangle,then $p_1^{-1} + p_2^{-1} - p_3^{-1}$ is equal to
A
$\frac{s - a}{\Delta}$
B
$\frac{s - b}{\Delta}$
C
$\frac{s - c}{\Delta}$
D
$\frac{s}{\Delta}$
Solution
(C) We know that the area of a triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
From this,we get $p_1 = \frac{2\Delta}{a}$,$p_2 = \frac{2\Delta}{b}$,and $p_3 = \frac{2\Delta}{c}$.
Therefore,$p_1^{-1} + p_2^{-1} - p_3^{-1} = \frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta}$.
This simplifies to $\frac{a + b - c}{2\Delta}$.
Since the semi-perimeter $s = \frac{a + b + c}{2}$,we have $a + b + c = 2s$,so $a + b - c = (a + b + c) - 2c = 2s - 2c = 2(s - c)$.
Substituting this into the expression,we get $\frac{2(s - c)}{2\Delta} = \frac{s - c}{\Delta}$.
From this,we get $p_1 = \frac{2\Delta}{a}$,$p_2 = \frac{2\Delta}{b}$,and $p_3 = \frac{2\Delta}{c}$.
Therefore,$p_1^{-1} + p_2^{-1} - p_3^{-1} = \frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta}$.
This simplifies to $\frac{a + b - c}{2\Delta}$.
Since the semi-perimeter $s = \frac{a + b + c}{2}$,we have $a + b + c = 2s$,so $a + b - c = (a + b + c) - 2c = 2s - 2c = 2(s - c)$.
Substituting this into the expression,we get $\frac{2(s - c)}{2\Delta} = \frac{s - c}{\Delta}$.
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MediumMCQ
Which of the following sets of data does not uniquely determine an acute-angled $\Delta ABC$ ($R$ = circum-radius)?
A
$a, \sin A, \sin B$
B
$a, b, c$
C
$a, \sin B, R$
D
$a, \sin A, R$
Solution
(D) In $\Delta ABC$,the Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
For option $(a)$: Given $a, \sin A, \sin B$. From $\frac{a}{\sin A} = 2R$,we find $R$. Then $b = 2R \sin B$ and $c = 2R \sin C = 2R \sin(180^{\circ} - (A+B)) = 2R \sin(A+B)$. Thus,the triangle is uniquely determined.
For option $(b)$: Given $a, b, c$. By the $SSS$ congruence criterion,the triangle is uniquely determined.
For option $(c)$: Given $a, \sin B, R$. We have $b = 2R \sin B$. Now we have two sides $a, b$ and the circum-radius $R$. Since $\sin A = \frac{a}{2R}$,angle $A$ is determined. Thus,the triangle is uniquely determined.
For option $(d)$: Given $a, \sin A, R$. We have $\frac{a}{\sin A} = 2R$. This is an identity that holds for any triangle with side $a$ and angle $A$ having circum-radius $R$. It does not provide enough information to fix the other sides or angles. Therefore,the triangle is not uniquely determined.
For option $(a)$: Given $a, \sin A, \sin B$. From $\frac{a}{\sin A} = 2R$,we find $R$. Then $b = 2R \sin B$ and $c = 2R \sin C = 2R \sin(180^{\circ} - (A+B)) = 2R \sin(A+B)$. Thus,the triangle is uniquely determined.
For option $(b)$: Given $a, b, c$. By the $SSS$ congruence criterion,the triangle is uniquely determined.
For option $(c)$: Given $a, \sin B, R$. We have $b = 2R \sin B$. Now we have two sides $a, b$ and the circum-radius $R$. Since $\sin A = \frac{a}{2R}$,angle $A$ is determined. Thus,the triangle is uniquely determined.
For option $(d)$: Given $a, \sin A, R$. We have $\frac{a}{\sin A} = 2R$. This is an identity that holds for any triangle with side $a$ and angle $A$ having circum-radius $R$. It does not provide enough information to fix the other sides or angles. Therefore,the triangle is not uniquely determined.
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DifficultMCQ
In the ambiguous case of a triangle,if $a, b$ and $A$ are given and there are two possible values for the third side,$c_1$ and $c_2$,then what is the value of $|c_1 - c_2|$?
A
$c_1 - c_2 = 2\sqrt{a^2 + b^2\sin^2 A}$
B
$c_1 - c_2 = 2\sqrt{a^2 - b^2\sin^2 A}$
C
$c_1 - c_2 = 4\sqrt{a^2 + b^2\sin^2 A}$
D
$c_1 - c_2 = 3\sqrt{a^2 - b^2\sin^2 A}$
Solution
(B) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
This can be rearranged as a quadratic equation in $c$: $c^2 - (2b\cos A)c + (b^2 - a^2) = 0$.
Let $c_1$ and $c_2$ be the two roots of this quadratic equation.
Then,the sum of roots is $c_1 + c_2 = 2b\cos A$ and the product of roots is $c_1 c_2 = b^2 - a^2$.
The difference between the roots is given by $|c_1 - c_2| = \sqrt{(c_1 + c_2)^2 - 4c_1 c_2}$.
Substituting the values: $|c_1 - c_2| = \sqrt{(2b\cos A)^2 - 4(b^2 - a^2)}$.
$|c_1 - c_2| = \sqrt{4b^2\cos^2 A - 4b^2 + 4a^2} = \sqrt{4a^2 - 4b^2(1 - \cos^2 A)}$.
Since $1 - \cos^2 A = \sin^2 A$,we get $|c_1 - c_2| = \sqrt{4a^2 - 4b^2\sin^2 A} = 2\sqrt{a^2 - b^2\sin^2 A}$.
This can be rearranged as a quadratic equation in $c$: $c^2 - (2b\cos A)c + (b^2 - a^2) = 0$.
Let $c_1$ and $c_2$ be the two roots of this quadratic equation.
Then,the sum of roots is $c_1 + c_2 = 2b\cos A$ and the product of roots is $c_1 c_2 = b^2 - a^2$.
The difference between the roots is given by $|c_1 - c_2| = \sqrt{(c_1 + c_2)^2 - 4c_1 c_2}$.
Substituting the values: $|c_1 - c_2| = \sqrt{(2b\cos A)^2 - 4(b^2 - a^2)}$.
$|c_1 - c_2| = \sqrt{4b^2\cos^2 A - 4b^2 + 4a^2} = \sqrt{4a^2 - 4b^2(1 - \cos^2 A)}$.
Since $1 - \cos^2 A = \sin^2 A$,we get $|c_1 - c_2| = \sqrt{4a^2 - 4b^2\sin^2 A} = 2\sqrt{a^2 - b^2\sin^2 A}$.
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MediumMCQ
Which of the following conditions allows for the existence of a triangle $ABC$?
A
$b \sin A = a, A < \frac{\pi}{2}$
B
$b \sin A > a, A > \frac{\pi}{2}$
C
$b \sin A > a, A < \frac{\pi}{2}$
D
None of these
Solution
(A) Using the Law of Sines,we have $\frac{\sin A}{a} = \frac{\sin B}{b}$,which implies $a \sin B = b \sin A$.
From option $(A)$,$b \sin A = a$,so $a \sin B = a$,which gives $\sin B = 1$,meaning $B = \frac{\pi}{2}$.
Since $A < \frac{\pi}{2}$,the sum of angles $A + B < \pi$,so the triangle $ABC$ is possible.
From option $(B)$,$b \sin A > a$,which implies $a \sin B > a$,or $\sin B > 1$,which is impossible.
Similarly,for option $(C)$,$b \sin A > a$ leads to $\sin B > 1$,which is also impossible.
From option $(A)$,$b \sin A = a$,so $a \sin B = a$,which gives $\sin B = 1$,meaning $B = \frac{\pi}{2}$.
Since $A < \frac{\pi}{2}$,the sum of angles $A + B < \pi$,so the triangle $ABC$ is possible.
From option $(B)$,$b \sin A > a$,which implies $a \sin B > a$,or $\sin B > 1$,which is impossible.
Similarly,for option $(C)$,$b \sin A > a$ leads to $\sin B > 1$,which is also impossible.
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AdvancedMCQ
If $A$ is the area and $2s$ is the sum of the $3$ sides of a triangle,then :
A
$A \le \frac{s^2}{3\sqrt{3}}$
B
$A = \frac{s^2}{2}$
C
$A > \frac{s^2}{\sqrt{3}}$
D
None
Solution
(A) For a given perimeter,an equilateral triangle has the maximum area.
Let the sides be $a, b, c$. The perimeter is $2s = a + b + c$,so the semi-perimeter is $s$.
For an equilateral triangle,$a = b = c = \frac{2s}{3}$.
The maximum area $A_{max} = \frac{\sqrt{3}}{4} \left(\frac{2s}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4s^2}{9} = \frac{s^2}{3\sqrt{3}}$.
Since any triangle with perimeter $2s$ has an area $A \le A_{max}$,we have $A \le \frac{s^2}{3\sqrt{3}}$.
Alternatively,using Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$.
By the $AM$-$GM$ inequality,$\frac{(s-a) + (s-b) + (s-c)}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$.
Since $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$,we have $\frac{s}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$.
Cubing both sides: $\frac{s^3}{27} \ge (s-a)(s-b)(s-c)$.
Multiplying by $s$: $A^2 = s(s-a)(s-b)(s-c) \le \frac{s^4}{27}$.
Taking the square root: $A \le \frac{s^2}{\sqrt{27}} = \frac{s^2}{3\sqrt{3}}$.
Let the sides be $a, b, c$. The perimeter is $2s = a + b + c$,so the semi-perimeter is $s$.
For an equilateral triangle,$a = b = c = \frac{2s}{3}$.
The maximum area $A_{max} = \frac{\sqrt{3}}{4} \left(\frac{2s}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4s^2}{9} = \frac{s^2}{3\sqrt{3}}$.
Since any triangle with perimeter $2s$ has an area $A \le A_{max}$,we have $A \le \frac{s^2}{3\sqrt{3}}$.
Alternatively,using Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$.
By the $AM$-$GM$ inequality,$\frac{(s-a) + (s-b) + (s-c)}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$.
Since $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$,we have $\frac{s}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$.
Cubing both sides: $\frac{s^3}{27} \ge (s-a)(s-b)(s-c)$.
Multiplying by $s$: $A^2 = s(s-a)(s-b)(s-c) \le \frac{s^4}{27}$.
Taking the square root: $A \le \frac{s^2}{\sqrt{27}} = \frac{s^2}{3\sqrt{3}}$.
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AdvancedMCQ
In any triangle $ABC$,$(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2} =$
A
$c^2$
B
$a^2$
C
$b^2$
D
$a^2 + b^2$
Solution
(A) Expanding the expression:
$(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2}$
$= (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2} + (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2}$
$= a^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + b^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$
$= a^2 + b^2 - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
$= a^2 + b^2 - 2ab \cos C$
Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$.
Therefore,the expression equals $c^2$.
$(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2}$
$= (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2} + (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2}$
$= a^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + b^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$
$= a^2 + b^2 - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
$= a^2 + b^2 - 2ab \cos C$
Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$.
Therefore,the expression equals $c^2$.
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AdvancedMCQ
In a $\Delta ABC$,the value of $\frac{a \cos A + b \cos B + c \cos C}{a + b + c}$ is equal to:
A
$\frac{r}{R}$
B
$\frac{R}{2r}$
C
$\frac{R}{r}$
D
$\frac{2r}{R}$
Solution
(A) Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the expression:
$LHS = \frac{2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C}{2R \sin A + 2R \sin B + 2R \sin C}$
$= \frac{\sin 2A + \sin 2B + \sin 2C}{2(\sin A + \sin B + \sin C)}$
Using the identities $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$:
$LHS = \frac{4 \sin A \sin B \sin C}{2(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2})}$
$= \frac{8 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}{8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$
$= \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{4R} \times 4 = \frac{r}{R}$.
Substituting these into the expression:
$LHS = \frac{2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C}{2R \sin A + 2R \sin B + 2R \sin C}$
$= \frac{\sin 2A + \sin 2B + \sin 2C}{2(\sin A + \sin B + \sin C)}$
Using the identities $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$:
$LHS = \frac{4 \sin A \sin B \sin C}{2(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2})}$
$= \frac{8 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}{8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$
$= \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{4R} \times 4 = \frac{r}{R}$.
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AdvancedMCQ
With usual notation in a $\Delta ABC$,if $R = k \frac{(r_1 + r_2)(r_2 + r_3)(r_3 + r_1)}{r_1 r_2 + r_2 r_3 + r_3 r_1}$,where $k$ has the value equal to:
A
$1$
B
$2$
C
$1/4$
D
$4$
Solution
(C) We know that $r_1 + r_2 = \frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{\Delta(s-b+s-a)}{(s-a)(s-b)} = \frac{\Delta c}{(s-a)(s-b)} = \frac{c}{r} \cdot \frac{\Delta^2}{s(s-a)(s-b)(s-c)} \cdot \frac{s(s-c)}{\Delta} = \frac{c}{r} \cdot \frac{s(s-c)}{\Delta} = \frac{c}{r} \cdot \frac{s(s-c)R}{abc/4R} = \dots$
Alternatively,using $r_1 = s \tan(A/2)$,$r_2 = s \tan(B/2)$,$r_3 = s \tan(C/2)$,we have $r_1 + r_2 = \frac{c}{\cos(C/2)} \cos(A/2) \cos(B/2) / \dots = \frac{4R \cos(C/2) \cos(A/2) \cos(B/2)}{\sin(C/2)} = \frac{4R \cos(C/2) \sin(C/2)}{\sin(C/2)} = 4R \cos^2(C/2)$ is not correct.
