Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance Questions in English

(A) In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^\circ$.
Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.
From $A + C = 180^\circ$,we have $A = 180^\circ - C$.
Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.
Similarly,from $B + D = 180^\circ$,we have $B = 180^\circ - D$.
Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.
Now,consider the expression $\cos A - \cos B + \cos C - \cos D$.
Rearranging the terms,we get $(\cos A + \cos C) - (\cos B + \cos D)$.
Substituting the values derived above,we get $0 - 0 = 0$.

(B) Given that $A, B, C$ are angles of a triangle,$A + B + C = \pi$,so $C = \pi - (A + B)$.
We know that $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$.
Starting with the expression: $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C$.
Using the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$,we substitute this into the expression:
$= (2 + 2\cos A \cos B \cos C) - 2\cos A \cos B \cos C$
$= 2$.

(C) We know that $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$.
Applying this to each term: $\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} = \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \sin^2 \frac{C}{2}$.
$= 1 - \frac{1}{2}(\cos A + \cos B) + \sin^2 \frac{C}{2}$.
$= 1 - \cos \frac{A+B}{2} \cos \frac{A-B}{2} + \sin^2 \frac{C}{2}$.
Since $A+B = 180^o - C$,then $\frac{A+B}{2} = 90^o - \frac{C}{2}$,so $\cos \frac{A+B}{2} = \sin \frac{C}{2}$.
$= 1 - \sin \frac{C}{2} \cos \frac{A-B}{2} + \sin^2 \frac{C}{2}$.
$= 1 - \sin \frac{C}{2} (\cos \frac{A-B}{2} - \sin \frac{C}{2})$.
$= 1 - \sin \frac{C}{2} (\cos \frac{A-B}{2} - \cos \frac{A+B}{2})$.
Using $\cos X - \cos Y = 2 \sin \frac{X+Y}{2} \sin \frac{Y-X}{2}$,we get $2 \sin \frac{A}{2} \sin \frac{B}{2}$.
Thus,the expression equals $1 - 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

(C) Given the equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$.
We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$.
Substituting this into the equation:
$1 - \frac{1}{2}\sin^2 2x + \sin 2x + \alpha = 0$.
Let $y = \sin 2x$,where $-1 \le y \le 1$.
The equation becomes $1 - \frac{1}{2}y^2 + y + \alpha = 0$,which simplifies to $y^2 - 2y - 2(1 + \alpha) = 0$.
For $y$ to have a real solution,the discriminant $D \ge 0$:
$D = (-2)^2 - 4(1)(-2(1 + \alpha)) = 4 + 8(1 + \alpha) = 12 + 8\alpha \ge 0 \Rightarrow \alpha \ge -\frac{3}{2}$.
The roots are $y = \frac{2 \pm \sqrt{4 + 8(1 + \alpha)}}{2} = 1 \pm \sqrt{1 + 2(1 + \alpha)} = 1 \pm \sqrt{3 + 2\alpha}$.
Since $-1 \le y \le 1$,we must have $-1 \le 1 - \sqrt{3 + 2\alpha} \le 1$ (the root $1 + \sqrt{3 + 2\alpha}$ will always be $\ge 1$).
$-1 \le 1 - \sqrt{3 + 2\alpha}$ $\Rightarrow \sqrt{3 + 2\alpha} \le 2$ $\Rightarrow 3 + 2\alpha \le 4$ $\Rightarrow \alpha \le \frac{1}{2}$.
Thus,the range is $-\frac{3}{2} \le \alpha \le \frac{1}{2}$.

(B) Given equations are:
$2\sin^2 x + \sin^2 2x = 2$ ... $(i)$
$\sin 2x + \cos 2x = \tan x$ ... $(ii)$
From $(i)$:
$2\sin^2 x + (2\sin x \cos x)^2 = 2$
$2\sin^2 x + 4\sin^2 x \cos^2 x = 2$
$2\sin^2 x + 4\sin^2 x (1 - \sin^2 x) = 2$
$2\sin^2 x + 4\sin^2 x - 4\sin^4 x = 2$
$4\sin^4 x - 6\sin^2 x + 2 = 0$
$2\sin^4 x - 3\sin^2 x + 1 = 0$
$(2\sin^2 x - 1)(\sin^2 x - 1) = 0$
So,$\sin^2 x = \frac{1}{2}$ or $\sin^2 x = 1$.
If $\sin^2 x = 1$,then $\cos^2 x = 0$,which makes $\tan x$ undefined.
If $\sin^2 x = \frac{1}{2}$,then $\cos^2 x = \frac{1}{2}$,so $\tan^2 x = 1$,which means $x = n\pi \pm \frac{\pi}{4}$.
From $(ii)$:
$\frac{2\tan x}{1 + \tan^2 x} + \frac{1 - \tan^2 x}{1 + \tan^2 x} = \tan x$
$2\tan x + 1 - \tan^2 x = \tan x + \tan^3 x$
$\tan^3 x + \tan^2 x - \tan x - 1 = 0$
$\tan^2 x(\tan x + 1) - 1(\tan x + 1) = 0$
$(\tan^2 x - 1)(\tan x + 1) = 0$
So,$\tan^2 x = 1$ or $\tan x = -1$.
Both imply $\tan x = \pm 1$,which gives $x = n\pi \pm \frac{\pi}{4} = (2n + 1)\frac{\pi}{4}$.
Thus,the common roots are $x = (2n + 1)\frac{\pi}{4}$.

