Relation between sides and angles, Solutions of triangles Questions in English

(C) Using the Sine Rule,we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = 2R,$ where $R$ is the circumradius of the triangle.
This implies $\sin A = \frac{a}{2R}$ and $\sin B = \frac{b}{2R}.$
Substituting these into the given equation $a \sin A = b \sin B$:
$a \left( \frac{a}{2R} \right) = b \left( \frac{b}{2R} \right)$
$\frac{a^2}{2R} = \frac{b^2}{2R}$
$a^2 = b^2$
Since $a$ and $b$ are side lengths of a triangle,they must be positive,so $a = b.$
Thus,the triangle is isosceles.

(B) We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{\circ}$.
Alternatively,using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{3}{2}$,we get $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{1}{8}$.
Since the maximum value of $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ is $\frac{1}{8}$ (by Jensen's inequality for concave functions),this equality holds only when $\frac{A}{2} = \frac{B}{2} = \frac{C}{2} = 30^{\circ}$,i.e.,$A = B = C = 60^{\circ}$.
Thus,the triangle is equilateral.

(D) Given that sides $AB$ and $AC$ are perpendicular,$\angle A = 90^{\circ}$ or $\frac{\pi}{2}$.
In any triangle $ABC$,the sum of angles is $A + B + C = 180^{\circ}$ or $\pi$.
Since $A = 90^{\circ}$,we have $B + C = 90^{\circ}$,which implies $C = 90^{\circ} - B$.
Taking the tangent on both sides,$\tan C = \tan(90^{\circ} - B)$.
Using the trigonometric identity $\tan(90^{\circ} - \theta) = \cot \theta$,we get $\tan C = \cot B$.
Since $\cot B = \frac{1}{\tan B}$,we have $\tan C = \frac{1}{\tan B}$.
Therefore,$\tan B \cdot \tan C = 1$.

(B) In a triangle $ABC$,the semi-perimeter $s$ is defined as $s = \frac{a+b+c}{2}$.
Using the half-angle formula for $\tan \left( \frac{A}{2} \right)$,we have:
$\tan \left( \frac{A}{2} \right) = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$.
Thus,the correct option is $B$.

(C) In $\triangle ABC$,$AD = b \sin C = c \sin B$.
Given $AD = \frac{abc}{b^2 - c^2}$,we have $AD(b^2 - c^2) = abc$.
Substituting $AD = b \sin C$ and $AD = c \sin B$:
$b \sin C (b^2 - c^2) = abc \implies \sin C (b^2 - c^2) = ac$.
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$:
$\sin C (4R^2 \sin^2 B - 4R^2 \sin^2 C) = (2R \sin A)(2R \sin C)$.
$\sin^2 B - \sin^2 C = \sin A = \sin(180^\circ - (B+C)) = \sin(B+C)$.
$(\sin B - \sin C)(\sin B + \sin C) = \sin B \cos C + \cos B \sin C$.
Since $\sin B - \sin C = 2 \sin(\frac{B-C}{2}) \cos(\frac{B+C}{2})$ and $\sin B + \sin C = 2 \sin(\frac{B+C}{2}) \cos(\frac{B-C}{2})$,
$2 \sin(\frac{B-C}{2}) \cos(\frac{B+C}{2}) \cdot 2 \sin(\frac{B+C}{2}) \cos(\frac{B-C}{2}) = \sin(B+C)$.
$2 \sin(B-C) \cdot 2 \sin(B+C) \cdot \frac{1}{2} = \sin(B+C)$.
$2 \sin(B-C) = 1 \implies \sin(B-C) = \frac{1}{2} = \sin 30^\circ$.
$B - C = 30^\circ \implies B = 30^\circ + 23^\circ = 53^\circ$ (Not in options).
Re-evaluating the condition $AD = \frac{abc}{b^2-c^2}$:
$AD = \frac{abc}{b^2-c^2} \implies \frac{1}{AD} = \frac{b^2-c^2}{abc} = \frac{b}{ac} - \frac{c}{ab} = \frac{\sin B}{a \sin C} - \frac{\sin C}{a \sin B} = \frac{\sin^2 B - \sin^2 C}{a \sin B \sin C}$.
Since $AD = c \sin B$,$\frac{1}{c \sin B} = \frac{\sin(B-C)\sin(B+C)}{a \sin B \sin C}$.
Using $a = \frac{c \sin A}{\sin C} = \frac{c \sin(B+C)}{\sin C}$,we get $\sin(B-C) = 1$.
$B - C = 90^\circ \implies B = 90^\circ + 23^\circ = 113^\circ$.

(D) Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.
Rearranging for $\sin B$: $\sin B = \frac{b \sin A}{a}$.
Substituting the given values: $\sin B = \frac{4\sqrt{3} \sin 60^\circ}{5} = \frac{4\sqrt{3} \times (\frac{\sqrt{3}}{2})}{5} = \frac{4 \times 3}{2 \times 5} = \frac{6}{5} = 1.2$.
Since the value of $\sin B$ must be $\le 1$,the value $1.2$ is inadmissible.
Therefore,no such triangle exists,and the correct option is $D$.

