Question 1: If in a $\triangle ABC, \angle A = 45^o, \angle B = 60^o$ and $\angle C = 75^o$; find the ratios of their sides.

Given in a $\triangle ABC, \angle A = 45^o, \angle B = 60^o$ and $\angle C = 75^o$;

Using Sine Rule $\frac{a}{\sin A}$ $=$ $\frac{b}{\sin B}$ $=$ $\frac{c}{\sin C}$ $= k$

$\Rightarrow$ $\frac{a}{\sin 45^o}$ $=$ $\frac{b}{\sin 60^o}$ $=$ $\frac{c}{\sin 75^o}$ $= k$

$\Rightarrow$ $\frac{a}{\frac{1}{\sqrt{2}}}$ $=$ $\frac{b}{\frac{\sqrt{3}}{2}}$ $=$ $\frac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}$ $= k$

$\Rightarrow a =$ $\frac{k}{\sqrt{2}}$ $, b =$ $\frac{\sqrt{3}}{2}$ $k , c =$ $\frac{\sqrt{3}+1}{2\sqrt{2}}$ $k$

$\therefore a : b: c =$ $\frac{1}{\sqrt{2}}$ $:$ $\frac{\sqrt{3}}{2}$ $:$ $\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\Rightarrow a : b: c = 2 : \sqrt{6} : (\sqrt{3} +1)$

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Question 2: If in a $\triangle ABC, \angle C = 105^o, \angle B = 45^o a = 2$, then find $b$.

Given in a $\triangle ABC, \angle C = 105^o, \angle B = 45^o a = 2$

$\therefore \angle A = 30^o$

Using sine rule,

$\frac{a}{\sin A}$ $=$ $\frac{b}{\sin B}$

$\Rightarrow$ $\frac{2}{\sin 30^o}$ $=$ $\frac{b}{\sin 45^o}$

$\Rightarrow b = 2 \times$ $\frac{\sin 45^o}{\sin 30^o}$ $= 2\sqrt{2}$

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Question 3: If in $\triangle ABC$, if $a = 18, b = 24$ and $c = 30$ and $\angle C = 90^o$, then find $\sin A, \sin B$ and $\sin C$

Given in $\triangle ABC$, if $a = 18, b = 24$ and $c = 30$ and $\angle C = 90^o$

Using sine rule,

$\frac{a}{\sin A}$ $=$ $\frac{b}{\sin B}$ $=$ $\frac{c}{\sin C}$

$\frac{18}{\sin A}$ $=$ $\frac{24}{\sin B}$ $=$ $\frac{30}{1}$

$\therefore \sin A =$ $\frac{18}{30}$ $=$ $\frac{3}{5}$

$\therefore \sin B =$ $\frac{24}{30}$ $=$ $\frac{4}{5}$

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In any triangle ABC, prove the following:

Question 4: $\frac{a-b}{a+b}$ $=$ $\frac{\tan ( \frac{A-B}{2}) }{\tan ( \frac{A+B}{2} )}$

LHS $=$ $\frac{a-b}{a+b}$

$=$ $\frac{k \sin A - k \sin B}{k \sin A + k \sin B}$

$=$ $\frac{ \sin A - \sin B}{ \sin A + \sin B}$

$=$ $\frac{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}}{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}$

$=$ $\frac{\tan \frac{A-B}{2}}{\tan \frac{A+B}{2}}$ $=$ RHS. Hence proved.

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Question 5: $(a-b) \cos$ $\frac{C}{2}$ $=$ $c \sin \Big($ $\frac{A-B}{2}$ $\Big)$

LHS $= (a-b) \cos$ $\frac{C}{2}$

$= k (\sin A - \sin B) \cos$ $\frac{C}{2}$

$= 2 k \cos$ $\frac{A+B}{2}$ $\sin$ $\frac{A-B}{2}$ $\cos$ $\frac{C}{2}$

$= 2 k \cos$ $\frac{\pi - C}{2}$ $\sin$ $\frac{A-B}{2}$ $\cos$ $\frac{C}{2}$

$= 2 k \sin$ $\frac{C}{2}$ $\sin$ $\frac{A-B}{2}$ $\cos$ $\frac{C}{2}$

$= k \sin C \sin$ $\frac{A-B}{2}$

$= c \sin$ $\frac{A-B}{2}$ $=$ RHS. Hence proved.