(C) We know that $\cot B + \cot C = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C} = \frac{\sin C \cos B + \cos C \sin B}{\sin B \sin C} = \frac{\sin(B+C)}{\sin B \sin C}$.
Since $A+B+C = \pi$,$\sin(B+C) = \sin A$.
So,$\cot B + \cot C = \frac{\sin A}{\sin B \sin C}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Thus,$\cot B + \cot C = \frac{a/2R}{(b/2R)(c/2R)} = \frac{2Ra}{bc}$.
Also,$\cot A = \frac{\cos A}{\sin A} = \frac{(b^2+c^2-a^2)/(2bc)}{a/(2R)} = \frac{R(b^2+c^2-a^2)}{abc}$.
Given $b^2+c^2 = 3a^2$,then $b^2+c^2-a^2 = 2a^2$.
So,$\cot A = \frac{R(2a^2)}{abc} = \frac{2Ra}{bc}$.
Therefore,$\cot B + \cot C - \cot A = \frac{2Ra}{bc} - \frac{2Ra}{bc} = 0$.

(C) Given that $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.$
Therefore,$\frac{1}{\sin^2 \frac{A}{2}}, \frac{1}{\sin^2 \frac{B}{2}}, \frac{1}{\sin^2 \frac{C}{2}}$ are in $A.P.$
Using the formula $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc},$ we have $\frac{bc}{(s-b)(s-c)}, \frac{ac}{(s-a)(s-c)}, \frac{ab}{(s-a)(s-b)}$ are in $A.P.$
This implies $\frac{ac}{(s-a)(s-c)} - \frac{bc}{(s-b)(s-c)} = \frac{ab}{(s-a)(s-b)} - \frac{ac}{(s-a)(s-c)}.$
Simplifying this expression leads to $ab + bc = 2ac.$
Dividing by $abc,$ we get $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}.$
This shows that $a, b, c$ are in $H.P.$

(C) Expand the expression: $(a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$
$= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$
$= (a^2 + b^2)(1) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
$= a^2 + b^2 - 2ab \cos C$
Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$,so the expression equals $c^2$.

(C) Using the cosine rule,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting these into the expression $2(bc \cos A + ca \cos B + ab \cos C)$:
$= 2 \left( bc \cdot \frac{b^2 + c^2 - a^2}{2bc} + ca \cdot \frac{a^2 + c^2 - b^2}{2ac} + ab \cdot \frac{a^2 + b^2 - c^2}{2ab} \right)$
$= (b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + (a^2 + b^2 - c^2)$
$= a^2 + b^2 + c^2$.

(B) Using the sine rule,$a = k \sin A, b = k \sin B, c = k \sin C.$
Substituting these into the expression:
$E = k^3 [\sin^3 A \cos(B-C) + \sin^3 B \cos(C-A) + \sin^3 C \cos(A-B)]$
Since $A+B+C = \pi,$ we have $\sin A = \sin(B+C).$
Thus,$\sin^3 A \cos(B-C) = \sin^2 A \sin(B+C) \cos(B-C) = \sin^2 A \cdot \frac{1}{2} (\sin 2B + \sin 2C).$
Expanding this for all terms and simplifying using the identity $\sin A \sin B \sin C = \frac{abc}{k^3},$ we get:
$E = 3k^3 \sin A \sin B \sin C = 3abc.$

(A) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the expression:
$E = \sum a^2(\cos^2 B - \cos^2 C) = \sum (2R \sin A)^2(\cos^2 B - \cos^2 C)$
Since $\cos^2 \theta = 1 - \sin^2 \theta$,we have $\cos^2 B - \cos^2 C = (1 - \sin^2 B) - (1 - \sin^2 C) = \sin^2 C - \sin^2 B$.
Thus,$E = 4R^2 \sum \sin^2 A(\sin^2 C - \sin^2 B)$.
Using $a = 2R \sin A$,we have $\sin A = \frac{a}{2R}$,so $\sin^2 A = \frac{a^2}{4R^2}$.
$E = 4R^2 \left[ \frac{a^2}{4R^2}(\sin^2 C - \sin^2 B) + \frac{b^2}{4R^2}(\sin^2 A - \sin^2 C) + \frac{c^2}{4R^2}(\sin^2 B - \sin^2 A) \right]$
$E = a^2(\sin^2 C - \sin^2 B) + b^2(\sin^2 A - \sin^2 C) + c^2(\sin^2 B - \sin^2 A)$
Substituting $\sin^2 A = \frac{a^2}{4R^2}$,$\sin^2 B = \frac{b^2}{4R^2}$,$\sin^2 C = \frac{c^2}{4R^2}$:
$E = a^2(\frac{c^2}{4R^2} - \frac{b^2}{4R^2}) + b^2(\frac{a^2}{4R^2} - \frac{c^2}{4R^2}) + c^2(\frac{b^2}{4R^2} - \frac{a^2}{4R^2})$
$E = \frac{1}{4R^2} [a^2c^2 - a^2b^2 + b^2a^2 - b^2c^2 + c^2b^2 - c^2a^2] = 0$.

(D) Given $2b^2 = a^2 + c^2$.
Using the identity $\sin 3B = 3\sin B - 4\sin^3 B$,we have $\frac{\sin 3B}{\sin B} = 3 - 4\sin^2 B$.
Using $\sin^2 B = 1 - \cos^2 B$,we get $3 - 4(1 - \cos^2 B) = 4\cos^2 B - 1$.
From the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Since $b^2 = \frac{a^2 + c^2}{2}$,we substitute this into the expression for $\cos B$:
$\cos B = \frac{a^2 + c^2 - \frac{a^2 + c^2}{2}}{2ac} = \frac{\frac{a^2 + c^2}{2}}{2ac} = \frac{a^2 + c^2}{4ac}$.
Now,substitute $\cos B$ back into $4\cos^2 B - 1$:
$4\left( \frac{a^2 + c^2}{4ac} \right)^2 - 1 = 4 \cdot \frac{(a^2 + c^2)^2}{16a^2c^2} - 1 = \frac{(a^2 + c^2)^2}{4a^2c^2} - 1$.
$= \frac{(a^2 + c^2)^2 - 4a^2c^2}{4a^2c^2} = \frac{(c^2 - a^2)^2}{4a^2c^2} = \left( \frac{c^2 - a^2}{2ac} \right)^2$.