(B) Given $\Delta = a^2 - (b - c)^2 = a^2 - b^2 - c^2 + 2bc$.
Since $a^2 = b^2 + c^2 - 2bc \cos A$,we have $a^2 - b^2 - c^2 = -2bc \cos A$.
Substituting this into the expression for $\Delta$:
$\Delta = 2bc - 2bc \cos A = 2bc(1 - \cos A) = 2bc(2 \sin^2 \frac{A}{2}) = 4bc \sin^2 \frac{A}{2}$ .....$(i)$
Also,the area of the triangle is given by $\Delta = \frac{1}{2} bc \sin A = bc \sin \frac{A}{2} \cos \frac{A}{2}$ .....$(ii)$
Dividing $(i)$ by $(ii)$:
$\frac{\Delta}{\Delta} = \frac{4bc \sin^2 \frac{A}{2}}{bc \sin \frac{A}{2} \cos \frac{A}{2}} \implies 1 = 4 \tan \frac{A}{2}$.
Thus,$\tan \frac{A}{2} = \frac{1}{4}$.
Using the formula $\tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^2 \frac{A}{2}}$:
$\tan A = \frac{2(\frac{1}{4})}{1 - (\frac{1}{4})^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{1}{2} \times \frac{16}{15} = \frac{8}{15}$.

(B) The area of triangle $ABC$ is given by $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
From this,we get $p_1 = \frac{2\Delta}{a}$,$p_2 = \frac{2\Delta}{b}$,and $p_3 = \frac{2\Delta}{c}$.
Therefore,${p_1}^{-2} + {p_2}^{-2} + {p_3}^{-2} = \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2}$.
Substituting the values,we get $\frac{a^2}{4\Delta^2} + \frac{b^2}{4\Delta^2} + \frac{c^2}{4\Delta^2} = \frac{a^2 + b^2 + c^2}{4\Delta^2}$.

(C) Let $AD$ be the median through $A$ to the side $BC$. We are given that $AD \perp AB$,so $\angle DAB = 90^{\circ}$.
Draw $CN \perp AB$ (produced). In $\Delta BCN$,$AD \parallel CN$ because both are perpendicular to $AB$.
Since $D$ is the midpoint of $BC$,by the intercept theorem,$A$ is the midpoint of $BN$,so $AB = AN$.
In $\Delta ACN$,$\angle CAN = \alpha$. Then $\angle CAB = 180^{\circ} - \alpha$,so $\tan A = \tan(180^{\circ} - \alpha) = -\tan \alpha$.
In $\Delta ACN$,$\tan \alpha = \frac{CN}{AN}$.
In $\Delta ABD$,$\tan B = \frac{AD}{AB}$.
Since $AD = \frac{1}{2}CN$ and $AB = AN$,we have $\tan \alpha = \frac{2AD}{AB} = 2\tan B$.
Therefore,$\tan A = -2\tan B$,which implies $\tan A + 2\tan B = 0$.

(B) Given $s - a = 3$ and $s - c = 2$,where $s = \frac{a+b+c}{2}$.
$s - a = \frac{b+c-a}{2} = 3 \Rightarrow b+c-a = 6$ ... $(i)$
$s - c = \frac{a+b-c}{2} = 2 \Rightarrow a+b-c = 4$ ... $(ii)$
Adding $(i)$ and $(ii)$,we get $2b = 10 \Rightarrow b = 5$.
Since the triangle is right-angled at $B$,by Pythagoras theorem,$a^2 + c^2 = b^2 = 25$ ... $(iii)$.
From $(i)$,$c - a = 6 - b = 6 - 5 = 1 \Rightarrow c = a + 1$.
Substituting into $(iii)$: $a^2 + (a+1)^2 = 25$ $\Rightarrow a^2 + a^2 + 2a + 1 = 25$ $\Rightarrow 2a^2 + 2a - 24 = 0$ $\Rightarrow a^2 + a - 12 = 0$.
$(a+4)(a-3) = 0$. Since $a > 0$,$a = 3$.
Then $c = a + 1 = 4$.
Thus,the values are $a = 3$ and $c = 4$.

(A) Using the Sine Rule,we have $\frac{\sin B}{b} = \frac{\sin C}{c}$.
Rearranging for $\sin C$,we get $\sin C = \frac{c}{b} \sin B$.
Given the condition $b < c \sin B$,we can rewrite this as $\frac{c}{b} \sin B > 1$.
Therefore,$\sin C > 1$,which is impossible because the value of the sine function cannot exceed $1$.
Hence,no triangle is possible.

(B) Using the Law of Cosines,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
This rearranges to the quadratic equation in $b$: $b^2 - (2c \cos A)b + (c^2 - a^2) = 0$.
Given that $b_1$ and $b_2$ are the roots of this equation,we have $b_1 + b_2 = 2c \cos A$ and $b_1 b_2 = c^2 - a^2$.
Since $b_2 = 2b_1$,we get $3b_1 = 2c \cos A \Rightarrow b_1 = \frac{2c \cos A}{3}$ and $2b_1^2 = c^2 - a^2$.
Substituting $b_1$ into the second equation: $2(\frac{2c \cos A}{3})^2 = c^2 - a^2$.
$2(\frac{4c^2 \cos^2 A}{9}) = c^2 - a^2 \Rightarrow 8c^2 \cos^2 A = 9(c^2 - a^2)$.
Using $\cos^2 A = 1 - \sin^2 A$,we get $8c^2(1 - \sin^2 A) = 9c^2 - 9a^2$.
$8c^2 - 8c^2 \sin^2 A = 9c^2 - 9a^2 \Rightarrow 8c^2 \sin^2 A = 9a^2 - c^2$.
Therefore,$\sin A = \sqrt{\frac{9a^2 - c^2}{8c^2}}$.