Correct identity: $r_1 + r_2 = \frac{c}{\sin C} \cos(A/2) \cos(B/2) \dots = 4R \cos(C/2) \cos(A/2) \cos(B/2) / \sin(C/2) = 4R \cos(C/2) \sin(C/2) / \sin(C/2) = 4R \cos^2(C/2)$ is wrong.
Actually,$r_1 + r_2 = \frac{c \cos(C/2)}{\sin(A/2) \sin(B/2)}$.
Using $r_1 = \frac{\Delta}{s-a}$,the expression simplifies to $R = k(4R)$.
Thus,$k = 1/4$.
Alternatively,using $r_1 = s \tan(A/2)$,$r_2 = s \tan(B/2)$,$r_3 = s \tan(C/2)$,we have $r_1 + r_2 = \frac{c}{\cos(C/2)} \cos(A/2) \cos(B/2) / \dots = \frac{4R \cos(C/2) \cos(A/2) \cos(B/2)}{\sin(C/2)} = \frac{4R \cos(C/2) \sin(C/2)}{\sin(C/2)} = 4R \cos^2(C/2)$ is not correct.
Correct identity: $r_1 + r_2 = \frac{c}{\sin C} \cos(A/2) \cos(B/2) \dots = 4R \cos(C/2) \cos(A/2) \cos(B/2) / \sin(C/2) = 4R \cos(C/2) \sin(C/2) / \sin(C/2) = 4R \cos^2(C/2)$ is wrong.
Actually,$r_1 + r_2 = \frac{c \cos(C/2)}{\sin(A/2) \sin(B/2)}$.
Using $r_1 = \frac{\Delta}{s-a}$,the expression simplifies to $R = k(4R)$.
Thus,$k = 1/4$.
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AdvancedMCQ
In a $\Delta ABC$,if $\cos 3A + \cos 3B + \cos 3C = 1$,then:
A
$\Delta ABC$ is right-angled
B
$\Delta ABC$ is acute-angled
C
$\Delta ABC$ is obtuse-angled
D
Nothing definite can be said about the nature of the $\Delta$
Solution
(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$.
Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$:
$\cos 3A + \cos 3B + \cos 3C = 1 - 4 \sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 1$.
This implies $4 \sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 0$.
Thus,$\sin \frac{3A}{2} = 0$ or $\sin \frac{3B}{2} = 0$ or $\sin \frac{3C}{2} = 0$.
For $\sin \frac{3A}{2} = 0$,we have $\frac{3A}{2} = \pi \Rightarrow A = \frac{2\pi}{3} = 120^\circ$.
Since one angle is $120^\circ$,which is greater than $90^\circ$,the triangle is obtuse-angled.
Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$:
$\cos 3A + \cos 3B + \cos 3C = 1 - 4 \sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 1$.
This implies $4 \sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 0$.
Thus,$\sin \frac{3A}{2} = 0$ or $\sin \frac{3B}{2} = 0$ or $\sin \frac{3C}{2} = 0$.
For $\sin \frac{3A}{2} = 0$,we have $\frac{3A}{2} = \pi \Rightarrow A = \frac{2\pi}{3} = 120^\circ$.
Since one angle is $120^\circ$,which is greater than $90^\circ$,the triangle is obtuse-angled.
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AdvancedMCQ
In a $\Delta ABC$,if $b + c = 3a$,then the value of $cot\, \frac{B}{2} \cdot cot\, \frac{C}{2}$ is equal to:
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(C) We know that $cot\, \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $cot\, \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get:
$cot\, \frac{B}{2} \cdot cot\, \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $b + c = 3a$,the perimeter $2s = a + b + c = a + 3a = 4a$,so $s = 2a$.
Substituting $s = 2a$ into the expression:
$\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.
Multiplying these,we get:
$cot\, \frac{B}{2} \cdot cot\, \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $b + c = 3a$,the perimeter $2s = a + b + c = a + 3a = 4a$,so $s = 2a$.
Substituting $s = 2a$ into the expression:
$\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.
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AdvancedMCQ
Let $f, g, h$ be the lengths of the perpendiculars from the circumcentre of the $\Delta ABC$ on the sides $a, b$ and $c$ respectively. If $\frac{a}{f} + \frac{b}{g} + \frac{c}{h} = \lambda \frac{abc}{fgh}$,then the value of $\lambda$ is:
A
$1/4$
B
$1/2$
C
$1$
D
$2$
Solution
(A) Let $R$ be the circumradius of $\Delta ABC$. The perpendicular distance from the circumcentre to the side $a$ is $f = R \cos A$.
Similarly,$g = R \cos B$ and $h = R \cos C$.
We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Thus,$\frac{a}{f} = \frac{2R \sin A}{R \cos A} = 2 \tan A$.
Similarly,$\frac{b}{g} = 2 \tan B$ and $\frac{c}{h} = 2 \tan C$.
Substituting these into the given equation: $2 \tan A + 2 \tan B + 2 \tan C = \lambda \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{(R \cos A)(R \cos B)(R \cos C)}$.
$2(\tan A + \tan B + \tan C) = \lambda \cdot 8 \tan A \tan B \tan C$.
In any triangle,$\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Therefore,$2(\tan A \tan B \tan C) = 8 \lambda (\tan A \tan B \tan C)$.
$2 = 8 \lambda \Rightarrow \lambda = 1/4$.
Similarly,$g = R \cos B$ and $h = R \cos C$.
We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Thus,$\frac{a}{f} = \frac{2R \sin A}{R \cos A} = 2 \tan A$.
Similarly,$\frac{b}{g} = 2 \tan B$ and $\frac{c}{h} = 2 \tan C$.
Substituting these into the given equation: $2 \tan A + 2 \tan B + 2 \tan C = \lambda \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{(R \cos A)(R \cos B)(R \cos C)}$.
$2(\tan A + \tan B + \tan C) = \lambda \cdot 8 \tan A \tan B \tan C$.
In any triangle,$\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Therefore,$2(\tan A \tan B \tan C) = 8 \lambda (\tan A \tan B \tan C)$.
$2 = 8 \lambda \Rightarrow \lambda = 1/4$.
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AdvancedMCQ
In a $\Delta ABC$,if $b = a(\sqrt{3} - 1)$ and $\angle C = 30^o$,then the measure of the angle $A$ is.....$^o$.
A
$15$
B
$45$
C
$75$
D
$105$
Solution
(D) Given $b = a(\sqrt{3} - 1)$ and $\angle C = 30^o$.
Since $\angle A + \angle B + \angle C = 180^o$,we have $\angle A + \angle B = 150^o$.
Using the Napier's Analogy: $\tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}\right)$.
Substituting $b = a(\sqrt{3} - 1)$:
$\frac{a - a(\sqrt{3} - 1)}{a + a(\sqrt{3} - 1)} = \frac{1 - \sqrt{3} + 1}{1 + \sqrt{3} - 1} = \frac{2 - \sqrt{3}}{\sqrt{3}}$.
Also,$\cot\left(\frac{30^o}{2}\right) = \cot(15^o) = 2 + \sqrt{3}$.
So,$\tan\left(\frac{A - B}{2}\right) = \left(\frac{2 - \sqrt{3}}{\sqrt{3}}\right)(2 + \sqrt{3}) = \frac{4 + 2\sqrt{3} - 2\sqrt{3} - 3}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Thus,$\frac{A - B}{2} = 30^o$,which implies $A - B = 60^o$.
Solving $A + B = 150^o$ and $A - B = 60^o$,we get $2A = 210^o$,so $A = 105^o$.
Since $\angle A + \angle B + \angle C = 180^o$,we have $\angle A + \angle B = 150^o$.
Using the Napier's Analogy: $\tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}\right)$.
Substituting $b = a(\sqrt{3} - 1)$:
$\frac{a - a(\sqrt{3} - 1)}{a + a(\sqrt{3} - 1)} = \frac{1 - \sqrt{3} + 1}{1 + \sqrt{3} - 1} = \frac{2 - \sqrt{3}}{\sqrt{3}}$.
Also,$\cot\left(\frac{30^o}{2}\right) = \cot(15^o) = 2 + \sqrt{3}$.
So,$\tan\left(\frac{A - B}{2}\right) = \left(\frac{2 - \sqrt{3}}{\sqrt{3}}\right)(2 + \sqrt{3}) = \frac{4 + 2\sqrt{3} - 2\sqrt{3} - 3}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Thus,$\frac{A - B}{2} = 30^o$,which implies $A - B = 60^o$.
Solving $A + B = 150^o$ and $A - B = 60^o$,we get $2A = 210^o$,so $A = 105^o$.
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View Solution163
AdvancedMCQ
If in a $\Delta ABC$,$\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is
A
right angled
B
isosceles
C
equilateral
D
obtuse
Solution
(C) Using the sine rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the given equation: $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.
This simplifies to $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,this implies $A = B = C$.
Therefore,the triangle is equilateral.
Substituting these into the given equation: $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.
This simplifies to $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,this implies $A = B = C$.
Therefore,the triangle is equilateral.
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View Solution164
AdvancedMCQ
If in a triangle $\sin A : \sin C = \sin (A - B) : \sin (B - C)$,then $a^2, b^2, c^2$:
A
are in $A.P.$
B
are in $G.P.$
C
are in $H.P.$
D
none of these
Solution
(A) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
In any triangle,$A + B + C = \pi$,so $A = \pi - (B + C)$,which implies $\sin A = \sin(B + C)$. Similarly,$C = \pi - (A + B)$,so $\sin C = \sin(A + B)$.
Substituting these into the given equation:
$\frac{\sin(B + C)}{\sin(A + B)} = \frac{\sin(A - B)}{\sin(B - C)}$
$\sin(B + C) \sin(B - C) = \sin(A + B) \sin(A - B)$
Using the identity $\sin(x+y)\sin(x-y) = \sin^2 x - \sin^2 y$:
$\sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$
$2 \sin^2 B = \sin^2 A + \sin^2 C$
Multiplying by $(2R)^2$:
$2b^2 = a^2 + c^2$
Therefore,$a^2, b^2, c^2$ are in $A.P.$
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
In any triangle,$A + B + C = \pi$,so $A = \pi - (B + C)$,which implies $\sin A = \sin(B + C)$. Similarly,$C = \pi - (A + B)$,so $\sin C = \sin(A + B)$.
Substituting these into the given equation:
$\frac{\sin(B + C)}{\sin(A + B)} = \frac{\sin(A - B)}{\sin(B - C)}$
$\sin(B + C) \sin(B - C) = \sin(A + B) \sin(A - B)$
Using the identity $\sin(x+y)\sin(x-y) = \sin^2 x - \sin^2 y$:
$\sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$
$2 \sin^2 B = \sin^2 A + \sin^2 C$
Multiplying by $(2R)^2$:
$2b^2 = a^2 + c^2$
Therefore,$a^2, b^2, c^2$ are in $A.P.$
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View Solution165
AdvancedMCQ
If in a triangle $ABC$,$\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then the value of the angle $A$ is:
A
$\frac{\pi}{8}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{2}$
Solution
(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting these into the given equation:
$\frac{2(b^2+c^2-a^2)}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{2(a^2+b^2-c^2)}{2abc} = \frac{a^2+b^2}{abc}$
Multiplying by $2abc$:
$2(b^2+c^2-a^2) + (a^2+c^2-b^2) + 2(a^2+b^2-c^2) = 2(a^2+b^2)$
$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = 2a^2 + 2b^2$
$3b^2 + c^2 + a^2 = 2a^2 + 2b^2$
$b^2 + c^2 = a^2$
Since $b^2 + c^2 = a^2$,by the converse of the Pythagorean theorem,$\angle A = 90^{\circ}$ or $\frac{\pi}{2}$.
Substituting these into the given equation:
$\frac{2(b^2+c^2-a^2)}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{2(a^2+b^2-c^2)}{2abc} = \frac{a^2+b^2}{abc}$
Multiplying by $2abc$:
$2(b^2+c^2-a^2) + (a^2+c^2-b^2) + 2(a^2+b^2-c^2) = 2(a^2+b^2)$
$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = 2a^2 + 2b^2$
$3b^2 + c^2 + a^2 = 2a^2 + 2b^2$
$b^2 + c^2 = a^2$
Since $b^2 + c^2 = a^2$,by the converse of the Pythagorean theorem,$\angle A = 90^{\circ}$ or $\frac{\pi}{2}$.
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View Solution166
AdvancedMCQ
With usual notations in a triangle $ABC$,if $r_1 = 2r_2 = 2r_3$,then:
A
$4a = 3b$
B
$3a = 2b$
C
$4b = 3a$
D
$2a = 3b$
Solution
(C) Given $r_1 = 2r_2 = 2r_3$.
Using the formulas for exradii,$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{2\Delta}{s-c}$.
From $2r_2 = 2r_3$,we get $s-b = s-c$,which implies $b = c$.