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Question 6: $\frac{c}{a-b}$ $=$ $\frac{ \tan \frac{A}{2} + \tan \frac{B}{2}}{ \tan \frac{A}{2} - \tan \frac{B}{2}}$

LHS $=$ $\frac{c}{a-b}$

$=$ $\frac{k\sin C}{k\sin A - k\sin B}$

$=$ $\frac{\sin C}{\sin A - \sin B}$

$=$ $\frac{2 \sin \frac{C}{2} \cos \frac{C}{2} }{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} }$

$= \frac{2 \sin \frac{C}{2} \cos \frac{\pi - ( A+B)}{2} }{2 \cos \frac{\pi - C}{2} \sin \frac{A-B}{2} }$

$=$ $\frac{ \sin \frac{C}{2} \sin \frac{ A+B}{2} }{ \sin \frac{ C}{2} \sin \frac{A-B}{2} }$

$=$ $\frac{ \sin \frac{ A+B}{2} }{ \sin \frac{A-B}{2} }$

$=$ $\frac{\sin \frac{A}{2} \cos \frac{B}{2} + \sin \frac{B}{2} \cos \frac{A}{2}}{\sin \frac{A}{2} \cos \frac{B}{2} - \sin \frac{B}{2} \cos \frac{A}{2}}$

Dividing Numerator and Denominator by $\cos$ $\frac{A}{2}$ $\cos$ $\frac{B}{2}$ we get

$=$ $\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\tan \frac{A}{2}-\tan \frac{B}{2}}$ $=$ RHS. Hence proved.

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Question 7: $\frac{c}{a+b}$ $=$ $\frac{ 1 - \tan \frac{A}{2} \tan \frac{B}{2}}{ 1+ \tan \frac{A}{2} \tan \frac{B}{2}}$

LHS $=$ $\frac{c}{a+b}$

$=$ $\frac{k\sin C}{k\sin A + k\sin B}$

$=$ $\frac{\sin C}{\sin A + \sin B}$

$=$ $\frac{2 \sin \frac{C}{2} \cos \frac{C}{2} }{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} }$

$= \frac{2 \sin \frac{\pi - ( A+B)}{2} \cos \frac{C}{2} }{2 \sin \frac{\pi - C}{2} \cos \frac{A-B}{2} }$

$=$ $\frac{ \cos \frac{A+B}{2} \cos \frac{ C}{2} }{ \cos \frac{ C}{2} \cos \frac{A-B}{2} }$

$=$ $\frac{ \cos \frac{ A+B}{2} }{ \cos \frac{A-B}{2} }$

$=$ $\frac{\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2}}{\cos \frac{A}{2} \cos \frac{B}{2} + \sin \frac{A}{2} \sin \frac{B}{2}}$

Dividing Numerator and Denominator by $\cos$ $\frac{A}{2}$ $\cos$ $\frac{B}{2}$ we get

$=$ $\frac{1 - \tan \frac{A}{2} \tan \frac{B}{2}}{1 + \tan \frac{A}{2} \tan \frac{B}{2}}$ $=$ RHS. Hence proved.

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Question 8: $\frac{a+b}{c}$ $=$ $\frac{\cos ( \frac{A-B}{2} ) }{\sin \frac{C}{2} }$

LHS $=$ $\frac{a+b}{c}$

$=$ $\frac{k\sin A + k\sin B}{k\sin C}$

$=$ $\frac{\sin A + \sin B}{\sin C}$

$=$ $\frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \sin \frac{C}{2} \cos \frac{C}{2} }$

$=$ $\frac{2 \sin \frac{\pi - C}{2} \cos \frac{A-B}{2}}{2 \sin \frac{C}{2} \cos \frac{C}{2} }$

$=$ $\frac{ \cos \frac{C}{2} \cos \frac{A-B}{2}}{ \sin \frac{C}{2} \cos \frac{C}{2} }$

$=$ $\frac{ \cos \frac{A-B}{2}}{ \sin \frac{C}{2} }$ $=$ RHS. Hence proved.

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Question 9:  $\sin ($ $\frac{B-C}{2}$ $) =$ $\frac{b-c}{a}$ $\cos$ $\frac{A}{2}$

LHS $=$ $\frac{b-c}{a}$ $\cos$ $\frac{A}{2}$

$=$ $\frac{k\sin B - k\sin C}{k\sin A}$ $\cos$ $\frac{A}{2}$

$=$ $\frac{\sin B - \sin C}{\sin A}$ $\cos$ $\frac{A}{2}$

$=$ $\frac{2 \cos \frac{B+C}{2} \sin \frac{B-C}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2} }$ $\cos$ $\frac{A}{2}$

$=$ $\frac{ \cos \frac{\pi - A}{2} \sin \frac{B-C}{2} }{ \sin \frac{A}{2} }$

$=$ $\frac{ \sin \frac{A}{2} \sin \frac{B-C}{2}}{ \sin \frac{A}{2} }$

$= \sin$ $\frac{B-C}{2}$ $=$ RHS. Hence proved.