(A) Using the Law of Cosines,we have $\cos A = \frac{c^2 + b^2 - a^2}{2bc}$.
This rearranges to the quadratic equation in $c$: $c^2 - (2b \cos A)c + (b^2 - a^2) = 0$.
Since $c_1$ and $c_2$ are the roots of this equation,we have $c_1 + c_2 = 2b \cos A$ and $c_1 c_2 = b^2 - a^2$.
The area of a triangle with sides $a, b$ and included angle $C$ is $\frac{1}{2}ab \sin C$.
The sum of the areas of the two triangles is $S = \frac{1}{2}ab \sin C_1 + \frac{1}{2}ab \sin C_2 = \frac{1}{2}ab(\sin C_1 + \sin C_2)$.
From the Law of Sines,$\frac{c}{\sin C} = \frac{b}{\sin B} = \frac{a}{\sin A} = 2R$,so $\sin C = \frac{c \sin A}{a}$.
Thus,$\sin C_1 + \sin C_2 = \frac{(c_1 + c_2) \sin A}{a} = \frac{(2b \cos A) \sin A}{a} = \frac{b \sin 2A}{a}$.
Substituting this into the sum of areas: $S = \frac{1}{2}ab \left( \frac{b \sin 2A}{a} \right) = \frac{1}{2}b^2 \sin 2A$.

(C) Given the equation: $2 \cos A = \sin B \csc C$
Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,so $\frac{\sin B}{\sin C} = \frac{b}{c}$.
Substituting this into the given equation: $2 \cos A = \frac{b}{c}$.
Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting $\cos A$ into the equation: $2 \left( \frac{b^2 + c^2 - a^2}{2bc} \right) = \frac{b}{c}$.
$\frac{b^2 + c^2 - a^2}{bc} = \frac{b}{c}$.
Multiplying both sides by $bc$: $b^2 + c^2 - a^2 = b^2$.
$c^2 - a^2 = 0 \Rightarrow c^2 = a^2$.
Since $a$ and $c$ are side lengths of a triangle,$c = a$.

(B) Let the sides be $a = 3$,$b = 5$,and $c = 7$.
To determine the nature of the triangle,we calculate the cosine of the largest angle,which is opposite the longest side $(c = 7)$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5} = \frac{9 + 25 - 49}{30} = \frac{34 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$.
Since $\cos C = -\frac{1}{2}$,the angle $C = 120^{\circ}$.
Because $C > 90^{\circ}$,the triangle has one obtuse angle.

(B) Given that $\sin P, \sin Q, \sin R$ are in $A.P.$
By the Sine Rule,$\frac{a}{\sin P} = \frac{b}{\sin Q} = \frac{c}{\sin R} = 2R$,where $R$ is the circumradius.
This implies $\sin P = \frac{a}{2R}, \sin Q = \frac{b}{2R}, \sin R = \frac{c}{2R}$.
Since $\sin P, \sin Q, \sin R$ are in $A.P.$,it follows that $a, b, c$ are in $A.P.$
Let $p_1, p_2, p_3$ be the altitudes from vertices $P, Q, R$ respectively.
The area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
Thus,$p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.
Since $a, b, c$ are in $A.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$
Multiplying by $2\Delta$,we get $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$
Therefore,$p_1, p_2, p_3$ are in $H.P.$

(A) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Substituting these,we get $\frac{a}{c} = \frac{\sin A \cos B - \cos A \sin B}{\sin B \cos C - \cos B \sin C}$.
Using the projection formula $a = b \cos C + c \cos B$,we have $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Cross-multiplying: $a(\sin B \cos C - \cos B \sin C) = c(\sin A \cos B - \cos A \sin B)$.
Using $\sin A = \frac{a}{2R}$,etc.,this simplifies to $a(b \cos C - c \cos B) = c(a \cos B - b \cos A)$.
$ab \cos C - ac \cos B = ac \cos B - bc \cos A$.
$ab \cos C + bc \cos A = 2ac \cos B$.
Substituting $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and $\cos A = \frac{b^2+c^2-a^2}{2bc}$ and $\cos B = \frac{a^2+c^2-b^2}{2ac}$:
$ab(\frac{a^2+b^2-c^2}{2ab}) + bc(\frac{b^2+c^2-a^2}{2bc}) = 2ac(\frac{a^2+c^2-b^2}{2ac})$.
$\frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} = a^2+c^2-b^2$.
$\frac{2b^2}{2} = a^2+c^2-b^2$.
$b^2 = a^2+c^2-b^2 \implies 2b^2 = a^2+c^2$.
Thus,$a^2, b^2, c^2$ are in $A.P.$

(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Rearranging the terms,we get the quadratic equation in $b$: $b^2 - (2c \cos A)b + (c^2 - a^2) = 0$.
Since $c \sin A < a < c$,the quadratic equation has two distinct positive roots,$b_1$ and $b_2$.
According to the properties of quadratic equations,the sum of the roots is given by the negative of the coefficient of $b$ divided by the coefficient of $b^2$.
Therefore,$b_1 + b_2 = -(-(2c \cos A)) / 1 = 2c \cos A$.