From $r_1 = 2r_2$,we have $\frac{\Delta}{s-a} = \frac{2\Delta}{s-b}$,so $s-b = 2(s-a)$.
Substituting $s = \frac{a+b+c}{2}$ and $b=c$:
$\frac{a+c-b}{2} = 2(\frac{b+c-a}{2})$.
Since $b=c$,this simplifies to $\frac{a}{2} = 2(\frac{2b-a}{2}) = 2b-a$.
$a = 4b - 2a$,which gives $3a = 4b$.
Using the formulas for exradii,$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{2\Delta}{s-c}$.
From $2r_2 = 2r_3$,we get $s-b = s-c$,which implies $b = c$.
From $r_1 = 2r_2$,we have $\frac{\Delta}{s-a} = \frac{2\Delta}{s-b}$,so $s-b = 2(s-a)$.
Substituting $s = \frac{a+b+c}{2}$ and $b=c$:
$\frac{a+c-b}{2} = 2(\frac{b+c-a}{2})$.
Since $b=c$,this simplifies to $\frac{a}{2} = 2(\frac{2b-a}{2}) = 2b-a$.
$a = 4b - 2a$,which gives $3a = 4b$.
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View Solution167
AdvancedMCQ
If $r_1, r_2, r_3$ are the radii of the excircles of triangle $ABC$,then $\frac{\sum r_1}{\sqrt{\sum r_1 r_2}}$ is equal to:
A
$\sum \cot \frac{A}{2}$
B
$\sum \cot \frac{A}{2} \cot \frac{B}{2}$
C
$\sum \tan \frac{A}{2}$
D
$\prod \tan \frac{A}{2}$
Solution
(C) We know that the radii of the excircles are given by $r_1 = s \tan \frac{A}{2}$,$r_2 = s \tan \frac{B}{2}$,and $r_3 = s \tan \frac{C}{2}$.
Sum of radii: $\sum r_1 = s(\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2})$.
Also,$\sum r_1 r_2 = s^2(\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2})$.
Using the identity $\sum \tan \frac{A}{2} \tan \frac{B}{2} = 1$ for any triangle $ABC$,we have $\sum r_1 r_2 = s^2(1) = s^2$.
Therefore,$\frac{\sum r_1}{\sqrt{\sum r_1 r_2}} = \frac{s \sum \tan \frac{A}{2}}{\sqrt{s^2}} = \sum \tan \frac{A}{2}$.
Sum of radii: $\sum r_1 = s(\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2})$.
Also,$\sum r_1 r_2 = s^2(\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2})$.
Using the identity $\sum \tan \frac{A}{2} \tan \frac{B}{2} = 1$ for any triangle $ABC$,we have $\sum r_1 r_2 = s^2(1) = s^2$.
Therefore,$\frac{\sum r_1}{\sqrt{\sum r_1 r_2}} = \frac{s \sum \tan \frac{A}{2}}{\sqrt{s^2}} = \sum \tan \frac{A}{2}$.
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View Solution168
AdvancedMCQ
In a $\Delta ABC$,the expression $\left( \frac{a^2}{\sin A} + \frac{b^2}{\sin B} + \frac{c^2}{\sin C} \right) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ simplifies to,where $\Delta$ is the area of the triangle.
A
$2\Delta$
B
$\Delta$
C
$\frac{\Delta}{2}$
D
$\frac{\Delta}{4}$
Solution
(B) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,the expression becomes $(a(2R) + b(2R) + c(2R)) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
$= 2R(a + b + c) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Since $a+b+c = 2s$,we have $2R(2s) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4Rs \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
We know that $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{4R}$,where $r$ is the inradius.
Substituting this,we get $4Rs \left( \frac{r}{4R} \right) = rs$.
Since $\Delta = rs$,the expression simplifies to $\Delta$.
Thus,the expression becomes $(a(2R) + b(2R) + c(2R)) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
$= 2R(a + b + c) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Since $a+b+c = 2s$,we have $2R(2s) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4Rs \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
We know that $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{4R}$,where $r$ is the inradius.
Substituting this,we get $4Rs \left( \frac{r}{4R} \right) = rs$.
Since $\Delta = rs$,the expression simplifies to $\Delta$.
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DifficultMCQ
The sines of two angles of a triangle are $\frac{5}{13}$ and $\frac{99}{101}$. The cosine of the third angle is: (in $/1313$)
A
$245$
B
$255$
C
$735$
D
$725$
Solution
(B) Let the angles of the triangle be $A, B,$ and $C$. Given $\sin A = \frac{5}{13}$ and $\sin B = \frac{99}{101}$.
Since $A$ and $B$ are angles of a triangle,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
Similarly,$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{9801}{10201}} = \sqrt{\frac{400}{10201}} = \frac{20}{101}$.
In a triangle,$A + B + C = 180^{\circ}$,so $C = 180^{\circ} - (A + B)$.
Therefore,$\cos C = \cos(180^{\circ} - (A + B)) = -\cos(A + B)$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$\cos C = -(\cos A \cos B - \sin A \sin B) = \sin A \sin B - \cos A \cos B$.
Substituting the values:
$\cos C = \left(\frac{5}{13} \times \frac{99}{101}\right) - \left(\frac{12}{13} \times \frac{20}{101}\right) = \frac{495}{1313} - \frac{240}{1313} = \frac{255}{1313}$.
Since $A$ and $B$ are angles of a triangle,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
Similarly,$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{9801}{10201}} = \sqrt{\frac{400}{10201}} = \frac{20}{101}$.
In a triangle,$A + B + C = 180^{\circ}$,so $C = 180^{\circ} - (A + B)$.
Therefore,$\cos C = \cos(180^{\circ} - (A + B)) = -\cos(A + B)$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$\cos C = -(\cos A \cos B - \sin A \sin B) = \sin A \sin B - \cos A \cos B$.
Substituting the values:
$\cos C = \left(\frac{5}{13} \times \frac{99}{101}\right) - \left(\frac{12}{13} \times \frac{20}{101}\right) = \frac{495}{1313} - \frac{240}{1313} = \frac{255}{1313}$.
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View Solution170
AdvancedMCQ
If in $\Delta ABC$,$AB = 4$,$BC = 6$ and $AC = 5$,and $h_1, h_2, h_3$ are the lengths of the altitudes from vertices $A, B, C$ respectively,then the value of $(\frac{1}{h_1} + \frac{1}{h_2} - \frac{1}{h_3})$ is equal to-
A
$\frac{\sqrt{7}}{15}$
B
$\frac{2\sqrt{7}}{15}$
C
$\frac{4\sqrt{7}}{15}$
D
$\frac{8\sqrt{7}}{15}$
Solution
(B) Let the sides opposite to vertices $A, B, C$ be $a, b, c$ respectively. Here $a = BC = 6$,$b = AC = 5$,and $c = AB = 4$.
The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$.
Thus,$h_1 = \frac{2\Delta}{a}$,$h_2 = \frac{2\Delta}{b}$,and $h_3 = \frac{2\Delta}{c}$.
Substituting these into the expression: $\frac{1}{h_1} + \frac{1}{h_2} - \frac{1}{h_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$.
Using Heron's formula,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+5+4}{2} = \frac{15}{2} = 7.5$.
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7.5(7.5-6)(7.5-5)(7.5-4)} = \sqrt{7.5 \times 1.5 \times 2.5 \times 3.5} = \sqrt{\frac{15}{2} \times \frac{3}{2} \times \frac{5}{2} \times \frac{7}{2}} = \frac{\sqrt{1575}}{4} = \frac{15\sqrt{7}}{4}$.
Now,$\frac{a+b-c}{2\Delta} = \frac{6+5-4}{2 \times \frac{15\sqrt{7}}{4}} = \frac{7}{\frac{15\sqrt{7}}{2}} = \frac{14}{15\sqrt{7}} = \frac{14\sqrt{7}}{15 \times 7} = \frac{2\sqrt{7}}{15}$.
The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$.
Thus,$h_1 = \frac{2\Delta}{a}$,$h_2 = \frac{2\Delta}{b}$,and $h_3 = \frac{2\Delta}{c}$.
Substituting these into the expression: $\frac{1}{h_1} + \frac{1}{h_2} - \frac{1}{h_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$.
Using Heron's formula,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+5+4}{2} = \frac{15}{2} = 7.5$.
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7.5(7.5-6)(7.5-5)(7.5-4)} = \sqrt{7.5 \times 1.5 \times 2.5 \times 3.5} = \sqrt{\frac{15}{2} \times \frac{3}{2} \times \frac{5}{2} \times \frac{7}{2}} = \frac{\sqrt{1575}}{4} = \frac{15\sqrt{7}}{4}$.
Now,$\frac{a+b-c}{2\Delta} = \frac{6+5-4}{2 \times \frac{15\sqrt{7}}{4}} = \frac{7}{\frac{15\sqrt{7}}{2}} = \frac{14}{15\sqrt{7}} = \frac{14\sqrt{7}}{15 \times 7} = \frac{2\sqrt{7}}{15}$.
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AdvancedMCQ
In a $\Delta ABC$,let $a, b$ and $c$ denote the lengths of sides opposite to vertices $A, B$ and $C$ respectively. If $b = 2, c = \sqrt{3}$ and $\angle BAC = \frac{\pi}{6}$,then the value of the circumradius of triangle $ABC$ is:
A
$1/2$
B
$1$
C
$2$
D
$1/4$
Solution
(B) First,we use the Law of Cosines to find side $a$:
$a^2 = b^2 + c^2 - 2bc \cos A$
$a^2 = 2^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos(\frac{\pi}{6})$
$a^2 = 4 + 3 - 4\sqrt{3} \times \frac{\sqrt{3}}{2}$
$a^2 = 7 - 6 = 1$
$a = 1$
Now,we use the formula for the circumradius $R$:
$R = \frac{a}{2 \sin A}$
$R = \frac{1}{2 \sin(\frac{\pi}{6})}$
$R = \frac{1}{2 \times (1/2)} = 1$
$a^2 = b^2 + c^2 - 2bc \cos A$
$a^2 = 2^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos(\frac{\pi}{6})$
$a^2 = 4 + 3 - 4\sqrt{3} \times \frac{\sqrt{3}}{2}$
$a^2 = 7 - 6 = 1$
$a = 1$
Now,we use the formula for the circumradius $R$:
$R = \frac{a}{2 \sin A}$
$R = \frac{1}{2 \sin(\frac{\pi}{6})}$
$R = \frac{1}{2 \times (1/2)} = 1$
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View Solution172
AdvancedMCQ
If two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^{\circ}$. If the third side is $3$,then the remaining fourth side is :-
A
$2$
B
$3$
C
$4$
D
$5$
Solution
(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB=5$,$AD=2$,$BC=3$,and $\angle DAB = 60^{\circ}$.
In $\triangle ABD$,by the Law of Cosines:
$BD^{2} = AB^{2} + AD^{2} - 2(AB)(AD)\cos(60^{\circ})$
$BD^{2} = 5^{2} + 2^{2} - 2(5)(2)(\frac{1}{2})$
$BD^{2} = 25 + 4 - 10 = 19$.
Since $ABCD$ is a cyclic quadrilateral,the opposite angles sum to $180^{\circ}$. Thus,$\angle BCD = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
In $\triangle BCD$,let $CD = x$. By the Law of Cosines:
$BD^{2} = BC^{2} + CD^{2} - 2(BC)(CD)\cos(120^{\circ})$
$19 = 3^{2} + x^{2} - 2(3)(x)(-\frac{1}{2})$
$19 = 9 + x^{2} + 3x$
$x^{2} + 3x - 10 = 0$
$(x+5)(x-2) = 0$
Since $x$ must be positive,$x = 2$.
In $\triangle ABD$,by the Law of Cosines:
$BD^{2} = AB^{2} + AD^{2} - 2(AB)(AD)\cos(60^{\circ})$
$BD^{2} = 5^{2} + 2^{2} - 2(5)(2)(\frac{1}{2})$
$BD^{2} = 25 + 4 - 10 = 19$.
Since $ABCD$ is a cyclic quadrilateral,the opposite angles sum to $180^{\circ}$. Thus,$\angle BCD = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
In $\triangle BCD$,let $CD = x$. By the Law of Cosines:
$BD^{2} = BC^{2} + CD^{2} - 2(BC)(CD)\cos(120^{\circ})$
$19 = 3^{2} + x^{2} - 2(3)(x)(-\frac{1}{2})$
$19 = 9 + x^{2} + 3x$
$x^{2} + 3x - 10 = 0$
$(x+5)(x-2) = 0$
Since $x$ must be positive,$x = 2$.
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AdvancedMCQ
In $\Delta ABC$,$8 \Delta = (b + c)(bc + 1)$,then the circumradius of $\Delta ABC$ is (where $\Delta$ denotes the area of the triangle and $b, c$ are the lengths of sides $AC$ and $AB$ respectively):
A
$\sqrt{\Delta}$
B
$\frac{1}{\sqrt{2\Delta}}$
C
$\sqrt{2}\Delta$
D
$\frac{1}{\sqrt{\Delta}}$
Solution
(A) Given $8 \Delta = (b + c)(bc + 1)$.
Since $\Delta = \frac{1}{2} bc \sin A$,we have $8 \cdot \frac{1}{2} bc \sin A = (b + c)(bc + 1)$.
$4 bc \sin A = (b + c)(bc + 1)$.