(A) Given: $a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$
Using the half-angle formula $\cos^2\frac{A}{2} = \frac{s(s-a)}{bc}$ and $\cos^2\frac{C}{2} = \frac{s(s-c)}{ab}$,where $s = \frac{a+b+c}{2}$ is the semi-perimeter:
$a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s-c + s-a) = \frac{3b}{2}$
Since $s-c+s-a = 2s - (a+c) = (a+b+c) - (a+c) = b$:
$\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$
Substituting $s = \frac{a+b+c}{2}$:
$\frac{a+b+c}{2} = \frac{3b}{2}$ $\Rightarrow a+b+c = 3b$ $\Rightarrow a+c = 2b$
Thus,$a, b, c$ are in $A.P.$

(D) Let the angles of the triangle be $A-d, A, A+d$. Since the sum of angles in a triangle is $180^o$,we have $(A-d) + A + (A+d) = 180^o$,which implies $3A = 180^o$,so $A = 60^o$.
Given $b:c = \sin B : \sin C = \sqrt{3} : \sqrt{2}$.
Since $B = A = 60^o$ is not necessarily true (the angles are $A-d, A, A+d$),let the angles be $B-d, B, B+d$. Then $B = 60^o$.
Using the sine rule,$\frac{\sin B}{\sin C} = \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$.
Substituting $B = 60^o$,we get $\frac{\sin 60^o}{\sin C} = \frac{\sqrt{3}/2}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$.
This simplifies to $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,so $C = 45^o$.
Since $A+B+C = 180^o$,we have $A + 60^o + 45^o = 180^o$,which gives $A = 75^o$.

(A) Given that $A, B, C$ are in $A.P.$,we have $A + C = 2B$. Since $A + B + C = 180^\circ$,we get $3B = 180^\circ$,so $B = 60^\circ$.
Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting $B = 60^\circ$ and $ac = b^2$: $\cos 60^\circ = \frac{a^2 + c^2 - b^2}{2b^2}$.
$\frac{1}{2} = \frac{a^2 + c^2 - b^2}{2b^2} \Rightarrow b^2 = a^2 + c^2 - b^2$.
Therefore,$a^2 + c^2 = 2b^2$,which implies that $a^2, b^2, c^2$ are in $A.P.$

(A) Given the relation $\cos A \cos B + \sin A \sin B \sin C = 1.$
Rearranging for $\sin C,$ we get $\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B}.$
Since $\sin C \le 1,$ we have $\frac{1 - \cos A \cos B}{\sin A \sin B} \le 1.$
This implies $1 - \cos A \cos B \le \sin A \sin B,$ or $1 \le \cos A \cos B + \sin A \sin B.$
Using the cosine difference formula,$1 \le \cos(A - B).$
Since the maximum value of $\cos \theta$ is $1,$ we must have $\cos(A - B) = 1,$ which implies $A - B = 0$ or $A = B.$
Substituting $A = B$ into the original equation,$\sin C = \frac{1 - \cos^2 A}{\sin^2 A} = \frac{\sin^2 A}{\sin^2 A} = 1.$
Thus,$C = 90^{\circ}.$ Since $A + B + C = 180^{\circ}$ and $A = B,$ we get $2A = 90^{\circ},$ so $A = B = 45^{\circ}.$
Using the sine rule,the sides are proportional to $\sin A : \sin B : \sin C = \sin 45^{\circ} : \sin 45^{\circ} : \sin 90^{\circ} = \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1 = 1 : 1 : \sqrt{2}.$

(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.
This simplifies to $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$.
Thus,$\Delta ABC$ is an equilateral triangle.
The area of an equilateral triangle is given by $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given $a = 2$,$\text{Area} = \frac{\sqrt{3}}{4} (2)^2 = \sqrt{3}$.

(B) Given,$\angle C = 45^o$.
Since $A + B + C = 180^o$,we have $A + B = 180^o - 45^o = 135^o$.
Using the formula $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$,we get:
$\cot(135^o) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$.
Since $\cot(135^o) = -1$,we have:
$-1 = \frac{\cot A \cot B - 1}{\cot A + \cot B}$.
$-(\cot A + \cot B) = \cot A \cot B - 1$.
$1 = \cot A \cot B + \cot A + \cot B$.
Adding $1$ to both sides:
$1 + 1 = 1 + \cot A + \cot B + \cot A \cot B$.
$2 = (1 + \cot A)(1 + \cot B)$.
Thus,$(1 + \cot A)(1 + \cot B) = 2$.

(A) Given that $\alpha + \beta + \gamma = \pi$,so $\gamma = \pi - (\alpha + \beta)$.
Then $\cos \gamma = -\cos(\alpha + \beta)$.
The expression is $E = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$E = (1 - \cos^2 \alpha) + (1 - \cos^2 \beta) + (1 - \cos^2 \gamma) - 2 \cos \alpha \cos \beta \cos \gamma$
$E = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 2 \cos \alpha \cos \beta \cos \gamma$.
Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 - 2 \cos \alpha \cos \beta \cos \gamma$,we substitute this:
$E = 3 - (1 - 2 \cos \alpha \cos \beta \cos \gamma) - 2 \cos \alpha \cos \beta \cos \gamma$
$E = 3 - 1 + 2 \cos \alpha \cos \beta \cos \gamma - 2 \cos \alpha \cos \beta \cos \gamma$
$E = 2$.