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Question 10: $\frac{a^2 - c^2}{b^2}$ $=$ $\frac{\sin (A-C)}{\sin (A+C)}$

LHS $=$ $\frac{a^2 - c^2}{b^2}$

$=$ $\frac{k^2 \sin^2 A - k^2 \sin^2 C }{k^2 \sin^2 B }$

$=$ $\frac{ \sin^2 A - \sin^2 C }{ \sin^2 B }$

$=$ $\frac{ \sin^2 A - \sin^2 C }{ \sin^2 (\pi - ( A+C)) }$

$=$ $\frac{ \sin^2 A - \sin^2 C }{ \sin^2 ( A+C) }$

$=$ $\frac{ \sin ( A + C) \sin ( A - C) }{ \sin^2 ( A+C) }$

$=$ $\frac{ \sin ( A - C) }{ \sin ( A+C) }$ $=$ RHS. Hence proved.

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Question 11: $b \sin B - c \sin C = a \sin (B-C)$

Using cosine rule

$\cos A = \Big($ $\frac{b^2 + c^2 - a^2}{2bc}$ $\Big)$     $\cos B = \Big($ $\frac{c^2 + a^2 - b^2}{2ca}$ $\Big)$     $\cos C = \Big($ $\frac{a^2 + b^2 - c^2}{2ab}$ $\Big)$

RHS $= a \sin (B-C)$

$= a ( \sin B \cos C - \cos B \sin C)$

$= a (bk) \Big($ $\frac{a^2 + b^2 - c^2}{2ab}$ $\Big) - a (ck) \Big($ $\frac{c^2 + a^2 - b^2}{2ca}$ $\Big)$

$= 2k \Big($ $\frac{b^2 - c^2}{2}$ $\Big)$

$= b(bk) - c(ck)$

$= b \sin B - c \sin C =$ LHS. Hence proved.

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Question 12: $a^2 \sin (B - C) = (b^2 - c^2) \sin A$

LHS $= a^2 \sin (B - C)$

$= a^2 ( \sin B \cos C - \cos B \sin C )$

$= a^2 \Big[ kb \Big($ $\frac{a^2 + b^2 - c^2}{2ab}$ $\Big) - kc \Big($ $\frac{a^2 + c^2 - b^2}{2ac}$ $\Big) \Big]$

$= a^2 k \Big[ \Big($ $\frac{a^2 + b^2 - c^2}{2a}$ $\Big) - \Big($ $\frac{a^2 + c^2 - b^2}{2a}$ $\Big) \Big]$

$= a^2 k \Big[$ $\frac{a^2 + b^2 - c^2 - a^2 - c^2 +b^2}{2a}$ $\Big]$

$= ak (b^2 - c^2)$

$= \sin A (b^2 - c^2) =$ RHS. Hence proved.

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Question 13: $\frac{\sqrt{\sin A}- \sqrt{\sin B} }{\sqrt{\sin A}+ \sqrt{\sin B}}$ $=$ $\frac{a+b-2\sqrt{ab}}{a-b}$

LHS $=$ $\frac{a+b-2\sqrt{ab}}{a-b}$

$=$ $\frac{( \sqrt{a} - \sqrt{b} )^2}{( \sqrt{a} - \sqrt{b} )( \sqrt{a} + \sqrt{b} )}$

$=$ $\frac{( \sqrt{a} - \sqrt{b} )}{( \sqrt{a} + \sqrt{b} )}$

$=$ $\frac{( \sqrt{k\sin A} - \sqrt{k \sin B} )}{( \sqrt{k \sin A} + \sqrt{k \sin B} )}$

$=$ $\frac{( \sqrt{\sin A} - \sqrt{ \sin B} )}{( \sqrt{ \sin A} + \sqrt{ \sin B} )}$ $=$ LHS. Hence proved.

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Question 14: $a( \sin B - \sin C) + b ( \sin C - \sin A) + c ( \sin A - \sin B) = 0$

LHS $= a( \sin A - \sin C) + b ( \sin C - \sin A) + c ( \sin A - \sin B)$

$= a ( kb - kc) + b( kc - ka) + c ( ka - kb)$

$= k ( ab -ac + bc - ab + ca - bc )$

$= k ( 0) = 0 =$ RHS. Hence proved.