(C) Given,${a^2} + {b^2} + {c^2} = ac + ab\sqrt{3}$
Rearranging the terms,we get: ${a^2} + {b^2} + {c^2} - ac - ab\sqrt{3} = 0$
We can rewrite this as: $\left(\frac{a^2}{4} - ac + c^2\right) + \left(\frac{3a^2}{4} - ab\sqrt{3} + b^2\right) = 0$
This simplifies to: ${\left(\frac{a}{2} - c\right)^2} + {\left(\frac{a\sqrt{3}}{2} - b\right)^2} = 0$
For the sum of squares to be zero,each term must be zero:
$\frac{a}{2} - c = 0 \implies a = 2c$
$\frac{a\sqrt{3}}{2} - b = 0 \implies b = \frac{a\sqrt{3}}{2}$
Now,check the relation between sides: ${b^2} + {c^2} = {\left(\frac{a\sqrt{3}}{2}\right)^2} + {\left(\frac{a}{2}\right)^2} = \frac{3a^2}{4} + \frac{a^2}{4} = \frac{4a^2}{4} = {a^2}$
Since ${b^2} + {c^2} = {a^2},$ the triangle is right-angled.

(C) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2}ab \sin C$.
Given $a = 1$,$b = 2$,and $\angle C = 60^\circ$.
Substituting these values into the formula:
$\text{Area} = \frac{1}{2} \times 1 \times 2 \times \sin 60^\circ$.
Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,we have:
$\text{Area} = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

(B) Given $b + c = 2a$ .....$(i)$
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Since $\angle A = 60^\circ$,we have $\frac{1}{2} = \frac{b^2 + c^2 - a^2}{2bc}$
$bc = b^2 + c^2 - a^2$
$a^2 = b^2 + c^2 - bc$
From $(i)$,$a = \frac{b+c}{2}$,so $a^2 = \frac{(b+c)^2}{4} = \frac{b^2 + c^2 + 2bc}{4}$
Equating the two expressions for $a^2$: $\frac{b^2 + c^2 + 2bc}{4} = b^2 + c^2 - bc$
$b^2 + c^2 + 2bc = 4b^2 + 4c^2 - 4bc$
$3b^2 + 3c^2 - 6bc = 0$
$3(b^2 + c^2 - 2bc) = 0$
$3(b - c)^2 = 0$
Thus,$b = c$.
Substituting $b = c$ into $(i)$,$2b = 2a \implies b = a$.
Since $a = b = c$,the triangle is equilateral.

(D) Let the altitudes from vertices $A, B, C$ be $h_a, h_b, h_c$ respectively.
We know that $h_a = \frac{2\Delta}{a}$,$h_b = \frac{2\Delta}{b}$,and $h_c = \frac{2\Delta}{c}$,where $\Delta$ is the area of the triangle.
Given that $h_a, h_b, h_c$ are in $H.P.$,we have $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$
This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$
Therefore,$a, b, c$ are in $A.P.$
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Since $a, b, c$ are in $A.P.$,it follows that $2R \sin A, 2R \sin B, 2R \sin C$ are in $A.P.$
Hence,$\sin A, \sin B, \sin C$ are in $A.P.$

(A) Given the equation: $a^4 + b^4 + c^4 = 2c^2(a^2 + b^2)$
Rearranging the terms: $a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 = 0$
Adding $2a^2b^2$ to both sides: $a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 + 2a^2b^2 = 2a^2b^2$
This simplifies to: $(a^2 + b^2 - c^2)^2 = 2a^2b^2$
Taking the square root: $a^2 + b^2 - c^2 = \pm \sqrt{2}ab$
Dividing by $2ab$: $\frac{a^2 + b^2 - c^2}{2ab} = \pm \frac{\sqrt{2}ab}{2ab} = \pm \frac{1}{\sqrt{2}}$
By the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,so $\cos C = \pm \frac{1}{\sqrt{2}}$
Therefore,$C = 45^\circ$ or $C = 135^\circ$.

(A) Given sides are $a = 3, b = 5, c = 4$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{3 + 5 + 4}{2} = \frac{12}{2} = 6$.
Using the half-angle formulas for a triangle:
$\sin \frac{B}{2} = \sqrt{\frac{(s - a)(s - c)}{ac}} = \sqrt{\frac{(6 - 3)(6 - 4)}{3 \times 4}} = \sqrt{\frac{3 \times 2}{12}} = \sqrt{\frac{6}{12}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
$\cos \frac{B}{2} = \sqrt{\frac{s(s - b)}{ac}} = \sqrt{\frac{6(6 - 5)}{3 \times 4}} = \sqrt{\frac{6 \times 1}{12}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\sin \frac{B}{2} + \cos \frac{B}{2} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k\sin A$,$b = k\sin B$,and $c = k\sin C$.
Consider the expression $\frac{b - c}{a} = \frac{\sin B - \sin C}{\sin A}$.
Using the sum-to-product formula,$\sin B - \sin C = 2\sin \frac{B - C}{2} \cos \frac{B + C}{2}$.
Since $A + B + C = \pi$,we have $\frac{B + C}{2} = \frac{\pi}{2} - \frac{A}{2}$,so $\cos \frac{B + C}{2} = \sin \frac{A}{2}$.
Also,$\sin A = 2\sin \frac{A}{2} \cos \frac{A}{2}$.
Substituting these into the expression:
$\frac{b - c}{a} = \frac{2\sin \frac{B - C}{2} \sin \frac{A}{2}}{2\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\sin \frac{B - C}{2}}{\cos \frac{A}{2}}$.
Rearranging gives $(b - c)\cos \frac{A}{2} = a\sin \frac{B - C}{2}$.