Dividing by $bc$,we get $4 \sin A = (b + c) \left(1 + \frac{1}{bc}\right) = b + \frac{1}{b} + c + \frac{1}{c}$.
By $AM$-$GM$ inequality,$b + \frac{1}{b} \ge 2$ and $c + \frac{1}{c} \ge 2$,so $b + \frac{1}{b} + c + \frac{1}{c} \ge 4$.
Since $\sin A \le 1$,$4 \sin A \le 4$.
Thus,equality holds only when $b = 1, c = 1$ and $\sin A = 1$,which means $A = 90^\circ$.
In this case,the sides are $b = 1, c = 1$,and $a = \sqrt{b^2 + c^2} = \sqrt{2}$.
The area $\Delta = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.
The circumradius $R = \frac{a}{2 \sin A} = \frac{\sqrt{2}}{2 \cdot 1} = \frac{1}{\sqrt{2}}$.
Since $\Delta = \frac{1}{2}$,we have $\sqrt{\Delta} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$R = \sqrt{\Delta}$.
Since $\Delta = \frac{1}{2} bc \sin A$,we have $8 \cdot \frac{1}{2} bc \sin A = (b + c)(bc + 1)$.
$4 bc \sin A = (b + c)(bc + 1)$.
Dividing by $bc$,we get $4 \sin A = (b + c) \left(1 + \frac{1}{bc}\right) = b + \frac{1}{b} + c + \frac{1}{c}$.
By $AM$-$GM$ inequality,$b + \frac{1}{b} \ge 2$ and $c + \frac{1}{c} \ge 2$,so $b + \frac{1}{b} + c + \frac{1}{c} \ge 4$.
Since $\sin A \le 1$,$4 \sin A \le 4$.
Thus,equality holds only when $b = 1, c = 1$ and $\sin A = 1$,which means $A = 90^\circ$.
In this case,the sides are $b = 1, c = 1$,and $a = \sqrt{b^2 + c^2} = \sqrt{2}$.
The area $\Delta = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.
The circumradius $R = \frac{a}{2 \sin A} = \frac{\sqrt{2}}{2 \cdot 1} = \frac{1}{\sqrt{2}}$.
Since $\Delta = \frac{1}{2}$,we have $\sqrt{\Delta} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$R = \sqrt{\Delta}$.
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View Solution174
AdvancedMCQ
The area of triangle $ABC$ is $84 \ sq. \ units$. If $AB = 13$ and $AC = 15$,then $BC$ can be ........ $units$.
A
$14$
B
$4\sqrt{14}$
C
$37$
D
$2\sqrt{37}$
Solution
(A) Let $AB = c = 13$,$AC = b = 15$,and $BC = a$. The area of the triangle is given by $\Delta = 84$.
Using the formula for the area of a triangle,$\Delta = \frac{1}{2} \times b \times c \times \sin(A)$,we have $84 = \frac{1}{2} \times 15 \times 13 \times \sin(A)$.
$\sin(A) = \frac{84 \times 2}{195} = \frac{168}{195} = \frac{56}{65}$.
Since $\sin^2(A) + \cos^2(A) = 1$,$\cos(A) = \pm \sqrt{1 - (\frac{56}{65})^2} = \pm \sqrt{\frac{4225 - 3136}{4225}} = \pm \sqrt{\frac{1089}{4225}} = \pm \frac{33}{65}$.
Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos(A) = 15^2 + 13^2 - 2(15)(13) \cos(A) = 225 + 169 - 390 \cos(A) = 394 - 390 \cos(A)$.
Case $1$: $\cos(A) = \frac{33}{65}$,$a^2 = 394 - 390(\frac{33}{65}) = 394 - 6(33) = 394 - 198 = 196$,so $a = 14$.
Case $2$: $\cos(A) = -\frac{33}{65}$,$a^2 = 394 - 390(-\frac{33}{65}) = 394 + 198 = 592$,so $a = \sqrt{592} = 4\sqrt{37}$.
Comparing with the options,$14$ is a valid value for $BC$.
Using the formula for the area of a triangle,$\Delta = \frac{1}{2} \times b \times c \times \sin(A)$,we have $84 = \frac{1}{2} \times 15 \times 13 \times \sin(A)$.
$\sin(A) = \frac{84 \times 2}{195} = \frac{168}{195} = \frac{56}{65}$.
Since $\sin^2(A) + \cos^2(A) = 1$,$\cos(A) = \pm \sqrt{1 - (\frac{56}{65})^2} = \pm \sqrt{\frac{4225 - 3136}{4225}} = \pm \sqrt{\frac{1089}{4225}} = \pm \frac{33}{65}$.
Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos(A) = 15^2 + 13^2 - 2(15)(13) \cos(A) = 225 + 169 - 390 \cos(A) = 394 - 390 \cos(A)$.
Case $1$: $\cos(A) = \frac{33}{65}$,$a^2 = 394 - 390(\frac{33}{65}) = 394 - 6(33) = 394 - 198 = 196$,so $a = 14$.
Case $2$: $\cos(A) = -\frac{33}{65}$,$a^2 = 394 - 390(-\frac{33}{65}) = 394 + 198 = 592$,so $a = \sqrt{592} = 4\sqrt{37}$.
Comparing with the options,$14$ is a valid value for $BC$.
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View Solution175
AdvancedMCQ
If in a $\Delta ABC$,$b = 2$,$c = \sqrt{3}$,$\angle A = \frac{\pi}{6}$,then $R = $?
A
$1/2$
B
$2$
C
$1$
D
None
Solution
(C) Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$
Given $b = 2$,$c = \sqrt{3}$,and $\angle A = \frac{\pi}{6}$:
$a^2 = (2)^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos(\frac{\pi}{6})$
$a^2 = 4 + 3 - 4\sqrt{3} \cdot \frac{\sqrt{3}}{2}$
$a^2 = 7 - 6 = 1$
$a = 1$
Using the Sine Rule: $R = \frac{a}{2 \sin A}$
$R = \frac{1}{2 \sin(\frac{\pi}{6})} = \frac{1}{2 \cdot (1/2)} = 1$
Given $b = 2$,$c = \sqrt{3}$,and $\angle A = \frac{\pi}{6}$:
$a^2 = (2)^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos(\frac{\pi}{6})$
$a^2 = 4 + 3 - 4\sqrt{3} \cdot \frac{\sqrt{3}}{2}$
$a^2 = 7 - 6 = 1$
$a = 1$
Using the Sine Rule: $R = \frac{a}{2 \sin A}$
$R = \frac{1}{2 \sin(\frac{\pi}{6})} = \frac{1}{2 \cdot (1/2)} = 1$
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View Solution176
AdvancedMCQ
The inradius of $\Delta ABC$ is $3$ units and the exradius opposite to vertex $A$ is $4$ units. Then,the length of the altitude from vertex $A$ is ............. $units$.
A
$24$
B
$23$
C
$26$
D
$22$
Solution
(A) Let $r$ be the inradius and $r_a$ be the exradius opposite to vertex $A$. We are given $r = 3$ and $r_a = 4$.
Using the formulas $r = \frac{\Delta}{s}$ and $r_a = \frac{\Delta}{s-a}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter,we have:
$\frac{\Delta}{s} = 3$ and $\frac{\Delta}{s-a} = 4$.
Dividing the two equations,we get $\frac{s-a}{s} = \frac{3}{4}$,which implies $1 - \frac{a}{s} = \frac{3}{4}$,so $\frac{a}{s} = \frac{1}{4}$.
Since $s = \frac{a+b+c}{2}$,we have $\frac{a}{s} = \frac{2a}{a+b+c} = \frac{1}{4}$,so $a+b+c = 8a$,or $b+c = 7a$.
Using the area formula $\Delta = \frac{1}{2} a h_a$,where $h_a$ is the altitude from $A$,we have $r = \frac{\Delta}{s} = \frac{a h_a}{2s} = 3$,so $h_a = \frac{6s}{a}$.
Since $\frac{s}{a} = 4$,we get $h_a = 6 \times 4 = 24$ units.
Using the formulas $r = \frac{\Delta}{s}$ and $r_a = \frac{\Delta}{s-a}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter,we have:
$\frac{\Delta}{s} = 3$ and $\frac{\Delta}{s-a} = 4$.
Dividing the two equations,we get $\frac{s-a}{s} = \frac{3}{4}$,which implies $1 - \frac{a}{s} = \frac{3}{4}$,so $\frac{a}{s} = \frac{1}{4}$.
Since $s = \frac{a+b+c}{2}$,we have $\frac{a}{s} = \frac{2a}{a+b+c} = \frac{1}{4}$,so $a+b+c = 8a$,or $b+c = 7a$.
Using the area formula $\Delta = \frac{1}{2} a h_a$,where $h_a$ is the altitude from $A$,we have $r = \frac{\Delta}{s} = \frac{a h_a}{2s} = 3$,so $h_a = \frac{6s}{a}$.
Since $\frac{s}{a} = 4$,we get $h_a = 6 \times 4 = 24$ units.
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View Solution177
AdvancedMCQ
If a triangle has sides $a, b, c$ and $r_1 > r_2 > r_3$ (where $r_1, r_2, r_3$ are the ex-radii),then:
A
$a > b > c$
B
$a < b < c$
C
$a > b$ and $b < c$
D
$a < b$ and $b > c$
Solution
(A) The ex-radii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-perimeter.
Given $r_1 > r_2 > r_3$,we have:
$\frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}$
Since $\Delta > 0$,taking the reciprocal reverses the inequality:
$s-a < s-b < s-c$
Subtracting $s$ from all parts:
$-a < -b < -c$
Multiplying by $-1$ reverses the inequality again:
$a > b > c$
Given $r_1 > r_2 > r_3$,we have:
$\frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}$
Since $\Delta > 0$,taking the reciprocal reverses the inequality:
$s-a < s-b < s-c$
Subtracting $s$ from all parts:
$-a < -b < -c$
Multiplying by $-1$ reverses the inequality again:
$a > b > c$
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View Solution178
AdvancedMCQ
In a $\Delta ABC$,$\angle A = 30^\circ$ and $a = 8 \text{ cm}$,then the distance of the orthocentre from vertex $A$ is equal to (with usual notations).
A
$4 \text{ cm}$
B
$8 \text{ cm}$
C
$8\sqrt{3} \text{ cm}$
D
$4\sqrt{3} \text{ cm}$
Solution
(C) The distance of the orthocentre $H$ from vertex $A$ in a triangle is given by the formula $AH = 2R \cos A$.
From the Sine Rule,we know that $\frac{a}{\sin A} = 2R$,which implies $2R = \frac{a}{\sin A}$.
Substituting this into the formula for $AH$,we get $AH = \frac{a}{\sin A} \times \cos A = a \cot A$.
Given $a = 8 \text{ cm}$ and $\angle A = 30^\circ$,we have $AH = 8 \times \cot(30^\circ)$.
Since $\cot(30^\circ) = \sqrt{3}$,we get $AH = 8\sqrt{3} \text{ cm}$.
From the Sine Rule,we know that $\frac{a}{\sin A} = 2R$,which implies $2R = \frac{a}{\sin A}$.
Substituting this into the formula for $AH$,we get $AH = \frac{a}{\sin A} \times \cos A = a \cot A$.
Given $a = 8 \text{ cm}$ and $\angle A = 30^\circ$,we have $AH = 8 \times \cot(30^\circ)$.
Since $\cot(30^\circ) = \sqrt{3}$,we get $AH = 8\sqrt{3} \text{ cm}$.
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View Solution179
AdvancedMCQ
In a $\Delta ABC$,if two sides $b$ and $c$ and an angle $B$ are given.
Statement $-1$: Let $b = 5 \ cm, c = 3 \ cm$ and $\angle B = 60^\circ$. The total number of possible triangles is $2$.
Statement $-2$: If $c \sin B < b < c$ and $B$ is an acute angle,then there are two possible values of $\angle C$.
Statement $-1$: Let $b = 5 \ cm, c = 3 \ cm$ and $\angle B = 60^\circ$. The total number of possible triangles is $2$.
Statement $-2$: If $c \sin B < b < c$ and $B$ is an acute angle,then there are two possible values of $\angle C$.
A
Statement $-1$ is true,statement $-2$ is true but statement $-1$ is not the correct explanation for statement $-2$
B
Statement $-1$ is true,statement $-2$ is false
C
Statement $-1$ is false,statement $-2$ is true
D
Both statements are true,and statement $-1$ is the true explanation of statement $-2$
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View Solution180
AdvancedMCQ
In a $\Delta ABC$,evaluate $\frac{b \sin(C - A)}{c^2 - a^2} + \frac{c \sin(A - B)}{a^2 - b^2}$.
A
$0$
B
$\frac{1}{R}$
C
$\frac{2}{R}$
D
None of these
Solution
(B) Given the expression: $E = \frac{b \sin(C - A)}{c^2 - a^2} + \frac{c \sin(A - B)}{a^2 - b^2}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the denominator $c^2 - a^2 = 4R^2(\sin^2 C - \sin^2 A) = 4R^2 \sin(C - A) \sin(C + A)$.
Since $A + B + C = \pi$,$\sin(C + A) = \sin(\pi - B) = \sin B$.
Thus,$c^2 - a^2 = 4R^2 \sin(C - A) \sin B$.