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Question 15: $\frac{a^2 \sin ( B - C)}{\sin A}$ $+$ $\frac{b^2 \sin ( C - A)}{\sin A}$ $+$ $\frac{c^2 \sin ( A - B)}{\sin A}$ $= 0$

LHS $=$ $\frac{a^2 \sin ( B - C)}{\sin A}$ $+$ $\frac{b^2 \sin ( C - A)}{\sin A}$ $+$ $\frac{c^2 \sin ( A - B)}{\sin A}$

$=$ $\frac{1}{k}$ $\Big[ a \sin ( B - C) + b \sin ( C - A) + c \sin ( A - B) \Big]$

$=$ $\frac{1}{k}$ $\Big[ k \sin A \sin ( B - C) + k \sin B \sin ( C - A) + k \sin C \sin ( A - B) \Big]$

$= \sin (\pi - (B+C)) \sin ( B - C) + \sin (\pi - (C+A)) \sin ( C - A) + \sin (\pi - (A+B) ) \sin ( A - B)$

$= \sin (B+C) \sin ( B - C) + \sin (C+A) \sin ( C - A) + \sin (A+B) \sin ( A - B)$

$= \sin^2 B - \sin^2 C + \sin^2 C - \sin^2 A + \sin^2 A - \sin^2 B = 0 =$ RHS. Hence proved.

$\\$

Question 16: $a^2 (\cos^2 B - \cos^2 C) +b^2 (\cos^2 C - \cos^2 A)+ C^2 (\cos^2 A - \cos^2 B) = 0$

LHS $= a^2 (\cos^2 B - \cos^2 C) +b^2 (\cos^2 C - \cos^2 A)+ C^2 (\cos^2 A - \cos^2 B)$

$= a^2 ( 1 - \sin^2 B - 1 + \sin^2 C) + b^2 ( 1 - \sin^2 C - 1 + \sin^2 A) + c^2 ( 1 - \sin^2 A - 1 + \sin^2 B)$

$= a^2 ( \sin^2 C - \sin^2 B ) + b^2 ( \sin^2 A - \sin^2 C ) + c^2 ( \sin^2 B - \sin^2 A)$

$= a^2 ( k^2c^2 - k^2b^2 ) + b^2 ( k^2a^2 - k^2c^2 ) + c^2 ( k^2b^2 - k^2a^2)$

$= k^2 (a^2c^2 - a^2b^2 + b^2a^2 -b^2c^2 + c^2b^2 - c^2a^2)$

$= k^2(0) = 0 =$ RHS. Hence proved.

$\\$

Question 17: $b \cos B + c \cos C = a \cos ( B - C)$

LHS $= b \cos B + c \cos C$

$= k \sin B \cos B + k \sin C \cos C$

$=$ $\frac{k}{2}$ $\Big[ 2\sin B \cos B + 2\sin C \cos C \Big]$

$=$ $\frac{k}{2}$ $\Big[ \sin 2B + \sin 2C \Big]$

$=$ $\frac{k}{2}$ $\Big[ 2 \sin ( B+C) \cos (B-C) \Big]$

$= k \sin ( B+C) \cos (B-C)$

$= k \sin (\pi - A) \cos (B-C)$

$= k \sin A \cos (B-C)$

$= a \cos (B-C) =$ RHS. Hence proved.

$\\$

Question 18: $\frac{\cos 2A}{a^2}$ $-$ $\frac{\cos 2B}{b^2}$ $=$ $\frac{1}{a^2}$ $-$ $\frac{1}{b^2}$

LHS $=$ $\frac{\cos 2A}{a^2}$ $-$ $\frac{\cos 2B}{b^2}$

$=$ $\frac{1 - 2\sin^2 A}{a^2}$ $-$ $\frac{1 - 2\sin^2 B}{b^2}$

$= \Big($ $\frac{1}{a^2}$ $-$ $\frac{1}{b^2}$ $\Big) - 2\Big($ $\frac{\sin^2 A}{a^2}$ $-$ $\frac{\sin^2 B}{b^2}$ $\Big)$

$= \Big($ $\frac{1}{a^2}$ $-$ $\frac{1}{b^2}$ $\Big) - 2\Big($ $\frac{k^2a^2}{a^2}$ $-$ $\frac{k^2b^2}{b^2}$ $\Big)$

$= \Big($ $\frac{1}{a^2}$ $-$ $\frac{1}{b^2}$ $\Big) - 2\Big( k^2- k^2 \Big)$

$= \Big($ $\frac{1}{a^2}$ $-$ $\frac{1}{b^2}$ $\Big) =$ RHS. Hence proved.