(A) In $\triangle ABC$,$\angle B = 90^\circ$ and it is isosceles,so $\angle CAB = \angle ACB = 45^\circ$.
Given $\angle DCB = 15^\circ$,then $\angle ACD = \angle ACB - \angle DCB = 45^\circ - 15^\circ = 30^\circ$.
In $\triangle DBC$,$\angle B = 90^\circ$ and $\angle DCB = 15^\circ$,so $\angle BDC = 75^\circ$.
Let $BC = h$. Since $\triangle ABC$ is isosceles,$AB = BC = h$.
Then $BD = AB - AD = h - 35$.
In $\triangle DBC$,$\tan(15^\circ) = \frac{BD}{BC} = \frac{h - 35}{h}$.
We know $\tan(15^\circ) = 2 - \sqrt{3}$.
$h(2 - \sqrt{3}) = h - 35 \implies h(1 - \sqrt{3}) = -35 \implies h = \frac{35}{\sqrt{3} - 1} = \frac{35(\sqrt{3} + 1)}{2}$.
Now,in $\triangle DBC$,$\cos(15^\circ) = \frac{BC}{CD} \implies CD = \frac{BC}{\cos(15^\circ)} = \frac{h}{\cos(15^\circ)}$.
Using $\cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$,we get $CD = \frac{35(\sqrt{3} + 1)}{2} \times \frac{4}{\sqrt{6} + \sqrt{2}} = \frac{70(\sqrt{3} + 1)}{\sqrt{2}(\sqrt{3} + 1)} = \frac{70}{\sqrt{2}} = 35\sqrt{2} \, cm$.

(D) Given $\frac{a}{\sin A} = \frac{b}{\sin B}$.
Substituting $a=5, b=4$ and $A = \frac{\pi}{2} + B$:
$\frac{5}{\sin(\frac{\pi}{2} + B)} = \frac{4}{\sin B} \implies \frac{5}{\cos B} = \frac{4}{\sin B}$.
Thus,$\tan B = \frac{4}{5}$.
Since $A = \frac{\pi}{2} + B$,$\tan A = \tan(\frac{\pi}{2} + B) = -\cot B = -\frac{5}{4}$.
In $\triangle ABC$,$C = \pi - (A + B)$.
$\tan C = \tan(\pi - (A + B)) = -\tan(A + B) = -\frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting values: $\tan C = -\frac{-\frac{5}{4} + \frac{4}{5}}{1 - (-\frac{5}{4})(\frac{4}{5})} = -\frac{-\frac{25}{20} + \frac{16}{20}}{1 + 1} = -\frac{-\frac{9}{20}}{2} = \frac{9}{40}$.
Using the identity $2\tan^{-1}(x) = \tan^{-1}(\frac{2x}{1-x^2})$,we check $2\tan^{-1}(\frac{1}{9}) = \tan^{-1}(\frac{2/9}{1 - 1/81}) = \tan^{-1}(\frac{2/9}{80/81}) = \tan^{-1}(\frac{2}{9} \times \frac{81}{80}) = \tan^{-1}(\frac{9}{40})$.
Therefore,$C = 2\tan^{-1}(\frac{1}{9})$.

(C) We know that by the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.
Therefore,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
Substituting these into the expression:
$2R^2 \sin A \sin B \sin C = 2R^2 \left( \frac{a}{2R} \right) \left( \frac{b}{2R} \right) \left( \frac{c}{2R} \right)$
$= 2R^2 \left( \frac{abc}{8R^3} \right)$
$= \frac{abc}{4R}$.
Since the area of a triangle $\Delta = \frac{abc}{4R}$,the expression equals $\Delta$.

(B) Let the sides of the triangle be $a = 3k, b = 7k, c = 8k$.
The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{3k + 7k + 8k}{2} = 9k$.
The circumradius $R$ is given by $R = \frac{abc}{4\Delta}$ and the inradius $r$ is given by $r = \frac{\Delta}{s}$,where $\Delta$ is the area of the triangle.
Thus,$\frac{R}{r} = \frac{abc}{4\Delta} \times \frac{s}{\Delta} = \frac{abc \cdot s}{4\Delta^2}$.
Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c) = 9k(9k-3k)(9k-7k)(9k-8k) = 9k(6k)(2k)(k) = 108k^4$.
Substituting these values: $\frac{R}{r} = \frac{(3k)(7k)(8k)(9k)}{4(108k^4)} = \frac{1512k^4}{432k^4} = \frac{7}{2}$.
Therefore,$R : r = 7 : 2$.

(A) The semi-perimeter $s$ of the triangle is given by $s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7$.
Using Heron's formula,the area $\Delta$ is $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7(7-3)(7-5)(7-6)} = \sqrt{7 \times 4 \times 2 \times 1} = \sqrt{56} = 2\sqrt{14}$.
The inradius $r$ is given by $r = \frac{\Delta}{s} = \frac{\sqrt{56}}{7} = \sqrt{\frac{56}{49}} = \sqrt{\frac{8}{7}}$.

(C) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these in the expression:
$a \cot A + b \cot B + c \cot C = 2R \sin A \cot A + 2R \sin B \cot B + 2R \sin C \cot C$
$= 2R (\cos A + \cos B + \cos C)$
Using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ and $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,we get:
$= 2R \left( 1 + \frac{r}{R} \right)$
$= 2(R + r)$.