Substituting this into the first term: $\frac{b \sin(C - A)}{4R^2 \sin(C - A) \sin B} = \frac{2R \sin B \sin(C - A)}{4R^2 \sin(C - A) \sin B} = \frac{1}{2R}$.
Similarly,for the second term: $\frac{c \sin(A - B)}{a^2 - b^2} = \frac{2R \sin C \sin(A - B)}{4R^2 \sin(A - B) \sin C} = \frac{1}{2R}$.
Adding them together: $\frac{1}{2R} + \frac{1}{2R} = \frac{1}{R}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the denominator $c^2 - a^2 = 4R^2(\sin^2 C - \sin^2 A) = 4R^2 \sin(C - A) \sin(C + A)$.
Since $A + B + C = \pi$,$\sin(C + A) = \sin(\pi - B) = \sin B$.
Thus,$c^2 - a^2 = 4R^2 \sin(C - A) \sin B$.
Substituting this into the first term: $\frac{b \sin(C - A)}{4R^2 \sin(C - A) \sin B} = \frac{2R \sin B \sin(C - A)}{4R^2 \sin(C - A) \sin B} = \frac{1}{2R}$.
Similarly,for the second term: $\frac{c \sin(A - B)}{a^2 - b^2} = \frac{2R \sin C \sin(A - B)}{4R^2 \sin(A - B) \sin C} = \frac{1}{2R}$.
Adding them together: $\frac{1}{2R} + \frac{1}{2R} = \frac{1}{R}$.
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View Solution181
AdvancedMCQ
The perimeter of $\Delta ABC$ is $6$ times the $A.M.$ of the sines of its angles. If $a = 1$,then $\angle A = $
A
$\frac{\pi}{3}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{6}$
Solution
(D) Given that the perimeter $a+b+c = 6 \times \frac{\sin A + \sin B + \sin C}{3}$.
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
This implies $2R = 2$,so $R = 1$.
Since $a = 2R \sin A$ and $a = 1$,we have $1 = 2(1) \sin A$.
Therefore,$\sin A = \frac{1}{2}$,which gives $\angle A = \frac{\pi}{6}$ (or $30^{\circ}$).
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
This implies $2R = 2$,so $R = 1$.
Since $a = 2R \sin A$ and $a = 1$,we have $1 = 2(1) \sin A$.
Therefore,$\sin A = \frac{1}{2}$,which gives $\angle A = \frac{\pi}{6}$ (or $30^{\circ}$).
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View Solution182
AdvancedMCQ
If in a $\Delta ABC$,$a, b, c$ are in $A.P.$,then $\tan \frac{A}{2} \tan \frac{C}{2} = $
A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$3$
D
$4$
Solution
(B) We know that $\tan \frac{A}{2} \tan \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \times \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} = \frac{s-b}{s}$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Adding $b$ to both sides,$3b = a + b + c = 2s$,so $b = \frac{2s}{3}$.
Substituting this into the expression:
$\tan \frac{A}{2} \tan \frac{C}{2} = \frac{s - \frac{2s}{3}}{s} = \frac{\frac{s}{3}}{s} = \frac{1}{3}$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Adding $b$ to both sides,$3b = a + b + c = 2s$,so $b = \frac{2s}{3}$.
Substituting this into the expression:
$\tan \frac{A}{2} \tan \frac{C}{2} = \frac{s - \frac{2s}{3}}{s} = \frac{\frac{s}{3}}{s} = \frac{1}{3}$.
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View Solution183
AdvancedMCQ
In a $\Delta ABC$,if $C = 90^{\circ}$,then $\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B)$ is equal to:
A
$2$
B
$1$
C
$0$
D
None of these
Solution
(B) Given $\angle C = 90^{\circ}$,in $\Delta ABC$,we have $a = c \sin A$ and $b = c \sin B$.
Substituting these into the expression:
$\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B) = \frac{c^{2} \sin^{2} A + c^{2} \sin^{2} B}{c^{2} \sin^{2} A - c^{2} \sin^{2} B} \sin(A - B)$
$= \frac{\sin^{2} A + \sin^{2} B}{\sin^{2} A - \sin^{2} B} \sin(A - B)$
Using the identity $\sin^{2} A - \sin^{2} B = \sin(A + B) \sin(A - B)$:
$= \frac{\sin^{2} A + \sin^{2} B}{\sin(A + B) \sin(A - B)} \sin(A - B)$
Since $A + B = 90^{\circ}$,$\sin(A + B) = \sin 90^{\circ} = 1$ and $\sin A = \cos B$:
$= \frac{\cos^{2} B + \sin^{2} B}{1} = 1$.
Substituting these into the expression:
$\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B) = \frac{c^{2} \sin^{2} A + c^{2} \sin^{2} B}{c^{2} \sin^{2} A - c^{2} \sin^{2} B} \sin(A - B)$
$= \frac{\sin^{2} A + \sin^{2} B}{\sin^{2} A - \sin^{2} B} \sin(A - B)$
Using the identity $\sin^{2} A - \sin^{2} B = \sin(A + B) \sin(A - B)$:
$= \frac{\sin^{2} A + \sin^{2} B}{\sin(A + B) \sin(A - B)} \sin(A - B)$
Since $A + B = 90^{\circ}$,$\sin(A + B) = \sin 90^{\circ} = 1$ and $\sin A = \cos B$:
$= \frac{\cos^{2} B + \sin^{2} B}{1} = 1$.
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View Solution184
AdvancedMCQ
In a $\Delta ABC$,the ratio of angles $A : B : C = 3 : 5 : 4$. Then $a + b + c \sqrt{2}$ is equal to
A
$2b$
B
$2c$
C
$3b$
D
$3a$
Solution
(C) Given $A : B : C = 3 : 5 : 4$. Since the sum of angles in a triangle is $180^{\circ}$,we have $3x + 5x + 4x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.
Thus,$A = 45^{\circ}, B = 75^{\circ}, C = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 75^{\circ}} = \frac{c}{\sin 60^{\circ}} = K$.
Then $a = K \sin 45^{\circ} = \frac{K}{\sqrt{2}}$,$b = K \sin 75^{\circ} = K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$,and $c = K \sin 60^{\circ} = K \frac{\sqrt{3}}{2}$.
Now,$a + b + c \sqrt{2} = \frac{K}{\sqrt{2}} + K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right) + K \frac{\sqrt{3}}{2} \cdot \sqrt{2}$.
$= \frac{K}{\sqrt{2}} \left( 1 + \frac{\sqrt{3}+1}{2} + \sqrt{3} \right) = \frac{K}{\sqrt{2}} \left( \frac{2 + \sqrt{3} + 1 + 2\sqrt{3}}{2} \right) = \frac{K}{\sqrt{2}} \left( \frac{3 + 3\sqrt{3}}{2} \right) = \frac{3K(\sqrt{3}+1)}{2\sqrt{2}}$.
Since $b = K \sin 75^{\circ} = K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$,we have $3b = 3K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$.
Therefore,$a + b + c \sqrt{2} = 3b$.
Thus,$A = 45^{\circ}, B = 75^{\circ}, C = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 75^{\circ}} = \frac{c}{\sin 60^{\circ}} = K$.
Then $a = K \sin 45^{\circ} = \frac{K}{\sqrt{2}}$,$b = K \sin 75^{\circ} = K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$,and $c = K \sin 60^{\circ} = K \frac{\sqrt{3}}{2}$.
Now,$a + b + c \sqrt{2} = \frac{K}{\sqrt{2}} + K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right) + K \frac{\sqrt{3}}{2} \cdot \sqrt{2}$.
$= \frac{K}{\sqrt{2}} \left( 1 + \frac{\sqrt{3}+1}{2} + \sqrt{3} \right) = \frac{K}{\sqrt{2}} \left( \frac{2 + \sqrt{3} + 1 + 2\sqrt{3}}{2} \right) = \frac{K}{\sqrt{2}} \left( \frac{3 + 3\sqrt{3}}{2} \right) = \frac{3K(\sqrt{3}+1)}{2\sqrt{2}}$.
Since $b = K \sin 75^{\circ} = K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$,we have $3b = 3K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$.
Therefore,$a + b + c \sqrt{2} = 3b$.
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View Solution185
AdvancedMCQ
In $\Delta ABC$,if $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$,then the triangle is:
A
Equilateral
B
Right-angled
C
Isosceles
D
None of these
Solution
(B) Given $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$.
Using $\sin^{2} \theta = 1 - \cos^{2} \theta$,we have:
$(1 - \cos^{2} A) + \sin^{2} B + \sin^{2} C = 2$
$\sin^{2} B + \sin^{2} C - \cos^{2} A = 1$
Using the identity $\sin^{2} B + \sin^{2} C = 1 - \cos(B+C)\cos(B-C)$ and $\cos A = -\cos(B+C)$:
$1 - \cos(B+C)\cos(B-C) - \cos^{2} A = 1$
$-\cos(B+C)\cos(B-C) - \cos^{2} A = 0$
$\cos A \cos(B-C) - \cos^{2} A = 0$
$\cos A [\cos(B-C) - \cos A] = 0$
$\cos A [\cos(B-C) + \cos(B+C)] = 0$
$\cos A [2 \cos B \cos C] = 0$
This implies $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.
Thus,one of the angles must be $90^{\circ}$,which means the triangle is right-angled.
Using $\sin^{2} \theta = 1 - \cos^{2} \theta$,we have:
$(1 - \cos^{2} A) + \sin^{2} B + \sin^{2} C = 2$
$\sin^{2} B + \sin^{2} C - \cos^{2} A = 1$
Using the identity $\sin^{2} B + \sin^{2} C = 1 - \cos(B+C)\cos(B-C)$ and $\cos A = -\cos(B+C)$:
$1 - \cos(B+C)\cos(B-C) - \cos^{2} A = 1$
$-\cos(B+C)\cos(B-C) - \cos^{2} A = 0$
$\cos A \cos(B-C) - \cos^{2} A = 0$
$\cos A [\cos(B-C) - \cos A] = 0$
$\cos A [\cos(B-C) + \cos(B+C)] = 0$
$\cos A [2 \cos B \cos C] = 0$
This implies $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.
Thus,one of the angles must be $90^{\circ}$,which means the triangle is right-angled.
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View Solution186
AdvancedMCQ
For the triangle shown in the figure,side $b = 10 \, m$,and angles $\angle C$ and $\angle A$ are equal. Find the lengths of sides $a$ and $c$.
A
side $a = \text{side } c = 10 \, m$
B
side $a \neq \text{side } c$
C
side $a = \text{side } c = \frac{10 \sqrt{3}}{3} \, m$
D
side $a = \text{side } c = \frac{10}{\sqrt{2}} \, m$
Solution
(C) In $\Delta ABC$,the sum of angles is $180^{\circ}$. Given $\angle B = 120^{\circ}$ and $\angle C = \angle A = \theta$,we have $120^{\circ} + 2\theta = 180^{\circ}$,so $2\theta = 60^{\circ}$,which means $\theta = 30^{\circ}$.
Draw a perpendicular $BM$ from $B$ to $AC$. Since $\Delta ABC$ is isosceles,$M$ is the midpoint of $AC$,so $CM = 5 \, m$.
In right-angled $\Delta BMC$,$\cos 30^{\circ} = \frac{CM}{BC}$.
$\frac{\sqrt{3}}{2} = \frac{5}{a}$.
$a = \frac{10}{\sqrt{3}} = \frac{10 \sqrt{3}}{3} \, m$.
Since the triangle is isosceles,$a = c = \frac{10 \sqrt{3}}{3} \, m$.
Draw a perpendicular $BM$ from $B$ to $AC$. Since $\Delta ABC$ is isosceles,$M$ is the midpoint of $AC$,so $CM = 5 \, m$.
In right-angled $\Delta BMC$,$\cos 30^{\circ} = \frac{CM}{BC}$.
$\frac{\sqrt{3}}{2} = \frac{5}{a}$.
$a = \frac{10}{\sqrt{3}} = \frac{10 \sqrt{3}}{3} \, m$.
Since the triangle is isosceles,$a = c = \frac{10 \sqrt{3}}{3} \, m$.
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DifficultMCQ
In a $\Delta ABC$,$\frac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$. Then the ordered pair $(\angle A, \angle B)$ is equal to
A
$(105^\circ, 15^\circ)$
B
$(75^\circ, 45^\circ)$
C
$(15^\circ, 105^\circ)$
D
$(45^\circ, 75^\circ)$
Solution
(A) Given $\frac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$.
Using the Law of Tangents: $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$.
$\frac{a-b}{a+b} = \frac{(2+\sqrt{3})b - b}{(2+\sqrt{3})b + b} = \frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$.
$\cot\left(\frac{60^\circ}{2}\right) = \cot(30^\circ) = \sqrt{3}$.
So,$\tan\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$.
$\frac{A-B}{2} = 45^\circ \implies A-B = 90^\circ$.
Since $A+B+C = 180^\circ$ and $C = 60^\circ$,$A+B = 120^\circ$.
Solving $A-B = 90^\circ$ and $A+B = 120^\circ$,we get $2A = 210^\circ \implies A = 105^\circ$ and $B = 15^\circ$.
Thus,the ordered pair is $(105^\circ, 15^\circ)$.
Using the Law of Tangents: $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$.