(C) Given: $a = 6, b = 3, \cos(A - B) = \frac{4}{5}.$
Using the formula $\cos(A - B) = \frac{1 - \tan^2(\frac{A - B}{2})}{1 + \tan^2(\frac{A - B}{2})},$
$\frac{4}{5} = \frac{1 - t^2}{1 + t^2} \implies 4 + 4t^2 = 5 - 5t^2 \implies 9t^2 = 1 \implies t = \frac{1}{3}.$
So,$\tan(\frac{A - B}{2}) = \frac{1}{3}.$
Using Napier's Analogy: $\tan(\frac{A - B}{2}) = \frac{a - b}{a + b} \cot(\frac{C}{2}).$
$\frac{1}{3} = \frac{6 - 3}{6 + 3} \cot(\frac{C}{2}) = \frac{3}{9} \cot(\frac{C}{2}) = \frac{1}{3} \cot(\frac{C}{2}).$
$\cot(\frac{C}{2}) = 1 \implies \frac{C}{2} = 45^\circ \implies C = 90^\circ.$
The area of $\Delta ABC = \frac{1}{2} ab \sin C = \frac{1}{2} \times 6 \times 3 \times \sin 90^\circ = 9 \times 1 = 9 \, square \, unit.$

(C) Given $\frac{a^2 - b^2}{a^2 + b^2} = \frac{\sin(A - B)}{\sin(A + B)}$.
Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$,we have $\frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B} = \frac{\sin(A - B)}{\sin(A + B)}$.
Since $\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B)$,the equation becomes $\frac{\sin(A - B)\sin(A + B)}{\sin^2 A + \sin^2 B} = \frac{\sin(A - B)}{\sin(A + B)}$.
This implies $\sin(A - B) = 0$ or $\frac{\sin(A + B)}{\sin^2 A + \sin^2 B} = \frac{1}{\sin(A + B)}$.
If $\sin(A - B) = 0$,then $A = B$,so the triangle is isosceles.
If $\sin^2(A + B) = \sin^2 A + \sin^2 B$,then $\sin^2 C = \sin^2 A + \sin^2 B$,which implies $c^2 = a^2 + b^2$,so the triangle is right-angled.
Thus,the triangle is right-angled or isosceles.

(B) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a < b < c$. Given $c = 7 \ cm$.
Since they are in $A.P.$,$2b = a + c$,so $a = 2b - 7$.
The angle opposite to the greatest side $c$ is $120^\circ$. Using the Law of Cosines:
$\cos(120^\circ) = \frac{a^2 + b^2 - c^2}{2ab}$
$-\frac{1}{2} = \frac{(2b-7)^2 + b^2 - 7^2}{2(2b-7)b}$
$-b(2b-7) = 4b^2 - 28b + 49 + b^2 - 49$
$-2b^2 + 7b = 5b^2 - 28b$
$7b^2 - 35b = 0$
Since $b \neq 0$,$b = 5 \ cm$.
Then $a = 2(5) - 7 = 3 \ cm$.
The area of the triangle is $\frac{1}{2}ab \sin(120^\circ) = \frac{1}{2} \times 3 \times 5 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4} \ cm^2$.

(C) We know that the area of triangle $ABC$ is given by $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B$.
Using the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, we have $a = 2R \sin A$ and $b = 2R \sin B$.
Now, consider the expression $E = a^2 \sin 2B + b^2 \sin 2A$.
$E = a^2 (2 \sin B \cos B) + b^2 (2 \sin A \cos A)$.
Substituting $a = 2R \sin A$ and $b = 2R \sin B$:
$E = (2R \sin A)^2 (2 \sin B \cos B) + (2R \sin B)^2 (2 \sin A \cos A)$.
$E = 8R^2 \sin A \sin B (\sin A \cos B + \sin B \cos A)$.
$E = 8R^2 \sin A \sin B \sin(A + B)$.
Since $A + B + C = 180^{\circ}$, $\sin(A + B) = \sin C$.
$E = 8R^2 \sin A \sin B \sin C$.
We know $\Delta = \frac{abc}{4R} = \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{4R} = 2R^2 \sin A \sin B \sin C$.
Therefore, $E = 4(2R^2 \sin A \sin B \sin C) = 4\Delta$.

(C) The given expression is $E = a(b^2 + c^2)\cos A + b(c^2 + a^2)\cos B + c(a^2 + b^2)\cos C$.
Expanding the terms,we get:
$E = ab^2\cos A + ac^2\cos A + bc^2\cos B + ba^2\cos B + ca^2\cos C + cb^2\cos C$.
Grouping the terms:
$E = ab(b\cos A + a\cos B) + bc(c\cos B + b\cos C) + ca(a\cos C + c\cos A)$.
Using the projection formula $c = b\cos A + a\cos B$,$a = c\cos B + b\cos C$,and $b = a\cos C + c\cos A$:
$E = ab(c) + bc(a) + ca(b) = abc + abc + abc = 3abc$.

(D) Given the condition ${c^2} = {a^2} + {b^2}$,the triangle is a right-angled triangle with the hypotenuse $c$ and $\angle C = 90^{\circ}$.
By Heron's formula,the area of the triangle $\Delta$ is given by $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
Squaring both sides,we get $\Delta^2 = s(s - a)(s - b)(s - c)$,which implies $4\Delta^2 = 4s(s - a)(s - b)(s - c)$.
For a right-angled triangle with legs $a$ and $b$,the area is $\Delta = \frac{1}{2}ab$.
Substituting this into the expression,we get $4\Delta^2 = 4(\frac{1}{2}ab)^2 = 4(\frac{1}{4}a^2b^2) = {a^2}{b^2}$.

(A) We have $2s = a + b + c$ and the area $A$ is given by Heron's formula: $A^2 = s(s - a)(s - b)(s - c)$.
By the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for the positive terms $(s-a), (s-b), (s-c)$:
$\frac{(s-a) + (s-b) + (s-c)}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$
Substituting $a+b+c = 2s$:
$\frac{3s - (a+b+c)}{3} \ge \sqrt[3]{\frac{A^2}{s}}$
$\frac{3s - 2s}{3} \ge \sqrt[3]{\frac{A^2}{s}}$
$\frac{s}{3} \ge \sqrt[3]{\frac{A^2}{s}}$
Cubing both sides:
$\frac{s^3}{27} \ge \frac{A^2}{s}$
$A^2 \le \frac{s^4}{27}$
Taking the square root:
$A \le \frac{s^2}{3\sqrt{3}}$.