$\\$

Question 19: $\frac{\cos^2 B - \cos^2 C}{b+c}$ $+$ $\frac{\cos^2 C - \cos^2 A}{c+a}$ $+$ $\frac{\cos^2 A - \cos^2 B}{a+b}$ $= 0$

LHS $=$ $\frac{\cos^2 B - \cos^2 C}{b+c}$ $+$ $\frac{\cos^2 C - \cos^2 A}{c+a}$ $+$ $\frac{\cos^2 A - \cos^2 B}{a+b}$

$=$ $\frac{1 - \sin^2 B - 1 + \sin^2 C}{b+c}$ $+$ $\frac{1 - \sin^2 C - 1 + \sin^2 A}{c+a}$ $+$ $\frac{1 - \sin^2 A - 1 + \sin^2 B}{a+b}$

$=$ $\frac{\sin^2 C - \sin^2 B}{b+c}$ $+$ $\frac{\sin^2 A - \sin^2 C}{c+a}$ $+$ $\frac{\sin^2 B - \sin^2 A}{a+b}$

$=$ $\frac{k^2(c^2 - b^2)}{b+c}$ $+$ $\frac{k^2(a^2 - c^2)}{c+a}$ $+$ $\frac{k^2(b^2 - a^2)}{a+b}$

$= k^2 ( c-b+a-c+b-a) = 0 =$ RHS. Hence proved.

$\\$

Question 20: $a \sin$ $\frac{A}{2}$ $\sin \Big($ $\frac{B-C}{2}$ $\Big) + b \sin$ $\frac{B}{2}$ $\sin \Big($ $\frac{C-A}{2}$ $\Big) + c \sin$ $\frac{C}{2}$ $\sin \Big($ $\frac{A-B}{2}$ $\Big) = 0$

LHS $= a \sin \frac{A}{2} \sin \frac{B-C}{2} + b \sin \frac{B}{2} \sin \frac{C-A}{2} + c \sin \frac{C}{2} \sin \frac{A-B}{2}$

$= a \sin \frac{\pi - (B+C)}{2} \sin \frac{B-C}{2} + b \sin \frac{\pi - (C+A)}{2} \sin \frac{C-A}{2} + c \sin \frac{\pi - (A+B)}{2} \sin \frac{A-B}{2}$

$= a \cos \frac{B+C}{2} \sin \frac{B-C}{2} + b \cos \frac{C+A}{2} \sin \frac{C-A}{2} + c \cos \frac{A+B}{2} \sin \frac{A-B}{2}$

$= a ( \sin B - \sin C) + b ( \sin C - \sin A) + c ( \sin A - \sin B)$

$= a ( kb - kc) + b ( kc - ka) + c ( ka - kb)$

$= k ( ab - ac + bc - ba + ca - cb) = k(0) = 0 =$ RHS. Hence proved.

$\\$

Question 21: $\frac{b \sec B + c \sec C}{\tan B + \tan C}$ $=$ $\frac{c \sec C + a \sec A}{\tan C + \tan A}$ $=$ $\frac{a \sec A + b \sec A}{\tan A + \tan B}$

LHS $=$ $\frac{b \sec B + c \sec C}{\tan B + \tan C}$

$=$ $\frac{k \sin B \sec B + k \sin C \sec C}{\tan B + \tan C}$

$=$ $\frac{k \tan B + k \tan C }{\tan B + \tan C}$

$= k$

Similarly. $\frac{c \sec C + a \sec A}{\tan C + \tan A}$ $= k$

and $\frac{a \sec A + b \sec A}{\tan A + \tan B}$ $= k$

Hence proved.

$\\$

Question 22: $a \cos A + b \cos B + c \cos C = 2a \sin B \sin C= 2b \sin A \sin C = 2 c \sin A \sin B$

LHS $= a \cos A + b \cos B + c \cos C$

$= k \sin A \cos A + k \sin B \cos B + k \sin C \cos C$

$=$ $\frac{k}{2}$ $\Big[ \sin 2A + \sin 2B + 2 \sin C \cos C \Big]$

$=$ $\frac{k}{2}$ $\Big[ 2 \sin ( A + B) \cos ( A - B) + 2 \sin C \cos C \Big]$

$=$ $\frac{k}{2}$ $\Big[ 2 \sin (\pi - C) \cos ( A - B) + 2 \sin C \cos C \Big]$

$=$ $\frac{k}{2}$ $\Big[ 2 \sin C \cos ( A - B) + 2 \sin C \cos C \Big]$

$= k \sin C \Big[ \cos ( A - B) + \cos C \Big]$

$= k \sin C \Big[ 2 \cos \frac{A-B + C}{2} \cos \frac{A-B - C}{2} \Big]$

$= k \sin C \Big[ 2 \cos \frac{\pi - 2B}{2} \cos \frac{2A-\pi}{2} \Big]$

$= k \sin C \Big[ 2 \sin B \sin A \Big]$

$= \sin C \Big[ 2 \sin B (k\sin A) \Big]$

$= 2a \sin B \sin C =$ RHS. Hence proved.