$\frac{a-b}{a+b} = \frac{(2+\sqrt{3})b - b}{(2+\sqrt{3})b + b} = \frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$.
$\cot\left(\frac{60^\circ}{2}\right) = \cot(30^\circ) = \sqrt{3}$.
So,$\tan\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$.
$\frac{A-B}{2} = 45^\circ \implies A-B = 90^\circ$.
Since $A+B+C = 180^\circ$ and $C = 60^\circ$,$A+B = 120^\circ$.
Solving $A-B = 90^\circ$ and $A+B = 120^\circ$,we get $2A = 210^\circ \implies A = 105^\circ$ and $B = 15^\circ$.
Thus,the ordered pair is $(105^\circ, 15^\circ)$.
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DifficultMCQ
If in a triangle $ABC$,$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$,then $\cos A$ is equal to
A
$5/7$
B
$1/5$
C
$35/19$
D
$19/35$
Solution
(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $ABC$.
$\Rightarrow b+c=11K, c+a=12K, a+b=13K$.
Adding these equations: $2(a+b+c) = 36K \Rightarrow a+b+c = 18K$.
Subtracting each equation from the sum:
$a = (a+b+c) - (b+c) = 18K - 11K = 7K$.
$b = (a+b+c) - (c+a) = 18K - 12K = 6K$.
$c = (a+b+c) - (a+b) = 18K - 13K = 5K$.
Using the cosine rule:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values:
$\cos A = \frac{(6K)^2 + (5K)^2 - (7K)^2}{2(6K)(5K)} = \frac{36K^2 + 25K^2 - 49K^2}{60K^2} = \frac{12K^2}{60K^2} = \frac{1}{5}$.
$\Rightarrow b+c=11K, c+a=12K, a+b=13K$.
Adding these equations: $2(a+b+c) = 36K \Rightarrow a+b+c = 18K$.
Subtracting each equation from the sum:
$a = (a+b+c) - (b+c) = 18K - 11K = 7K$.
$b = (a+b+c) - (c+a) = 18K - 12K = 6K$.
$c = (a+b+c) - (a+b) = 18K - 13K = 5K$.
Using the cosine rule:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values:
$\cos A = \frac{(6K)^2 + (5K)^2 - (7K)^2}{2(6K)(5K)} = \frac{36K^2 + 25K^2 - 49K^2}{60K^2} = \frac{12K^2}{60K^2} = \frac{1}{5}$.
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DifficultMCQ
With the usual notation,in $\Delta ABC$,if $\angle A + \angle B = 120^{\circ}$ and $a : b = (\sqrt{3} + 1) : (\sqrt{3} - 1)$,then the ratio $\angle A : \angle B$ is
A
$7 : 1$
B
$5 : 3$
C
$9 : 7$
D
$3 : 1$
Solution
(A) Given $\angle A + \angle B = 120^{\circ}$ and $\frac{a}{b} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.
Using the Sine Rule,$\frac{\sin A}{\sin B} = \frac{a}{b} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.
Since $\angle B = 120^{\circ} - A$,we have $\frac{\sin A}{\sin(120^{\circ}-A)} = 2+\sqrt{3}$.
$\frac{\sin A}{\sin 120^{\circ} \cos A - \cos 120^{\circ} \sin A} = 2+\sqrt{3}$.
$\frac{\sin A}{\frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A} = 2+\sqrt{3}$.
Dividing numerator and denominator by $\sin A$,we get $\frac{1}{\frac{\sqrt{3}}{2} \cot A + \frac{1}{2}} = 2+\sqrt{3}$.
$\frac{2}{\sqrt{3} \cot A + 1} = 2+\sqrt{3} \implies \sqrt{3} \cot A + 1 = \frac{2}{2+\sqrt{3}} = 2(2-\sqrt{3}) = 4-2\sqrt{3}$.
$\sqrt{3} \cot A = 3-2\sqrt{3} \implies \cot A = \sqrt{3}-2$.
Since $\cot 105^{\circ} = \cot(60^{\circ}+45^{\circ}) = \frac{\cot 60^{\circ} \cot 45^{\circ}-1}{\cot 60^{\circ}+\cot 45^{\circ}} = \frac{\frac{1}{\sqrt{3}}-1}{\frac{1}{\sqrt{3}}+1} = \frac{1-\sqrt{3}}{1+\sqrt{3}} = \frac{(1-\sqrt{3})^2}{1-3} = \frac{4-2\sqrt{3}}{-2} = \sqrt{3}-2$.
Thus,$A = 105^{\circ}$ and $B = 120^{\circ} - 105^{\circ} = 15^{\circ}$.
The ratio $A : B = 105^{\circ} : 15^{\circ} = 7 : 1$.
Using the Sine Rule,$\frac{\sin A}{\sin B} = \frac{a}{b} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.
Since $\angle B = 120^{\circ} - A$,we have $\frac{\sin A}{\sin(120^{\circ}-A)} = 2+\sqrt{3}$.
$\frac{\sin A}{\sin 120^{\circ} \cos A - \cos 120^{\circ} \sin A} = 2+\sqrt{3}$.
$\frac{\sin A}{\frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A} = 2+\sqrt{3}$.
Dividing numerator and denominator by $\sin A$,we get $\frac{1}{\frac{\sqrt{3}}{2} \cot A + \frac{1}{2}} = 2+\sqrt{3}$.
$\frac{2}{\sqrt{3} \cot A + 1} = 2+\sqrt{3} \implies \sqrt{3} \cot A + 1 = \frac{2}{2+\sqrt{3}} = 2(2-\sqrt{3}) = 4-2\sqrt{3}$.
$\sqrt{3} \cot A = 3-2\sqrt{3} \implies \cot A = \sqrt{3}-2$.
Since $\cot 105^{\circ} = \cot(60^{\circ}+45^{\circ}) = \frac{\cot 60^{\circ} \cot 45^{\circ}-1}{\cot 60^{\circ}+\cot 45^{\circ}} = \frac{\frac{1}{\sqrt{3}}-1}{\frac{1}{\sqrt{3}}+1} = \frac{1-\sqrt{3}}{1+\sqrt{3}} = \frac{(1-\sqrt{3})^2}{1-3} = \frac{4-2\sqrt{3}}{-2} = \sqrt{3}-2$.
Thus,$A = 105^{\circ}$ and $B = 120^{\circ} - 105^{\circ} = 15^{\circ}$.
The ratio $A : B = 105^{\circ} : 15^{\circ} = 7 : 1$.
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DifficultMCQ
In a triangle,the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$. If $x^2 - c^2 = y$,where $c$ is the length of the third side of the triangle,then the circumradius of the triangle is
A
$\frac{3}{2}y$
B
$\frac{c}{\sqrt{3}}$
C
$\frac{c}{3}$
D
$\frac{y}{\sqrt{3}}$
Solution
(B) Let the sides of the triangle be $a, b,$ and $c$. Given $a + b = x$ and $ab = y$.
Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:
$(a + b)^2 - c^2 = ab$
$a^2 + b^2 + 2ab - c^2 = ab$
$a^2 + b^2 - c^2 = -ab$
Dividing by $2ab$,we get:
$\frac{a^2 + b^2 - c^2}{2ab} = -\frac{1}{2}$
By the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,so $\cos C = -\frac{1}{2}$.
Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.
Then $\sin C = \sin(120^\circ) = \frac{\sqrt{3}}{2}$.
Using the Sine Rule,$\frac{c}{\sin C} = 2R$,where $R$ is the circumradius:
$R = \frac{c}{2 \sin C} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.
Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:
$(a + b)^2 - c^2 = ab$
$a^2 + b^2 + 2ab - c^2 = ab$
$a^2 + b^2 - c^2 = -ab$
Dividing by $2ab$,we get:
$\frac{a^2 + b^2 - c^2}{2ab} = -\frac{1}{2}$
By the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,so $\cos C = -\frac{1}{2}$.
Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.
Then $\sin C = \sin(120^\circ) = \frac{\sqrt{3}}{2}$.
Using the Sine Rule,$\frac{c}{\sin C} = 2R$,where $R$ is the circumradius:
$R = \frac{c}{2 \sin C} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.
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DifficultMCQ
Given $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$,then the ordered triple $(\alpha, \beta, \gamma)$ is proportional to
A
$(7, 19, 25)$
B
$(3, 4, 5)$
C
$(5, 12, 13)$
D
$(19, 7, 25)$
Solution
(A) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$.
Summing these gives $2(a+b+c) = 36k$,so $a+b+c = 18k$.
Subtracting each original equation from this sum:
$a = (a+b+c) - (b+c) = 18k - 11k = 7k$.
$b = (a+b+c) - (c+a) = 18k - 12k = 6k$.
$c = (a+b+c) - (a+b) = 18k - 13k = 5k$.
Using the cosine rule $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{36k^2+25k^2-49k^2}{2(6k)(5k)} = \frac{12k^2}{60k^2} = \frac{1}{5}$.
Similarly,$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{49k^2+25k^2-36k^2}{2(7k)(5k)} = \frac{38k^2}{70k^2} = \frac{19}{35}$.
And $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{49k^2+36k^2-25k^2}{2(7k)(6k)} = \frac{60k^2}{84k^2} = \frac{5}{7}$.
Given $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma} = \lambda$,we have $\alpha = \frac{\cos A}{\lambda} = \frac{1}{5\lambda}$,$\beta = \frac{\cos B}{\lambda} = \frac{19}{35\lambda}$,$\gamma = \frac{\cos C}{\lambda} = \frac{5}{7\lambda}$.
Multiplying by $35\lambda$,we get $\alpha : \beta : \gamma = 7 : 19 : 25$.
Summing these gives $2(a+b+c) = 36k$,so $a+b+c = 18k$.
Subtracting each original equation from this sum:
$a = (a+b+c) - (b+c) = 18k - 11k = 7k$.
$b = (a+b+c) - (c+a) = 18k - 12k = 6k$.
$c = (a+b+c) - (a+b) = 18k - 13k = 5k$.
Using the cosine rule $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{36k^2+25k^2-49k^2}{2(6k)(5k)} = \frac{12k^2}{60k^2} = \frac{1}{5}$.
Similarly,$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{49k^2+25k^2-36k^2}{2(7k)(5k)} = \frac{38k^2}{70k^2} = \frac{19}{35}$.
And $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{49k^2+36k^2-25k^2}{2(7k)(6k)} = \frac{60k^2}{84k^2} = \frac{5}{7}$.
Given $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma} = \lambda$,we have $\alpha = \frac{\cos A}{\lambda} = \frac{1}{5\lambda}$,$\beta = \frac{\cos B}{\lambda} = \frac{19}{35\lambda}$,$\gamma = \frac{\cos C}{\lambda} = \frac{5}{7\lambda}$.
Multiplying by $35\lambda$,we get $\alpha : \beta : \gamma = 7 : 19 : 25$.
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Medium
Prove that in any triangle $ABC$,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$,where $a, b, c$ are the lengths of the sides opposite to the vertices $A, B, C$ respectively.
Solution
(N/A) Consider a triangle $ABC$ with sides $a, b, c$ opposite to vertices $A, B, C$. Draw a perpendicular $BD$ from vertex $B$ to side $AC$. Let $D$ be the point on $AC$ such that $BD \perp AC$.
In the right-angled triangle $ABD$,we have:
$AD = c \cos A$
$BD = c \sin A$
Since $AC = b$,the length $CD = AC - AD = b - c \cos A$.
Now,in the right-angled triangle $BDC$,by the Pythagorean theorem:
$BC^{2} = BD^{2} + CD^{2}$
$a^{2} = (c \sin A)^{2} + (b - c \cos A)^{2}$
Expanding the terms:
$a^{2} = c^{2} \sin^{2} A + b^{2} + c^{2} \cos^{2} A - 2bc \cos A$
Using the identity $\sin^{2} A + \cos^{2} A = 1$:
$a^{2} = b^{2} + c^{2}(\sin^{2} A + \cos^{2} A) - 2bc \cos A$
$a^{2} = b^{2} + c^{2} - 2bc \cos A$
Rearranging the terms to solve for $\cos A$:
$2bc \cos A = b^{2} + c^{2} - a^{2}$
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$
In the right-angled triangle $ABD$,we have:
$AD = c \cos A$
$BD = c \sin A$
Since $AC = b$,the length $CD = AC - AD = b - c \cos A$.
Now,in the right-angled triangle $BDC$,by the Pythagorean theorem:
$BC^{2} = BD^{2} + CD^{2}$
$a^{2} = (c \sin A)^{2} + (b - c \cos A)^{2}$
Expanding the terms:
$a^{2} = c^{2} \sin^{2} A + b^{2} + c^{2} \cos^{2} A - 2bc \cos A$
Using the identity $\sin^{2} A + \cos^{2} A = 1$:
$a^{2} = b^{2} + c^{2}(\sin^{2} A + \cos^{2} A) - 2bc \cos A$
$a^{2} = b^{2} + c^{2} - 2bc \cos A$
Rearranging the terms to solve for $\cos A$:
$2bc \cos A = b^{2} + c^{2} - a^{2}$
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$
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DifficultMCQ
Let $a, b$ and $c$ be the lengths of the sides of a triangle $ABC$ such that $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}$. If $r$ and $R$ are the inradius and circumradius of the triangle $ABC$,respectively,then the value of $\frac{R}{r}$ is equal to
A
$\frac{5}{2}$
B
$2$
C
$\frac{3}{2}$
D
$1$
Solution
(A) Given $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = \lambda$.