(D) Given expression: ${a^2}\sin 2C + {c^2}\sin 2A$
$= {a^2}(2\sin C \cos C) + {c^2}(2\sin A \cos A)$
Using the area formula $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A$,we have $\sin C = \frac{2\Delta}{ab}$ and $\sin A = \frac{2\Delta}{bc}$.
Substituting these values:
$= 2{a^2}\left( \frac{2\Delta}{ab} \cos C \right) + 2{c^2}\left( \frac{2\Delta}{bc} \cos A \right)$
$= 4\Delta \left( \frac{a \cos C + c \cos A}{b} \right)$
Since $a \cos C + c \cos A = b$ (Projection Rule):
$= 4\Delta \left( \frac{b}{b} \right) = 4\Delta$.

(C) Using the sine rule,we have $a = 2R\sin A$,$b = 2R\sin B$,and $c = 2R\sin C$.
Substituting these into the expression:
$a\cos A + b\cos B + c\cos C = (2R\sin A)\cos A + (2R\sin B)\cos B + (2R\sin C)\cos C$
$= R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C)$
$= R(\sin 2A + \sin 2B + \sin 2C)$
Using the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ for a triangle $ABC$:
$= R(4\sin A\sin B\sin C)$
$= 4R\sin A\sin B\sin C$.

(A) Let $\Delta$ be the area of the triangle. The perpendicular distances from the circumcenter to the sides $a, b, c$ are $x = R \cos A$,$y = R \cos B$,and $z = R \cos C$.
However,the problem states $x, y, z$ are perpendiculars to sides $a, b, c$. In a triangle,the perpendicular distance from the circumcenter to side $a$ is $R \cos A$.
Given $\Delta = \frac{1}{2}ax = \frac{1}{2}by = \frac{1}{2}cz$,we have $x = \frac{2\Delta}{a}$,$y = \frac{2\Delta}{b}$,and $z = \frac{2\Delta}{c}$.
Substituting these into the expression:
$\frac{bx}{c} + \frac{cy}{a} + \frac{az}{b} = \frac{b}{c}(\frac{2\Delta}{a}) + \frac{c}{a}(\frac{2\Delta}{b}) + \frac{a}{b}(\frac{2\Delta}{c})$
$= 2\Delta \left( \frac{b^2 + c^2 + a^2}{abc} \right)$
Using $\Delta = \frac{abc}{4R}$,we get:
$= 2(\frac{abc}{4R}) \left( \frac{a^2 + b^2 + c^2}{abc} \right) = \frac{a^2 + b^2 + c^2}{2R}$.

(D) Let $AB = h$. Since $C$ is the midpoint of $AB$,$AC = \frac{h}{2}$ and $CB = \frac{h}{2}$.
Given $AP = n \cdot AB = nh$.
In $\triangle PAC$,$\tan(\angle APC) = \frac{AC}{AP} = \frac{h/2}{nh} = \frac{1}{2n}$.
In $\triangle PAB$,$\tan(\angle APB) = \frac{AB}{AP} = \frac{h}{nh} = \frac{1}{n}$.
We know $\alpha = \angle APB - \angle APC$.
Therefore,$\tan \alpha = \tan(\angle APB - \angle APC) = \frac{\tan(\angle APB) - \tan(\angle APC)}{1 + \tan(\angle APB) \cdot \tan(\angle APC)}$.
Substituting the values: $\tan \alpha = \frac{\frac{1}{n} - \frac{1}{2n}}{1 + (\frac{1}{n})(\frac{1}{2n})} = \frac{\frac{1}{2n}}{1 + \frac{1}{2n^2}} = \frac{\frac{1}{2n}}{\frac{2n^2 + 1}{2n^2}} = \frac{1}{2n} \cdot \frac{2n^2}{2n^2 + 1} = \frac{n}{2n^2 + 1}$.
Thus,$n = (2n^2 + 1)\tan \alpha$.

(C) We know that $\cos^2\frac{A}{2} = \frac{s(s-a)}{bc}$.
Substituting this into the expression:
$\sum \frac{1}{a} \cos^2\frac{A}{2} = \sum \frac{1}{a} \cdot \frac{s(s-a)}{bc}$
$= \sum \frac{s(s-a)}{abc}$
$= \frac{s}{abc} [(s-a) + (s-b) + (s-c)]$
$= \frac{s}{abc} [3s - (a+b+c)]$
Since $a+b+c = 2s$,we have:
$= \frac{s}{abc} [3s - 2s] = \frac{s}{abc} \cdot s = \frac{s^2}{abc}$.

(D) We know that $\tan \frac{A}{2} \tan \frac{C}{2} = \frac{s-b}{s}$.
Given $\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,we have $\frac{5}{6} \times \frac{2}{5} = \frac{s-b}{s}$.
$\frac{1}{3} = \frac{s-b}{s}$ $\Rightarrow s = 3s - 3b$ $\Rightarrow 2s = 3b$.
Since $2s = a + b + c$,we have $a + b + c = 3b$,which implies $a + c = 2b$.
Thus,$a, b, c$ are in $A.P.$
By the Sine Rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.
Since $a, b, c$ are in $A.P.$,$2R \sin A, 2R \sin B, 2R \sin C$ are in $A.P.$
Therefore,$\sin A, \sin B, \sin C$ are also in $A.P.$

(A) Let the triangle be $\Delta ABC$ with base $BC = x$ and height $AD = h$. The angles at the base are $\angle B = 22.5^o$ and $\angle ACB = 112.5^o$.
Since $\angle ACB$ is an exterior angle to $\Delta ABD$,the angle $\angle ACD = 180^o - 112.5^o = 67.5^o$.
In right-angled $\Delta ACD$,$\angle CAD = 90^o - 67.5^o = 22.5^o$.
Thus,$\angle BAC = 180^o - (22.5^o + 112.5^o) = 45^o$.
Using the sine rule in $\Delta ABC$,$\frac{x}{\sin 45^o} = \frac{AC}{\sin 22.5^o} \Rightarrow AC = \frac{x \sin 22.5^o}{\sin 45^o}$.
In $\Delta ACD$,$h = AC \sin 67.5^o = \frac{x \sin 22.5^o \sin 67.5^o}{\sin 45^o}$.
Using $\sin 67.5^o = \cos 22.5^o$,we get $h = \frac{x \sin 22.5^o \cos 22.5^o}{\sin 45^o} = \frac{x \sin 45^o}{2 \sin 45^o} = \frac{x}{2}$.
Therefore,the ratio $\frac{h}{x} = \frac{1}{2}$.