Similarly, $a \cos A + b \cos B + c \cos C = 2b \sin A \sin C = 2 c \sin A \sin B$

$\\$

Question 23: $a( \cos B \cos C + \cos A) = b ( \cos C \cos A + \cos B)= c (\cos A \cos B + \cos C)$

$a( \cos B \cos C + \cos A)$

$= a \big( \cos B \cos C + \cos (\pi - (B+C)) \big)$

$= a( \cos B \cos C - \cos (B+C) )$

$= a( \cos B \cos C - \cos B \cos C + \sin B \sin C )$

$= a\sin B \sin C$

$= k \sin A \sin B \sin C$

Similarly, $b( \cos A \cos C + \cos B) = k \sin A \sin B \sin C$

$c( \cos A \cos B + \cos C) = k \sin A \sin B \sin C$

$\\$

Question 24: $a( \cos C - \cos B) = 2 (b-c) \cos^2$ $\frac{A}{2}$

LHS $= a( \cos C - \cos B)$

$= a \Big[ 2 \sin \frac{C+B}{2} \sin \frac{B-C}{2} \Big]$

$= \Big[ 2 k \sin A \sin \frac{\pi -A}{2} \sin \frac{B-C}{2} \Big]$

$= 2k \Big[ 2 \sin \frac{A}{2} \cos \frac{A}{2} \cos \frac{A}{2} \sin \frac{B-C}{2} \Big]$

$= 2k \cos^2 \frac{A}{2} \Big[ 2 \sin \frac{A}{2} \sin \frac{B-C}{2} \Big]$

$= 2k \cos^2 \frac{A}{2} \Big[ 2 \sin \frac{\pi - (B+C)}{2} \sin \frac{B-C}{2} \Big]$

$= 2k \cos^2 \frac{A}{2} \Big[ 2 \sin \frac{B-C}{2} \cos \frac{B+C}{2} \Big]$

$= 2k \cos^2 \frac{A}{2} \Big[ \sin B - \sin C \Big]$

$= 2 \cos^2 \frac{A}{2} \Big[ k\sin B - k\sin C \Big]$

$= 2 (b-c) \cos^2 \frac{A}{2} =$ RHS. Hence proved.

$\\$

Question 25: In $\triangle ABC$ prove that, if $\theta$ be any angle, then $b \cos \theta = c \cos ( A - \theta) + a \cos (C + \theta)$

RHS $= c \cos ( A - \theta) + a \cos (C + \theta)$

$= c \cos A \cos \theta + c \sin A \sin \theta + a \cos C \cos \theta - a \sin C \sin \theta$

$= k \sin C \cos A \cos \theta + k \sin C \sin A \sin \theta + k \sin A \cos C \cos \theta - k \sin A \sin C \sin \theta$

$= k \sin C \cos A \cos \theta + k \sin A \cos C \cos \theta$

$= k \cos \theta ( \sin C \cos A + \sin A \cos C)$

$= k \cos \theta \sin ( C + A)$

$= k \cos \theta \sin ( \pi - B)$

$= k \cos \theta \sin B$

$= b \cos \theta =$ LHS. Hence proved.

$\\$

Question 26: In $\triangle ABC$, if  $\sin^2 A + \sin^2 B = \sin^2 C$, show that the triangle is a right angled.

Let $\sin A = ak, \sin B = bk$,  and $\sin C = ck$

Given  $\sin^2 A + \sin^2 B = \sin^2 C$

$\Rightarrow k^2a^2 + k^2b^2 = k^2c^2$

$\Rightarrow a^2 + b^2 = c^2$

Since the triangle satisfies Pythagoras Theorem, the triangle is a right angles triangle.

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Question 27: In any $\triangle ABC$, if $a^2 , b^2, c^2$ are in AP, prove that $\cot A, \cot B$ and $\cot C$ are also in AP.

Given $a^2 , b^2, c^2$ are in AP

$\Rightarrow -2a^2 , -2b^2, -2c^2$ are in AP

$\Rightarrow (a^2+b^2+c^2) -2a^2 , (a^2+b^2+c^2) -2b^2, (a^2+b^2+c^2) -2c^2$ are in AP

$\Rightarrow b^2+c^2 -a^2 , a^2+c^2-b^2, a^2+b^2 -c^2$ are in AP

$\Rightarrow$ $\frac{b^2+c^2 -a^2}{2abc}$ $,$ $\frac{a^2+c^2-b^2}{2abc}$ $,$ $\frac{a^2+b^2 -c^2}{2abc}$ are in AP