(D) Let $R$ be the circumradius of $\Delta PQR$. Given $R = PQ = PR$.
By the sine rule,$R = \frac{PQ}{2 \sin R} = \frac{PR}{2 \sin Q} = \frac{QR}{2 \sin P}$.
Since $R = PQ$,we have $PQ = \frac{PQ}{2 \sin R}$,which implies $\sin R = \frac{1}{2}$.
Thus,$\angle R = \frac{\pi}{6}$.
Since $PQ = PR$,the triangle is isosceles,so $\angle Q = \angle R = \frac{\pi}{6}$.
The sum of angles in a triangle is $\pi$. Therefore,$\angle P = \pi - (\angle Q + \angle R) = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(B) For a regular $n$-sided polygon with side length $a$,the inradius $r$ and circumradius $R$ are given by:
$r = \frac{a}{2} \cot \left( \frac{\pi}{n} \right)$
$R = \frac{a}{2} \csc \left( \frac{\pi}{n} \right)$
Sum $= r + R = \frac{a}{2} \left( \cot \frac{\pi}{n} + \csc \frac{\pi}{n} \right)$
Using the identity $\cot \theta + \csc \theta = \cot \left( \frac{\theta}{2} \right)$:
Sum $= \frac{a}{2} \cot \left( \frac{\pi}{2n} \right)$.

(A) Given that in a $\Delta ABC$,${r_1} < {r_2} < {r_3}$.
Taking the reciprocal of the inequality,we get:
$\frac{1}{r_1} > \frac{1}{r_2} > \frac{1}{r_3}$.
Using the formula for the exradii,$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,and $\frac{1}{r_3} = \frac{s-c}{\Delta}$,where $s$ is the semi-perimeter and $\Delta$ is the area of the triangle:
$\frac{s-a}{\Delta} > \frac{s-b}{\Delta} > \frac{s-c}{\Delta}$.
Multiplying by $\Delta$ (since $\Delta > 0$):
$s - a > s - b > s - c$.
Subtracting $s$ from all parts:
$-a > -b > -c$.
Multiplying by $-1$ reverses the inequality signs:
$a < b < c$.

(B) We know that $a = 2R\sin A$,$b = 2R\sin B$,and $c = 2R\sin C$.
Substituting these in the expression:
$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = \frac{2R\sin A \cos A + 2R\sin B \cos B + 2R\sin C \cos C}{2R\sin A + 2R\sin B + 2R\sin C}$
$= \frac{\sin 2A + \sin 2B + \sin 2C}{2(\sin A + \sin B + \sin C)}$
Using the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 4\cos(A/2)\cos(B/2)\cos(C/2)$:
$= \frac{4\sin A \sin B \sin C}{2(4\cos(A/2)\cos(B/2)\cos(C/2))}$
$= \frac{4(2\sin(A/2)\cos(A/2))(2\sin(B/2)\cos(B/2))(2\sin(C/2)\cos(C/2))}{8\cos(A/2)\cos(B/2)\cos(C/2)}$
$= 8\sin(A/2)\sin(B/2)\sin(C/2)$
Since $r = 4R\sin(A/2)\sin(B/2)\sin(C/2)$,we have $4\sin(A/2)\sin(B/2)\sin(C/2) = r/R$.
Thus,the expression equals $r/R$.

(D) To determine which inequality is not true,we can test with a specific triangle. Let the sides be $a = 2, b = 3, c = 4$. These satisfy the triangle inequality $(2+3 > 4)$.
For option $A$: $8(2)(3)(4) = 192$ and $(2+3)(3+4)(4+2) = 5 \times 7 \times 6 = 210$. Since $192 \le 210$,this is true.
For option $B$: $3(2)(3)(4) = 72$ and $2^3 + 3^3 + 4^3 = 8 + 27 + 64 = 99$. Since $72 \le 99$,this is true.
For option $C$: $6(2)(3)(4) = 144$ and $3(4)(3+4) + 4(2)(4+2) + 2(3)(2+3) = 12(7) + 8(6) + 6(5) = 84 + 48 + 30 = 162$. Since $144 \le 162$,this is true.
For option $D$: $abc = 24$ and $(2+3-4)(3+4-2)(4+2-3) = (1)(5)(3) = 15$. Since $24 \not\le 15$,this inequality is not true.

(B) Using the projection formula,we know that $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting these values into the expression:
$a(b\cos C - c\cos B) = a \left( b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - c \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \right)$
$= a \left( \frac{a^2 + b^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \right)$
$= \frac{a^2 + b^2 - c^2 - (a^2 + c^2 - b^2)}{2}$
$= \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}$
$= \frac{2b^2 - 2c^2}{2}$
$= b^2 - c^2$.

(A) We know the formula for $\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$.
Given $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,we equate the two:
$\sqrt{\frac{s(s - a)}{bc}} = \sqrt{\frac{b + c}{2c}}$
Squaring both sides:
$\frac{s(s - a)}{bc} = \frac{b + c}{2c}$
$\frac{s(s - a)}{b} = \frac{b + c}{2}$
$2s(s - a) = b(b + c)$
Substitute $2s = a + b + c$ and $s - a = \frac{b + c - a}{2}$:
$(a + b + c) \left( \frac{b + c - a}{2} \right) = b^2 + bc$
$((b + c) + a)((b + c) - a) = 2(b^2 + bc)$
$(b + c)^2 - a^2 = 2b^2 + 2bc$
$b^2 + c^2 + 2bc - a^2 = 2b^2 + 2bc$
$c^2 - a^2 = b^2$
$c^2 = a^2 + b^2$.