Then $a+b = 7\lambda$,$b+c = 8\lambda$,and $c+a = 9\lambda$.
Adding these equations,$2(a+b+c) = 24\lambda$,so $a+b+c = 12\lambda$.
Subtracting the given equations from $a+b+c = 12\lambda$,we get $c = 5\lambda$,$a = 4\lambda$,and $b = 3\lambda$.
Since $a^2 + b^2 = (4\lambda)^2 + (3\lambda)^2 = 25\lambda^2 = c^2$,the triangle is a right-angled triangle with $\angle C = 90^{\circ}$.
For a right-angled triangle,the circumradius $R = \frac{c}{2} = \frac{5\lambda}{2}$ and the inradius $r = \frac{a+b-c}{2} = \frac{4\lambda+3\lambda-5\lambda}{2} = \lambda$.
Therefore,$\frac{R}{r} = \frac{5\lambda/2}{\lambda} = \frac{5}{2}$.
Then $a+b = 7\lambda$,$b+c = 8\lambda$,and $c+a = 9\lambda$.
Adding these equations,$2(a+b+c) = 24\lambda$,so $a+b+c = 12\lambda$.
Subtracting the given equations from $a+b+c = 12\lambda$,we get $c = 5\lambda$,$a = 4\lambda$,and $b = 3\lambda$.
Since $a^2 + b^2 = (4\lambda)^2 + (3\lambda)^2 = 25\lambda^2 = c^2$,the triangle is a right-angled triangle with $\angle C = 90^{\circ}$.
For a right-angled triangle,the circumradius $R = \frac{c}{2} = \frac{5\lambda}{2}$ and the inradius $r = \frac{a+b-c}{2} = \frac{4\lambda+3\lambda-5\lambda}{2} = \lambda$.
Therefore,$\frac{R}{r} = \frac{5\lambda/2}{\lambda} = \frac{5}{2}$.
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AdvancedMCQ
Let $ABC$ be a triangle such that $AB=15$ and $AC=9$. The bisector of $\angle BAC$ meets $BC$ in $D$. If $\angle ACB=2\angle ABC$,then $BD$ is
A
$8$
B
$9$
C
$10$
D
$12$
Solution
(C) Let $\angle ABC = \theta$,then $\angle ACB = 2\theta$. Let $\angle BAC = 180^{\circ} - 3\theta$. Let $BC = x$.
By the sine rule in $\triangle ABC$:
$\frac{9}{\sin \theta} = \frac{15}{\sin 2\theta} = \frac{x}{\sin 3\theta}$
From $\frac{9}{\sin \theta} = \frac{15}{2 \sin \theta \cos \theta}$,we get $\cos \theta = \frac{15}{18} = \frac{5}{6}$.
Using $\frac{x}{\sin 3\theta} = \frac{9}{\sin \theta}$,we have $x = 9 \cdot \frac{\sin 3\theta}{\sin \theta} = 9(3 - 4 \sin^2 \theta)$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{25}{36} = \frac{11}{36}$,we get $x = 9(3 - 4 \cdot \frac{11}{36}) = 9(3 - \frac{11}{9}) = 27 - 11 = 16$.
By the Angle Bisector Theorem,$D$ divides $BC$ in the ratio $AB:AC = 15:9 = 5:3$.
Thus,$BD = \frac{5}{5+3} \cdot BC = \frac{5}{8} \cdot 16 = 10$.
By the sine rule in $\triangle ABC$:
$\frac{9}{\sin \theta} = \frac{15}{\sin 2\theta} = \frac{x}{\sin 3\theta}$
From $\frac{9}{\sin \theta} = \frac{15}{2 \sin \theta \cos \theta}$,we get $\cos \theta = \frac{15}{18} = \frac{5}{6}$.
Using $\frac{x}{\sin 3\theta} = \frac{9}{\sin \theta}$,we have $x = 9 \cdot \frac{\sin 3\theta}{\sin \theta} = 9(3 - 4 \sin^2 \theta)$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{25}{36} = \frac{11}{36}$,we get $x = 9(3 - 4 \cdot \frac{11}{36}) = 9(3 - \frac{11}{9}) = 27 - 11 = 16$.
By the Angle Bisector Theorem,$D$ divides $BC$ in the ratio $AB:AC = 15:9 = 5:3$.
Thus,$BD = \frac{5}{5+3} \cdot BC = \frac{5}{8} \cdot 16 = 10$.
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AdvancedMCQ
In a $\triangle ABC$,the angle bisector $BD$ of $\angle B$ intersects $AC$ in $D$. Suppose $BC=2$,$CD=1$ and $BD=\frac{3}{\sqrt{2}}$. The perimeter of the $\triangle ABC$ is
A
$\frac{17}{2}$
B
$\frac{15}{2}$
C
$\frac{17}{4}$
D
$\frac{15}{4}$
Solution
(B) Let $AB = c$ and $AD = x$. By the Angle Bisector Theorem,$\frac{AB}{BC} = \frac{AD}{CD}$,so $\frac{c}{2} = \frac{x}{1}$,which implies $x = \frac{c}{2}$.
In $\triangle BCD$,by the Law of Cosines,$BD^2 = BC^2 + CD^2 - 2(BC)(CD) \cos C$.
$\left(\frac{3}{\sqrt{2}}\right)^2 = 2^2 + 1^2 - 2(2)(1) \cos C \implies \frac{9}{2} = 5 - 4 \cos C \implies 4 \cos C = \frac{1}{2} \implies \cos C = \frac{1}{8}$.
In $\triangle ABC$,by the Law of Cosines,$c^2 = AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos C$.
$c^2 = 2^2 + (x+1)^2 - 2(2)(x+1) \left(\frac{1}{8}\right)$.
Substituting $x = \frac{c}{2}$,$c^2 = 4 + (\frac{c}{2}+1)^2 - (\frac{c}{2}+1) = 4 + \frac{c^2}{4} + c + 1 - \frac{c}{2} - 1 = \frac{c^2}{4} + \frac{c}{2} + 4$.
$\frac{3c^2}{4} - \frac{c}{2} - 4 = 0 \implies 3c^2 - 2c - 16 = 0$.
$(3c - 8)(c + 2) = 0$. Since $c > 0$,$c = \frac{8}{3}$.
Then $AD = x = \frac{c}{2} = \frac{4}{3}$.
Perimeter $= AB + BC + AC = c + 2 + (x+1) = \frac{8}{3} + 2 + \frac{4}{3} + 1 = 4 + 3 = 7$.
Wait,re-evaluating the provided solution logic: The formula $BD = \frac{2ac}{a+c} \cos \frac{B}{2}$ is standard. Using $a=2, CD=1, AD=x, c=AB$,we have $x/1 = c/2 \implies x = c/2$. $AC = c/2 + 1$. Using Stewart's Theorem on $\triangle ABC$ with cevian $BD$: $c^2(1) + 2^2(c/2) = (c/2+1)(BD^2 + (c/2)(1))$.
$c^2 + 2c = (c/2+1)(9/2 + c/2) = (c/2+1)(c+9)/2 = (c^2 + 9c + 2c + 18)/4$.
$4c^2 + 8c = c^2 + 11c + 18 \implies 3c^2 - 3c - 18 = 0 \implies c^2 - c - 6 = 0 \implies (c-3)(c+2) = 0$. So $c=3$.
Then $AD = c/2 = 3/2$. Perimeter $= 3 + 2 + (3/2 + 1) = 5 + 2.5 = 7.5 = \frac{15}{2}$.
In $\triangle BCD$,by the Law of Cosines,$BD^2 = BC^2 + CD^2 - 2(BC)(CD) \cos C$.
$\left(\frac{3}{\sqrt{2}}\right)^2 = 2^2 + 1^2 - 2(2)(1) \cos C \implies \frac{9}{2} = 5 - 4 \cos C \implies 4 \cos C = \frac{1}{2} \implies \cos C = \frac{1}{8}$.
In $\triangle ABC$,by the Law of Cosines,$c^2 = AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos C$.
$c^2 = 2^2 + (x+1)^2 - 2(2)(x+1) \left(\frac{1}{8}\right)$.
Substituting $x = \frac{c}{2}$,$c^2 = 4 + (\frac{c}{2}+1)^2 - (\frac{c}{2}+1) = 4 + \frac{c^2}{4} + c + 1 - \frac{c}{2} - 1 = \frac{c^2}{4} + \frac{c}{2} + 4$.
$\frac{3c^2}{4} - \frac{c}{2} - 4 = 0 \implies 3c^2 - 2c - 16 = 0$.
$(3c - 8)(c + 2) = 0$. Since $c > 0$,$c = \frac{8}{3}$.
Then $AD = x = \frac{c}{2} = \frac{4}{3}$.
Perimeter $= AB + BC + AC = c + 2 + (x+1) = \frac{8}{3} + 2 + \frac{4}{3} + 1 = 4 + 3 = 7$.
Wait,re-evaluating the provided solution logic: The formula $BD = \frac{2ac}{a+c} \cos \frac{B}{2}$ is standard. Using $a=2, CD=1, AD=x, c=AB$,we have $x/1 = c/2 \implies x = c/2$. $AC = c/2 + 1$. Using Stewart's Theorem on $\triangle ABC$ with cevian $BD$: $c^2(1) + 2^2(c/2) = (c/2+1)(BD^2 + (c/2)(1))$.
$c^2 + 2c = (c/2+1)(9/2 + c/2) = (c/2+1)(c+9)/2 = (c^2 + 9c + 2c + 18)/4$.
$4c^2 + 8c = c^2 + 11c + 18 \implies 3c^2 - 3c - 18 = 0 \implies c^2 - c - 6 = 0 \implies (c-3)(c+2) = 0$. So $c=3$.
Then $AD = c/2 = 3/2$. Perimeter $= 3 + 2 + (3/2 + 1) = 5 + 2.5 = 7.5 = \frac{15}{2}$.
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AdvancedMCQ
Let $a=BC, b=CA, c=AB$ be the side lengths of a $\triangle ABC$ and $m$ be the length of the median through $A$. If $a=8, b-c=2, m=6$,then the nearest integer to $b$ is
A
$7$
B
$8$
C
$9$
D
$10$
Solution
(B) Let $AD$ be the median to side $BC$. By Apollonius Theorem,we have:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given $a = BC = 8$,so $BD = DC = 4$. Also $m = AD = 6$.
Let $AC = b$ and $AB = c$. Given $b - c = 2$,so $c = b - 2$.
Substituting these values into the theorem:
$(b - 2)^2 + b^2 = 2(6^2 + 4^2)$
$b^2 - 4b + 4 + b^2 = 2(36 + 16)$
$2b^2 - 4b + 4 = 2(52)$
$2b^2 - 4b + 4 = 104$
$2b^2 - 4b - 100 = 0$
$b^2 - 2b - 50 = 0$
Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-50)}}{2(1)}$:
$b = \frac{2 \pm \sqrt{4 + 200}}{2} = \frac{2 \pm \sqrt{204}}{2} = 1 \pm \sqrt{51}$
Since $b$ must be positive,$b = 1 + \sqrt{51} \approx 1 + 7.14 = 8.14$.
The nearest integer to $b$ is $8$.
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given $a = BC = 8$,so $BD = DC = 4$. Also $m = AD = 6$.
Let $AC = b$ and $AB = c$. Given $b - c = 2$,so $c = b - 2$.
Substituting these values into the theorem:
$(b - 2)^2 + b^2 = 2(6^2 + 4^2)$
$b^2 - 4b + 4 + b^2 = 2(36 + 16)$
$2b^2 - 4b + 4 = 2(52)$
$2b^2 - 4b + 4 = 104$
$2b^2 - 4b - 100 = 0$
$b^2 - 2b - 50 = 0$
Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-50)}}{2(1)}$:
$b = \frac{2 \pm \sqrt{4 + 200}}{2} = \frac{2 \pm \sqrt{204}}{2} = 1 \pm \sqrt{51}$
Since $b$ must be positive,$b = 1 + \sqrt{51} \approx 1 + 7.14 = 8.14$.
The nearest integer to $b$ is $8$.
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View Solution197
AdvancedMCQ
Let $ABC$ be a triangle such that $AB = BC$. Let $F$ be the mid-point of $AB$ and $X$ be a point on $BC$ such that $FX$ is perpendicular to $AB$. If $BX = 3XC$,then the ratio $BC / AC$ equals
A
$\sqrt{3}$
B
$\sqrt{2}$
C
$\sqrt{\frac{3}{2}}$
D
$1$
Solution
(C) Let $BC = 4x$. Since $BX = 3XC$ and $BC = BX + XC$,we have $BX = 3x$ and $XC = x$.
Given $AB = BC$,so $AB = 4x$.
Since $F$ is the mid-point of $AB$,$BF = AF = 2x$.
In $\triangle BFX$,$\angle BFX = 90^\circ$. Thus,$\cos B = \frac{BF}{BX} = \frac{2x}{3x} = \frac{2}{3}$.