$\Rightarrow$ $\frac{1}{a}$ $\Big[$ $\frac{b^2+c^2 -a^2}{2bc}$ $\Big] ,$ $\frac{1}{b}$ $\Big[$ $\frac{a^2+c^2-b^2}{2ac}$ $\Big],$ $\frac{1}{c}$ $\Big[$ $\frac{a^2+b^2 -c^2}{2ab}$ $\Big]$ are in AP

$\Rightarrow$ $\frac{1}{a}$ $\cos A ,$ $\frac{1}{b}$ $\cos B,$ $\frac{1}{c}$ $\cos C$ are in AP

$\Rightarrow$ $\frac{k}{a}$ $\cos A ,$ $\frac{k}{b}$ $\cos B,$ $\frac{k}{c}$ $\cos C$ are in AP

$\Rightarrow$ $\frac{\cos A}{\sin A}$ $,$ $\frac{\cos B}{\sin B}$ $,$ $\frac{\cos C}{\sin C}$ are in AP

$\Rightarrow \cot A , \cot B, \cot C$ are in AP

$\\$

Question 28: The upper part of a tree broken over by the wind makes an angle of $30^o$ with the ground and the distance from the root to the point where the top of the tree touches the ground is $15$ m. Using sine rule, find the height of the tree.

Using sine rule,

$\frac{\sin A}{15}$ $=$ $\frac{\sin C}{h}$

$\frac{\sin 60^o}{15}$ $=$ $\frac{\sin 30^o}{h}$

$\frac{\sqrt{3}}{2 \times 15}$ $=$ $\frac{1}{2 \times h}$

$\Rightarrow h =$ $\frac{15}{\sqrt{3}}$ $= 5\sqrt{3}$ m

$\\$

Question 29: At the foot of the mountain the elevation of the peak is $45^o$, after ascending $1000$ m towards the mountain up the slope of 30^o inclination, the elevation is found to be $60^o$. Find the height of the mountain.

$DE = 1000 \sin 30^o = 1000 \times$ $\frac{1}{2}$ $= 500 = FB$

$EC = 1000 \cos 30^o = 1000 \times$ $\frac{\sqrt{3}}{2}$ $= 500\sqrt{3}$ m

Let $AF = x$ m

$DF =$ $\frac{x}{\sqrt{3}}$ $=BE$

In $\triangle ABC, \tan 45^o =$ $\frac{AB}{BC}$

$\Rightarrow 1 =$ $\frac{AF + FB}{BE + EC}$

$\Rightarrow 1 =$ $\frac{x + 500}{\frac{x}{\sqrt{3}} +500\sqrt{3}}$

$\Rightarrow$ $\frac{x}{\sqrt{3}}$ $+ 500\sqrt{3} = x + 500$

$\Rightarrow 1500 - 500\sqrt{3} = x\sqrt{3} - x$

$\Rightarrow 500\sqrt{3}(\sqrt{3}-1)= x(\sqrt{3}-1)$

$\Rightarrow x = 500\sqrt{3}$ m

Therefore the height of triangle is $AB = AF + FB = 500\sqrt{3} + 500 = 500(\sqrt{3}+1)$ m

$\\$

Question 30: A person observes the angle of elevation of the peak of a hill from a station to be $\alpha$. He walks c meters along a slope inclined at an angle $\beta$ and finds the angle of elevation of the peak of the hill as $\gamma$. Show that the height of the peak above the ground is $\frac{c \sin \alpha \sin (\gamma - \beta)}{(\sin \gamma - \alpha)}$

The person is observing the peak from point $Q$

Distance traveled is $QT=c$ and the angle of inclination is $\beta$

Observing the peak from point $T$, angle of inclination is $\gamma$

Now consider $\triangle QUT$

$\angle TQU = \beta - \alpha$

$\Rightarrow \sin ( \beta - \alpha) =$ $\frac{a}{c}$

$\Rightarrow a = c \sin ( \beta - \alpha)$     … … … … … i)

Now consider $\triangle PQR$

We know, $\angle QPR = 90-\alpha$

In $\triangle PTS, \angle TPS = 90^o - \gamma$

$\therefore \angle TPU = \angle QPR - \angle TPS$

$\Rightarrow \angle TPU = ( 90-\alpha) - ( 90-\gamma) = \gamma - \alpha$

Now consider $\triangle TPU$,

$\sin ( \gamma - \alpha) =$ $\frac{a}{b}$

$b =$ $\frac{a}{\sin (\gamma - \alpha)}$

Substituting the value of $a$ from i)

$b =$ $\frac{c \times \sin ( \alpha - \beta)}{\sin (\gamma - \alpha)}$    … … … … … ii)

We need to find the total height of the peak $PR$

$PR = PS + SR$    … … … … … iii)