(B) Given: $(\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3\sin A\sin B$
Using the identity $(x+y)(x-y) = x^2 - y^2$,we get:
$(\sin A + \sin B)^2 - \sin^2 C = 3\sin A\sin B$
$\sin^2 A + \sin^2 B + 2\sin A\sin B - \sin^2 C = 3\sin A\sin B$
$\sin^2 A + \sin^2 B - \sin^2 C = \sin A\sin B$
By the Sine Rule,$\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$. Substituting these:
$\frac{a^2}{4R^2} + \frac{b^2}{4R^2} - \frac{c^2}{4R^2} = \frac{a}{2R} \cdot \frac{b}{2R}$
$a^2 + b^2 - c^2 = ab$
Dividing by $2ab$:
$\frac{a^2 + b^2 - c^2}{2ab} = \frac{ab}{2ab} = \frac{1}{2}$
By the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Therefore,$\cos C = \frac{1}{2} = \cos \frac{\pi}{3}$.
Thus,$\angle C = \frac{\pi}{3}$.

(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
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 given equation:
$\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$
$\cot A = \cot B = \cot C$
Since $A, B, C$ are angles of a triangle,$A = B = C$.
Therefore,the triangle is equilateral.

(C) Let the angles of the triangle be $A, B, C$ in $A.P.$ such that $A \le B \le C$. Since $A + B + C = 180^{\circ}$ and $2B = A + C$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Given the sides are $10 \, cm$ and $9 \, cm$,and $B = 60^{\circ}$ is the middle angle,the side opposite to $B$ (let it be $b$) must be the middle side. Since $10$ and $9$ are the two larger sides,$b$ must be $9 \, cm$ and the largest side $a = 10 \, cm$.
Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting the values: $\cos 60^{\circ} = \frac{10^2 + c^2 - 9^2}{2(10)(c)} \Rightarrow \frac{1}{2} = \frac{100 + c^2 - 81}{20c}$.
$10c = 19 + c^2 \Rightarrow c^2 - 10c + 19 = 0$.
Solving for $c$ using the quadratic formula: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = \frac{10 \pm \sqrt{24}}{2} = 5 \pm \sqrt{6}$.
Since both values are positive and satisfy the triangle inequality,the third side can be $5 - \sqrt{6}$ or $5 + \sqrt{6}$.

(D) In $\triangle ABC$,the sum of angles is $180^\circ$.
$\angle C = 180^\circ - 45^\circ - 75^\circ = 60^\circ$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin 45^\circ = k \frac{1}{\sqrt{2}}$ and $c = k \sin 60^\circ = k \frac{\sqrt{3}}{2}$.
We need to evaluate $a + c\sqrt{2} = k \frac{1}{\sqrt{2}} + k \frac{\sqrt{3}}{2} \sqrt{2} = k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} \right) = k \frac{1 + \sqrt{3}}{\sqrt{2}}$.
Since $b = k \sin 75^\circ = k \sin(45^\circ + 30^\circ) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = k \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Therefore,$2b = 2k \frac{\sqrt{3} + 1}{2\sqrt{2}} = k \frac{\sqrt{3} + 1}{\sqrt{2}}$.
Hence,$a + c\sqrt{2} = 2b$.

(A) Given $\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6} = k$ (where $k$ is a constant).
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,$a = 4(2Rk)$,$b = 5(2Rk)$,and $c = 6(2Rk)$.
Let $a = 4\lambda$,$b = 5\lambda$,and $c = 6\lambda$.
Using the Cosine Rule:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5\lambda)^2 + (6\lambda)^2 - (4\lambda)^2}{2(5\lambda)(6\lambda)} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$.
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4\lambda)^2 + (6\lambda)^2 - (5\lambda)^2}{2(4\lambda)(6\lambda)} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = \frac{9}{16}$.
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4\lambda)^2 + (5\lambda)^2 - (6\lambda)^2}{2(4\lambda)(5\lambda)} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$.
Summing these: $\cos A + \cos B + \cos C = \frac{3}{4} + \frac{9}{16} + \frac{1}{8} = \frac{12 + 9 + 2}{16} = \frac{23}{16}$.
Since $\frac{23}{16} = \frac{69}{48}$,the correct option is $A$.

(B) The area of a triangle is given by the formula $\Delta = \frac{1}{2}ab \sin C.$
Given $a = 2x,$ $b = 2y,$ and $\angle C = 120^\circ.$
Substituting these values into the formula:
$\Delta = \frac{1}{2} \times (2x) \times (2y) \times \sin(120^\circ)$
Since $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2},$
$\Delta = \frac{1}{2} \times 4xy \times \frac{\sqrt{3}}{2}$
$\Delta = 2xy \times \frac{\sqrt{3}}{2} = xy\sqrt{3}.$

(B) Let the sides be $a = x^2 + x + 1$,$b = 2x + 1$,and $c = x^2 - 1$.
The greatest side subtends the greatest angle. Comparing the sides for $x > 1$,$x^2 + x + 1$ is the largest.
Using the Law of Cosines,$\cos \theta = \frac{b^2 + c^2 - a^2}{2bc}$.
$\cos \theta = \frac{(2x + 1)^2 + (x^2 - 1)^2 - (x^2 + x + 1)^2}{2(2x + 1)(x^2 - 1)}$.
Expanding the terms:
$(2x + 1)^2 = 4x^2 + 4x + 1$
$(x^2 - 1)^2 = x^4 - 2x^2 + 1$
$(x^2 + x + 1)^2 = x^4 + x^2 + 1 + 2x^3 + 2x^2 + 2x = x^4 + 2x^3 + 3x^2 + 2x + 1$.
Numerator $= (4x^2 + 4x + 1) + (x^4 - 2x^2 + 1) - (x^4 + 2x^3 + 3x^2 + 2x + 1) = -2x^3 - x^2 + 2x + 1$.
Factorizing the numerator: $-(2x^3 + x^2 - 2x - 1) = -[x^2(2x + 1) - 1(2x + 1)] = -(x^2 - 1)(2x + 1)$.
$\cos \theta = \frac{-(x^2 - 1)(2x + 1)}{2(2x + 1)(x^2 - 1)} = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,$\theta = 120^o$.