(C) Given the equation $a \cos x + b \sin x = c.$
Using the half-angle formulas $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$ and $\sin x = \frac{2 \tan(x/2)}{1 + \tan^2(x/2)},$ we get:
$a \left( \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \right) + b \left( \frac{2 \tan(x/2)}{1 + \tan^2(x/2)} \right) = c$
$a(1 - \tan^2(x/2)) + 2b \tan(x/2) = c(1 + \tan^2(x/2))$
$(a + c) \tan^2(x/2) - 2b \tan(x/2) + (c - a) = 0$
Let $t = \tan(x/2).$ The roots of this quadratic equation are $t_1 = \tan(\alpha/2)$ and $t_2 = \tan(\beta/2).$
From the properties of quadratic equations,the sum of roots is $t_1 + t_2 = \frac{2b}{a + c}$ and the product of roots is $t_1 t_2 = \frac{c - a}{a + c}.$
We need to find $\tan \left( \frac{\alpha + \beta}{2} \right) = \tan \left( \frac{\alpha}{2} + \frac{\beta}{2} \right).$
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},$
$\tan \left( \frac{\alpha + \beta}{2} \right) = \frac{t_1 + t_2}{1 - t_1 t_2} = \frac{\frac{2b}{a + c}}{1 - \frac{c - a}{a + c}} = \frac{2b}{a + c - c + a} = \frac{2b}{2a} = \frac{b}{a}.$

(C) Let $\angle BAD = \alpha$ and $\angle CAD = \beta$. Since $AD \perp BC$,we have $\tan \alpha = \frac{BD}{AD} = \frac{2}{6} = \frac{1}{3}$ and $\tan \beta = \frac{CD}{AD} = \frac{3}{6} = \frac{1}{2}$.
We need to find $\angle A = \alpha + \beta$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$,we get:
$\tan A = \frac{\frac{1}{3} + \frac{1}{2}}{1 - (\frac{1}{3} \times \frac{1}{2})} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$.
Since $\tan A = 1$,we have $A = \frac{\pi}{4}$.

(B) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.
Since $A + B + C = \pi$,we have $A = \pi - (B + C)$,so $\sin A = \sin (B + C)$.
Similarly,$C = \pi - (A + B)$,so $\sin C = \sin (A + B)$.
Substituting these,we get $\frac{\sin (B + C)}{\sin (A + B)} = \frac{\sin (A - B)}{\sin (B - C)}$.
Cross-multiplying gives $\sin (B + C) \sin (B - C) = \sin (A + B) \sin (A - B)$.
Using the identity $\sin (x+y) \sin (x-y) = \sin^2 x - \sin^2 y$,we get $\sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$.
Rearranging,$2 \sin^2 B = \sin^2 A + \sin^2 C$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Substituting these,$2(\frac{b}{2R})^2 = (\frac{a}{2R})^2 + (\frac{c}{2R})^2$.
This simplifies to $2b^2 = a^2 + c^2$.
Thus,$a^2, b^2, c^2$ are in $A.P.$

(A) In a right-angled triangle $\Delta ABC$ with $\angle C = \frac{\pi}{2},$ the hypotenuse is $c = AB.$
The circumradius $R$ is half of the hypotenuse,so $R = \frac{c}{2}.$
The inradius $r$ is given by $r = \frac{\Delta}{s},$ where $\Delta = \frac{1}{2}ab$ and $s = \frac{a + b + c}{2}.$
Thus,$r = \frac{\frac{1}{2}ab}{\frac{1}{2}(a + b + c)} = \frac{ab}{a + b + c}.$
Now,$r + R = \frac{ab}{a + b + c} + \frac{c}{2} = \frac{2ab + c(a + b + c)}{2(a + b + c)}.$
Since $c^2 = a^2 + b^2,$ we have $2ab + c(a + b + c) = 2ab + ca + cb + c^2 = 2ab + ca + cb + a^2 + b^2 = (a + b)^2 + c(a + b) = (a + b)(a + b + c).$
Therefore,$r + R = \frac{(a + b)(a + b + c)}{2(a + b + c)} = \frac{a + b}{2}.$
Hence,$2(r + R) = a + b.$

(C) Let the tower be $AB$ with height $h$. Let the two points on the horizontal plane be $C$ and $D$ such that $BC = b$ and $BD = a$.
Given that the angles of elevation from $C$ and $D$ are $\alpha$ and $90^\circ - \alpha$ respectively.
In $\Delta ABC$,$\tan \alpha = \frac{AB}{BC} = \frac{h}{b} \implies h = b \tan \alpha$ ... $(i)$
In $\Delta ABD$,$\tan(90^\circ - \alpha) = \frac{AB}{BD} = \frac{h}{a} \implies \cot \alpha = \frac{h}{a} \implies h = a \cot \alpha$ ... (ii)
Multiplying equations $(i)$ and (ii):
$h^2 = (b \tan \alpha)(a \cot \alpha) = ab \tan \alpha \cdot \frac{1}{\tan \alpha} = ab$
Therefore,$h = \sqrt{ab}$.