In $\triangle ABC$,by the Law of Cosines:
$\cos B = \frac{AB^2 + BC^2 - AC^2}{2(AB)(BC)}$
$\frac{2}{3} = \frac{(4x)^2 + (4x)^2 - AC^2}{2(4x)(4x)}$
$\frac{2}{3} = \frac{32x^2 - AC^2}{32x^2}$
$64x^2 = 3(32x^2 - AC^2)$
$64x^2 = 96x^2 - 3AC^2$
$3AC^2 = 32x^2$
$AC^2 = \frac{32x^2}{3}$
$AC = \sqrt{\frac{32}{3}}x = 4x \sqrt{\frac{2}{3}}$
Therefore,$\frac{BC}{AC} = \frac{4x}{4x \sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}}$.
Given $AB = BC$,so $AB = 4x$.
Since $F$ is the mid-point of $AB$,$BF = AF = 2x$.
In $\triangle BFX$,$\angle BFX = 90^\circ$. Thus,$\cos B = \frac{BF}{BX} = \frac{2x}{3x} = \frac{2}{3}$.
In $\triangle ABC$,by the Law of Cosines:
$\cos B = \frac{AB^2 + BC^2 - AC^2}{2(AB)(BC)}$
$\frac{2}{3} = \frac{(4x)^2 + (4x)^2 - AC^2}{2(4x)(4x)}$
$\frac{2}{3} = \frac{32x^2 - AC^2}{32x^2}$
$64x^2 = 3(32x^2 - AC^2)$
$64x^2 = 96x^2 - 3AC^2$
$3AC^2 = 32x^2$
$AC^2 = \frac{32x^2}{3}$
$AC = \sqrt{\frac{32}{3}}x = 4x \sqrt{\frac{2}{3}}$
Therefore,$\frac{BC}{AC} = \frac{4x}{4x \sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}}$.
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View Solution198
AdvancedMCQ
Let $ABC$ be an acute-angled triangle and let $D$ be the mid-point of $BC$. If $AB = AD$,then $\tan(B) / \tan(C)$ equals
A
$\sqrt{2}$
B
$\sqrt{3}$
C
$2$
D
$3$
Solution
(D) Given,in $\triangle ABC$,$D$ is the mid-point of $BC$ and $AB = AD$.
Since $AB = AD$,we have $\angle B = \angle ADB$.
Let $\angle ADB = B$. Then $\angle ADC = \pi - B$. Let $\angle ADC = \theta$,so $\theta = \pi - B$.
Since $D$ is the mid-point of $BC$,$BD = DC = 1$ (taking ratio $m:n = 1:1$).
Applying the cotangent theorem in $\triangle ABC$ with cevian $AD$:
$(m+n) \cot \theta = n \cot B - m \cot C$
Substituting $m=1, n=1$ and $\theta = \pi - B$:
$(1+1) \cot(\pi - B) = 1 \cdot \cot B - 1 \cdot \cot C$
$2(-\cot B) = \cot B - \cot C$
$-2 \cot B = \cot B - \cot C$
$\cot C = 3 \cot B$
$\frac{1}{\tan C} = \frac{3}{\tan B}$
$\frac{\tan B}{\tan C} = 3$.
Since $AB = AD$,we have $\angle B = \angle ADB$.
Let $\angle ADB = B$. Then $\angle ADC = \pi - B$. Let $\angle ADC = \theta$,so $\theta = \pi - B$.
Since $D$ is the mid-point of $BC$,$BD = DC = 1$ (taking ratio $m:n = 1:1$).
Applying the cotangent theorem in $\triangle ABC$ with cevian $AD$:
$(m+n) \cot \theta = n \cot B - m \cot C$
Substituting $m=1, n=1$ and $\theta = \pi - B$:
$(1+1) \cot(\pi - B) = 1 \cdot \cot B - 1 \cdot \cot C$
$2(-\cot B) = \cot B - \cot C$
$-2 \cot B = \cot B - \cot C$
$\cot C = 3 \cot B$
$\frac{1}{\tan C} = \frac{3}{\tan B}$
$\frac{\tan B}{\tan C} = 3$.
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View Solution199
AdvancedMCQ
In a $\triangle ABC$,the altitude $AD$ and the median $AE$ divide $\angle A$ into three equal parts. If $BC=28$,then the nearest integer to $AB+AC$ is
A
$38$
B
$37$
C
$36$
D
$33$
Solution
(A) Let $\angle BAD = \angle DAE = \angle EAC = \theta$. Thus,$\angle A = 3\theta$.
Since $AD$ is the altitude,$\triangle ABD$ and $\triangle ADC$ are right-angled at $D$.
In $\triangle ABD$,$\tan \theta = \frac{BD}{AD}$.
In $\triangle ADE$,$\tan \theta = \frac{DE}{AD}$.
Since $\tan \theta = \tan \theta$,we have $BD = DE$. Let $BD = DE = x$.
Since $AE$ is the median,$BE = EC = 14$. Thus,$DE = BE - BD = 14 - x$.
Equating the two expressions for $DE$: $x = 14 - x$ $\Rightarrow 2x = 14$ $\Rightarrow x = 7$.
So,$BD = 7$ and $DE = 7$.
In $\triangle ADC$,$\angle DAC = 2\theta$,so $\tan 2\theta = \frac{DC}{AD} = \frac{DE+EC}{AD} = \frac{7+14}{AD} = \frac{21}{AD}$.
In $\triangle ABD$,$\tan \theta = \frac{BD}{AD} = \frac{7}{AD}$.
Dividing the two equations: $\frac{\tan 2\theta}{\tan \theta} = \frac{21}{7} = 3$.
Using the identity $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we get $\frac{2}{1-\tan^2 \theta} = 3$.
$2 = 3 - 3\tan^2 \theta$ $\Rightarrow 3\tan^2 \theta = 1$ $\Rightarrow \tan \theta = \frac{1}{\sqrt{3}}$.
Thus,$\theta = 30^{\circ}$.
In $\triangle ABD$,$\sin 30^{\circ} = \frac{BD}{AB} = \frac{7}{AB} \Rightarrow AB = \frac{7}{1/2} = 14$.
In $\triangle ADC$,$\angle DAC = 60^{\circ}$,so $\sin 60^{\circ} = \frac{DC}{AC} = \frac{21}{AC}$ $\Rightarrow AC = \frac{21}{\sqrt{3}/2} = \frac{42}{\sqrt{3}} = 14\sqrt{3}$.
$AB + AC = 14 + 14\sqrt{3} = 14(1 + 1.732) = 14(2.732) = 38.248$.
The nearest integer is $38$.
Since $AD$ is the altitude,$\triangle ABD$ and $\triangle ADC$ are right-angled at $D$.
In $\triangle ABD$,$\tan \theta = \frac{BD}{AD}$.
In $\triangle ADE$,$\tan \theta = \frac{DE}{AD}$.
Since $\tan \theta = \tan \theta$,we have $BD = DE$. Let $BD = DE = x$.
Since $AE$ is the median,$BE = EC = 14$. Thus,$DE = BE - BD = 14 - x$.
Equating the two expressions for $DE$: $x = 14 - x$ $\Rightarrow 2x = 14$ $\Rightarrow x = 7$.
So,$BD = 7$ and $DE = 7$.
In $\triangle ADC$,$\angle DAC = 2\theta$,so $\tan 2\theta = \frac{DC}{AD} = \frac{DE+EC}{AD} = \frac{7+14}{AD} = \frac{21}{AD}$.
In $\triangle ABD$,$\tan \theta = \frac{BD}{AD} = \frac{7}{AD}$.
Dividing the two equations: $\frac{\tan 2\theta}{\tan \theta} = \frac{21}{7} = 3$.
Using the identity $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we get $\frac{2}{1-\tan^2 \theta} = 3$.
$2 = 3 - 3\tan^2 \theta$ $\Rightarrow 3\tan^2 \theta = 1$ $\Rightarrow \tan \theta = \frac{1}{\sqrt{3}}$.
Thus,$\theta = 30^{\circ}$.
In $\triangle ABD$,$\sin 30^{\circ} = \frac{BD}{AB} = \frac{7}{AB} \Rightarrow AB = \frac{7}{1/2} = 14$.
In $\triangle ADC$,$\angle DAC = 60^{\circ}$,so $\sin 60^{\circ} = \frac{DC}{AC} = \frac{21}{AC}$ $\Rightarrow AC = \frac{21}{\sqrt{3}/2} = \frac{42}{\sqrt{3}} = 14\sqrt{3}$.
$AB + AC = 14 + 14\sqrt{3} = 14(1 + 1.732) = 14(2.732) = 38.248$.
The nearest integer is $38$.
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AdvancedMCQ
Let $ABC$ be a triangle such that $AB=4, BC=5$ and $CA=6$. Choose points $D, E, F$ on $AB, BC, CA$ respectively,such that $AD=2, BE=2, CF=2$. Then find the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$.
A
$\frac{1}{4}$
B
$\frac{3}{15}$
C
$\frac{4}{15}$
D
$\frac{7}{30}$
Solution
(C) Given $AB=c=4, BC=a=5, CA=b=6$. Points $D, E, F$ are on $AB, BC, CA$ such that $AD=2, BE=2, CF=2$.
Note that $BD = AB - AD = 4 - 2 = 2$,$CE = BC - BE = 5 - 2 = 3$,$AF = AC - CF = 6 - 2 = 4$.
The area of $\triangle ABC = \Delta$.
The area of $\triangle ADF = \frac{1}{2} \cdot AD \cdot AF \cdot \sin A = \frac{1}{2} \cdot 2 \cdot 4 \cdot \sin A = 4 \sin A$.
Since $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 6 \cdot 4 \cdot \sin A = 12 \sin A$,we have $\sin A = \frac{\Delta}{12}$.
Thus,Area$(\triangle ADF)$ = $4 \cdot \frac{\Delta}{12} = \frac{1}{3} \Delta$.
Similarly,Area$(\triangle BED)$ = $\frac{1}{2} \cdot BE \cdot BD \cdot \sin B = \frac{1}{2} \cdot 2 \cdot 2 \cdot \sin B = 2 \sin B$.
Since $\Delta = \frac{1}{2} ac \sin B = \frac{1}{2} \cdot 5 \cdot 4 \cdot \sin B = 10 \sin B$,we have $\sin B = \frac{\Delta}{10}$.
Thus,Area$(\triangle BED)$ = $2 \cdot \frac{\Delta}{10} = \frac{1}{5} \Delta$.
Similarly,Area$(\triangle CFE)$ = $\frac{1}{2} \cdot CF \cdot CE \cdot \sin C = \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin C = 3 \sin C$.
Since $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} \cdot 5 \cdot 6 \cdot \sin C = 15 \sin C$,we have $\sin C = \frac{\Delta}{15}$.
Thus,Area$(\triangle CFE)$ = $3 \cdot \frac{\Delta}{15} = \frac{1}{5} \Delta$.
Area$(\triangle DEF)$ = $\Delta - (\text{Area}(\triangle ADF) + \text{Area}(\triangle BED) + \text{Area}(\triangle CFE)) = \Delta - (\frac{1}{3} + \frac{1}{5} + \frac{1}{5}) \Delta = \Delta - \frac{11}{15} \Delta = \frac{4}{15} \Delta$.
Therefore,the ratio is $\frac{4}{15}$.
Note that $BD = AB - AD = 4 - 2 = 2$,$CE = BC - BE = 5 - 2 = 3$,$AF = AC - CF = 6 - 2 = 4$.
The area of $\triangle ABC = \Delta$.
The area of $\triangle ADF = \frac{1}{2} \cdot AD \cdot AF \cdot \sin A = \frac{1}{2} \cdot 2 \cdot 4 \cdot \sin A = 4 \sin A$.
Since $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 6 \cdot 4 \cdot \sin A = 12 \sin A$,we have $\sin A = \frac{\Delta}{12}$.
Thus,Area$(\triangle ADF)$ = $4 \cdot \frac{\Delta}{12} = \frac{1}{3} \Delta$.
Similarly,Area$(\triangle BED)$ = $\frac{1}{2} \cdot BE \cdot BD \cdot \sin B = \frac{1}{2} \cdot 2 \cdot 2 \cdot \sin B = 2 \sin B$.
Since $\Delta = \frac{1}{2} ac \sin B = \frac{1}{2} \cdot 5 \cdot 4 \cdot \sin B = 10 \sin B$,we have $\sin B = \frac{\Delta}{10}$.
Thus,Area$(\triangle BED)$ = $2 \cdot \frac{\Delta}{10} = \frac{1}{5} \Delta$.
Similarly,Area$(\triangle CFE)$ = $\frac{1}{2} \cdot CF \cdot CE \cdot \sin C = \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin C = 3 \sin C$.
Since $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} \cdot 5 \cdot 6 \cdot \sin C = 15 \sin C$,we have $\sin C = \frac{\Delta}{15}$.
Thus,Area$(\triangle CFE)$ = $3 \cdot \frac{\Delta}{15} = \frac{1}{5} \Delta$.
Area$(\triangle DEF)$ = $\Delta - (\text{Area}(\triangle ADF) + \text{Area}(\triangle BED) + \text{Area}(\triangle CFE)) = \Delta - (\frac{1}{3} + \frac{1}{5} + \frac{1}{5}) \Delta = \Delta - \frac{11}{15} \Delta = \frac{4}{15} \Delta$.
Therefore,the ratio is $\frac{4}{15}$.
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