(B) 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 $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \times \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}.$
Given $3a = b + c,$ we add $a$ to both sides to get $a + b + c = 4a,$ which implies $2s = 4a,$ so $s = 2a.$
Substituting $s = 2a$ into the expression,we get $\frac{2a}{2a - a} = \frac{2a}{a} = 2.$

(A) We know that $a = 2R \sin A,$ $b = 2R \sin B,$ and $c = 2R \sin C.$
Substituting these in the given equation $8R^2 = a^2 + b^2 + c^2,$
$8R^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2$
$8R^2 = 4R^2(\sin^2 A + \sin^2 B + \sin^2 C)$
$2 = \sin^2 A + \sin^2 B + \sin^2 C$
Using $\sin^2 \theta = 1 - \cos^2 \theta,$
$2 = (1 - \cos^2 A) + (1 - \cos^2 B) + \sin^2 C$
$2 = 2 - \cos^2 A - \cos^2 B + \sin^2 C$
$\cos^2 A + \cos^2 B = \sin^2 C = 1 - \cos^2 C$
$\cos^2 A + \cos^2 B + \cos^2 C = 1$
Using the identity $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C,$
$1 = 1 - 2 \cos A \cos B \cos C$
$2 \cos A \cos B \cos C = 0$
This implies $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0.$
Therefore,$A = \frac{\pi}{2}$ or $B = \frac{\pi}{2}$ or $C = \frac{\pi}{2}.$
Thus,the triangle is a right-angled triangle.

(B) Let $\Delta$ be the area of $\Delta ABC$ and $p_1, p_2, p_3$ be the altitudes from $A, B, C$ respectively.
We know that $\Delta = \frac{1}{2} p_1 a = \frac{1}{2} p_2 b = \frac{1}{2} p_3 c$.
Therefore,$p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.
Given that $p_1, p_2, p_3$ are in harmonic progression $(H.P.)$,
$\frac{1}{p_1}, \frac{1}{p_2}, \frac{1}{p_3}$ are in arithmetic progression $(A.P.)$.
Substituting the values,$\frac{a}{2\Delta}, \frac{b}{2\Delta}, \frac{c}{2\Delta}$ are in $A.P.$
This implies $a, b, c$ are in $A.P.$
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.
Thus,$2R \sin A, 2R \sin B, 2R \sin C$ are in $A.P.$
Therefore,$\sin A, \sin B, \sin C$ are in arithmetic progression.

(B) Given equations are:
$3 \sin P + 4 \cos Q = 6$ $(1)$
$4 \sin Q + 3 \cos P = 1$ $(2)$
Squaring and adding $(1)$ and $(2)$:
$(3 \sin P + 4 \cos Q)^2 + (4 \sin Q + 3 \cos P)^2 = 6^2 + 1^2$
$9 \sin^2 P + 16 \cos^2 Q + 24 \sin P \cos Q + 16 \sin^2 Q + 9 \cos^2 P + 24 \sin Q \cos P = 37$
$9(\sin^2 P + \cos^2 P) + 16(\sin^2 Q + \cos^2 Q) + 24(\sin P \cos Q + \cos P \sin Q) = 37$
$9(1) + 16(1) + 24 \sin(P + Q) = 37$
$25 + 24 \sin(P + Q) = 37$
$24 \sin(P + Q) = 12$
$\sin(P + Q) = \frac{1}{2}$
Since $P + Q + R = \pi$,we have $\sin(P + Q) = \sin(\pi - R) = \sin R$.
Thus,$\sin R = \frac{1}{2}$,which implies $R = \frac{\pi}{6}$ or $R = \frac{5\pi}{6}$.
If $R = \frac{5\pi}{6}$,then $P + Q = \frac{\pi}{6}$. Since $P, Q > 0$,$\sin P < \sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos Q < 1$. This contradicts $3 \sin P + 4 \cos Q = 6$.
Therefore,$R = \frac{\pi}{6}$.

(C) Let the sides be $a = \sin \alpha$,$b = \cos \alpha$,and $c = \sqrt{1 + \sin \alpha \cos \alpha}$.
Since $0 < \alpha < \frac{\pi}{2}$,both $\sin \alpha$ and $\cos \alpha$ are positive,and $c$ is clearly the longest side because $c^2 = 1 + \sin \alpha \cos \alpha > \sin^2 \alpha + \cos^2 \alpha = 1$.
Using the Law of Cosines for the angle $C$ opposite to side $c$:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos C = \frac{\sin^2 \alpha + \cos^2 \alpha - (1 + \sin \alpha \cos \alpha)}{2 \sin \alpha \cos \alpha}$
$\cos C = \frac{1 - 1 - \sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha}$
$\cos C = \frac{-\sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha} = -\frac{1}{2}$
Since $\cos C = -\frac{1}{2}$,the angle $C = 120^\circ$.