(B) Let $DP = h$ be the height of the clock tower standing at the mid-point $D$ of $BC$.
Given $\angle PAD = \alpha = \cot^{-1} 3.2 \Rightarrow \cot \alpha = 3.2$.
Given $\angle PBD = \beta = \csc^{-1} 2.6 \Rightarrow \csc \beta = 2.6$.
Then $\cot \beta = \sqrt{\csc^2 \beta - 1} = \sqrt{(2.6)^2 - 1} = \sqrt{6.76 - 1} = \sqrt{5.76} = 2.4$.
In right-angled triangles $\triangle PAD$ and $\triangle PBD$:
$AD = h \cot \alpha = 3.2h$ and $BD = h \cot \beta = 2.4h$.
In the right-angled $\triangle ABD$ (since $AD \perp BC$ in isosceles $\triangle ABC$ where $AB=AC$):
$AB^2 = AD^2 + BD^2$
$100^2 = (3.2h)^2 + (2.4h)^2$
$10000 = (10.24 + 5.76)h^2$
$10000 = 16h^2$
$h^2 = \frac{10000}{16} = 625$
$h = 25 \, m$.

(D) Let $H = 150 \, m$ be the height of the second chimney.
Let $d$ be the distance between the two chimneys.
From the geometry of the problem,the angle of depression to the foot of the first chimney is $\phi$,so $\tan \phi = \frac{H}{d}$.
Given $\tan \phi = \frac{5}{2}$,we have $\frac{150}{d} = \frac{5}{2}$,which gives $d = \frac{150 \times 2}{5} = 60 \, m$.
Let $h$ be the height of the first chimney. The angle of depression to the top of the first chimney is $\theta$,so $\tan \theta = \frac{H - h}{d}$.
Given $\tan \theta = \frac{4}{3}$,we have $\frac{150 - h}{60} = \frac{4}{3}$.
$150 - h = 60 \times \frac{4}{3} = 80$.
$h = 150 - 80 = 70 \, m$.
The vertical distance between the tops is $H - h = 150 - 70 = 80 \, m$.
The horizontal distance between the tops is $d = 60 \, m$.
The distance between the tops is $\sqrt{(H - h)^2 + d^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, m$.

(A) Let $AB = x$. Drop a perpendicular from $D$ to $AB$ at point $M$. Then $DM = BC = p$ and $MB = CD = q$. Thus,$AM = AB - MB = x - q$.
In $\triangle BCD$,$\tan \alpha = \frac{p}{q}$.
In $\triangle ADM$,$\angle DAM = \pi - (\theta + \alpha)$.
Therefore,$\tan(\pi - (\theta + \alpha)) = \frac{DM}{AM} = \frac{p}{x - q}$.
This implies $-\tan(\theta + \alpha) = \frac{p}{x - q}$,or $\tan(\theta + \alpha) = \frac{p}{q - x}$.
$q - x = p \cot(\theta + \alpha) = p \left( \frac{\cot \theta \cot \alpha - 1}{\cot \alpha + \cot \theta} \right)$.
Since $\cot \alpha = \frac{q}{p}$,we have $q - x = p \left( \frac{\frac{q}{p} \cot \theta - 1}{\frac{q}{p} + \cot \theta} \right) = p \left( \frac{q \cot \theta - p}{q + p \cot \theta} \right) = p \left( \frac{q \cos \theta - p \sin \theta}{q \sin \theta + p \cos \theta} \right)$.
$x = q - \frac{pq \cos \theta - p^2 \sin \theta}{q \sin \theta + p \cos \theta} = \frac{q^2 \sin \theta + pq \cos \theta - pq \cos \theta + p^2 \sin \theta}{p \cos \theta + q \sin \theta}$.
$x = \frac{(p^2 + q^2) \sin \theta}{p \cos \theta + q \sin \theta}$.

(A) Given the expression $\sum \frac{\cot A + \cot B}{\tan A + \tan B}$.
We know that $\cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin B \cos A + \cos B \sin A}{\sin A \sin B} = \frac{\sin(A+B)}{\sin A \sin B}$.
Also,$\tan A + \tan B = \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \frac{\sin(A+B)}{\cos A \cos B}$.
Therefore,$\frac{\cot A + \cot B}{\tan A + \tan B} = \frac{\sin(A+B)}{\sin A \sin B} \times \frac{\cos A \cos B}{\sin(A+B)} = \cot A \cot B$.
Summing over the cyclic permutations,we get $\sum \cot A \cot B = \cot A \cot B + \cot B \cot C + \cot C \cot A$.
For any triangle $ABC$,$A+B+C = \pi$,which implies $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.

(C) Let the right-angled triangle have sides $a$ and $b$ and hypotenuse $c$. Let $p$ be the length of the perpendicular from the right-angle vertex to the hypotenuse.
We know that $p = \frac{ab}{c}$.
Given $c = 2\sqrt{2}p$,we have $c = 2\sqrt{2} \frac{ab}{c}$,so $c^2 = 2\sqrt{2}ab$.
Since $c^2 = a^2 + b^2$,we have $a^2 + b^2 = 2\sqrt{2}ab$.
Dividing by $ab$,we get $\frac{a}{b} + \frac{b}{a} = 2\sqrt{2}$.
Let $\theta$ be one of the acute angles,so $\tan \theta = \frac{b}{a}$.
Then $\cot \theta + \tan \theta = 2\sqrt{2}$.
$\frac{1}{\tan \theta} + \tan \theta = 2\sqrt{2} \Rightarrow \frac{1 + \tan^2 \theta}{\tan \theta} = 2\sqrt{2}$.
$\frac{\sec^2 \theta}{\tan \theta} = 2\sqrt{2}$ $\Rightarrow \frac{1}{\sin \theta \cos \theta} = 2\sqrt{2}$ $\Rightarrow \frac{2}{\sin 2\theta} = 2\sqrt{2}$.
$\sin 2\theta = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$.
Thus,$2\theta = \frac{\pi}{4}$ or $2\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
So,$\theta = \frac{\pi}{8}$ or $\theta = \frac{3\pi}{8}$.
The two acute angles are $\frac{\pi}{8}$ and $\frac{3\pi}{8}$.

(D) Given: $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$.
For a right-angled isosceles triangle,the angles are $90^{\circ}, 45^{\circ}, 45^{\circ}$.
Let $A = 90^{\circ}, B = 45^{\circ}, C = 45^{\circ}$.
Then $\sin 90^{\circ} + \sin 45^{\circ} + \sin 45^{\circ} = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 1 + \frac{2}{\sqrt{2}} = 1 + \sqrt{2}$.
Also,$\cos 90^{\circ} + \cos 45^{\circ} + \cos 45^{\circ} = 0 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Since these values satisfy the given equations,the triangle is a right-angled isosceles triangle.

(C) Let $\angle BAD = \alpha$ and $\angle CAD = \beta$.
In $\Delta ADB$,applying the sine rule:
$\frac{BD}{\sin \alpha} = \frac{AD}{\sin B} \implies \frac{x}{\sin \alpha} = \frac{AD}{\sin(\pi/3)}$ .....$(i)$
In $\Delta ADC$,applying the sine rule:
$\frac{CD}{\sin \beta} = \frac{AD}{\sin C} \implies \frac{3x}{\sin \beta} = \frac{AD}{\sin(\pi/4)}$ .....(ii)
Dividing $(i)$ by (ii):
$\frac{x}{\sin \alpha} \times \frac{\sin \beta}{3x} = \frac{AD}{\sin(\pi/3)} \times \frac{\sin(\pi/4)}{AD}$
$\frac{\sin \beta}{3 \sin \alpha} = \frac{1/\sqrt{2}}{\sqrt{3}/2} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$
$\frac{\sin \beta}{\sin \alpha} = 3 \times \sqrt{\frac{2}{3}} = \sqrt{3} \times \sqrt{2} = \sqrt{6}$
Therefore,$\frac{\sin \angle BAD}{\sin \angle CAD} = \frac{\sin \alpha}{\sin \beta} = \frac{1}{\sqrt{6}}$.

(C) Given the equation $3\sin x - 4\sin^3 x = k$,we know that $\sin(3x) = k$.
Since angles $A$ and $B$ satisfy this,we have $\sin(3A) = k$ and $\sin(3B) = k$.
This implies $\sin(3A) - \sin(3B) = 0$,which gives $2\cos\left(\frac{3A+3B}{2}\right)\sin\left(\frac{3A-3B}{2}\right) = 0$.
Since $A > B$,$\sin\left(\frac{3A-3B}{2}\right) \neq 0$,so we must have $\cos\left(\frac{3A+3B}{2}\right) = 0$.
Thus,$\frac{3(A+B)}{2} = \frac{\pi}{2} + n\pi$. For a triangle,$A+B < \pi$,so $\frac{3(A+B)}{2} = \frac{\pi}{2}$ or $\frac{3\pi}{2}$.
If $\frac{3(A+B)}{2} = \frac{\pi}{2}$,then $A+B = \frac{\pi}{3}$,which gives $C = \pi - (A+B) = \frac{2\pi}{3}$.
If $\frac{3(A+B)}{2} = \frac{3\pi}{2}$,then $A+B = \pi$,which is impossible for a triangle as $C$ would be $0$.

(B) Let the total height of the pole be $h$. The pole is divided into two parts: the upper part of length $\frac{3h}{4}$ and the lower part of length $\frac{h}{4}$.
Let the point on the ground be $C$ and the foot of the pole be $B$. Given $BC = 40 \ m$.
Let $\theta$ be the angle of elevation of the top of the pole $(A)$ and $\alpha$ be the angle of elevation of the point $D$ (where the pole is divided).
Then $\tan \theta = \frac{h}{40}$ and $\tan \alpha = \frac{h/4}{40} = \frac{h}{160}$.
The angle subtended by the upper part $AD$ at $C$ is $\beta = \theta - \alpha = \tan^{-1}\left(\frac{3}{5}\right)$.
Thus,$\tan \beta = \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha} = \frac{3}{5}$.
Substituting the values: $\frac{\frac{h}{40} - \frac{h}{160}}{1 + \left(\frac{h}{40}\right)\left(\frac{h}{160}\right)} = \frac{3}{5}$.
$\frac{\frac{3h}{160}}{1 + \frac{h^2}{6400}} = \frac{3}{5}$ $\Rightarrow \frac{3h}{160} \times \frac{6400}{6400 + h^2} = \frac{3}{5}$.
$\frac{120h}{6400 + h^2} = \frac{3}{5}$ $\Rightarrow 600h = 3(6400 + h^2)$ $\Rightarrow 200h = 6400 + h^2$.
$h^2 - 200h + 6400 = 0 \Rightarrow (h - 160)(h - 40) = 0$.
So,$h = 40 \ m$ or $h = 160 \ m$.

(A) Let $P$ be the eye of the observer and $B$ be the centre of the spherical balloon. Let $Q$ be a point of tangency on the balloon from $P$. Then $\angle QPB = \frac{\alpha}{2}$.
In the right-angled triangle $\triangle PQB$,we have $\sin\left(\frac{\alpha}{2}\right) = \frac{BQ}{PB} = \frac{r}{l}$,where $l = PB$ is the distance from the observer to the centre of the balloon.
Thus,$l = r \csc\left(\frac{\alpha}{2}\right)$.
In the right-angled triangle formed by the height $H$ of the centre $B$ from the ground,we have $\sin \beta = \frac{H}{l}$.
Therefore,$H = l \sin \beta = r \csc\left(\frac{\alpha}{2}\right) \sin \beta$.