From $\triangle PST$,

$\sin \gamma =$ $\frac{PS}{PT}$ $=$ $\frac{PS}{b}$

$\Rightarrow PS = b \sin \gamma$    … … … … … iv)

From $\triangle QTW$

$\sin \beta =$ $\frac{TW}{QT}$ $=$ $\frac{TW}{c}$

$\Rightarrow TW = SR = c \sin \beta$  … … … … … v)

$\therefore PR = PS + SR$

$= b \sin \gamma + c \sin \beta$

$=$ $\frac{c \sin (\alpha - \beta)}{\sin (\gamma - \alpha)}$ $\sin \gamma + c \sin \beta$

$=$ $\frac{c \sin (\alpha - \beta)\sin \gamma +c \sin \beta \sin (\gamma - \alpha) }{\sin (\gamma - \alpha)}$

$= c \Big[$ $\frac{\sin \alpha \cos \beta \sin \gamma - \cos \alpha \sin \beta \sin \gamma + \sin \beta \sin \gamma \cos \alpha - \sin \beta \sin \alpha \cos \gamma }{\sin (\gamma - \alpha)}$ $\Big]$

$= c \Big[$ $\frac{\sin \alpha \cos \beta \sin \gamma - \sin \beta \sin \alpha \cos \gamma }{\sin (\gamma - \alpha)}$ $\Big]$

$= c \sin \alpha \Big[$ $\frac{ \cos \beta \sin \gamma - \sin \beta \cos \gamma }{\sin (\gamma - \alpha)}$ $\Big]$

$= c \sin \alpha \Big[$ $\frac{\sin (\gamma - \beta) }{\sin (\gamma - \alpha)}$ $\Big]$

Hence proved.

$\\$

Question 31: If the sides $a, b, c$ of $\triangle ABC$ are in HP, prove that $\sin^2$ $\frac{A}{2}$ $, \sin^2$ $\frac{B}{2}$ $, \sin^2$ $\frac{C}{2}$ are in HP too.

If sides $a, b, c$ are in HP

$\Rightarrow$ $\frac{1}{a}$ $,$ $\frac{1}{b}$ $,$ $\frac{1}{c}$ are in AP

$\Rightarrow$ $\frac{1}{b}$ $-$ $\frac{1}{a}$ $=$ $\frac{1}{c}$ $-$ $\frac{1}{b}$

$\Rightarrow$ $\frac{a-b}{ab}$ $=$ $\frac{b-c}{bc}$

Using sine rule,

$\Rightarrow$ $\frac{\sin A-\sin B}{\sin A \sin B}$ $=$ $\frac{\sin B-\sin C}{\sin B\sin C}$

$\Rightarrow$ $\frac{\sin A-\sin B}{\sin A }$ $=$ $\frac{\sin B-\sin C}{\sin C}$

$\Rightarrow$ $\frac{2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}}{\sin A }$ $=$ $\frac{2 \sin \frac{B-C}{2} \cos \frac{B+C}{2}}{\sin C}$

$\Rightarrow$ $\frac{2 \sin \frac{A-B}{2} \cos \frac{\pi - C}{2}}{\sin A }$ $=$ $\frac{2 \sin \frac{B-C}{2} \cos \frac{\pi - A}{2}}{\sin C}$

$\Rightarrow$ $\frac{\sin \frac{A-B}{2} \sin \frac{C}{2}}{\sin A }$ $=$ $\frac{ \sin \frac{B-C}{2} \sin \frac{ A}{2}}{\sin C}$

$\Rightarrow 2 \sin \frac{A-B}{2} \sin^2 \frac{C}{2} \cos \frac{C}{2}= 2 \sin \frac{B-C}{2} \sin^2 \frac{A}{2} \cos \frac{A}{2}$

$\Rightarrow \sin^2 \frac{C}{2} \sin \frac{A-B}{2} \cos \frac{\pi - (A+B)}{2}= \sin^2 \frac{A}{2} \sin \frac{B-C}{2} \cos \frac{\pi - ( B + C) }{2}$

$\Rightarrow \sin^2 \frac{C}{2} \sin \frac{A-B}{2} \sin \frac{A+B}{2}= \sin^2 \frac{A}{2} \sin \frac{B-C}{2} \sin \frac{ B + C }{2}$

$\Rightarrow \sin^2 \frac{C}{2} \Big[ \sin^2 \frac{A}{2} - \sin^2 \frac{B}{2} \Big] = \sin^2 \frac{A}{2} \Big[ \sin^2 \frac{B}{2} - \sin^2 \frac{C}{2} \Big]$

$\Rightarrow \sin^2 \frac{C}{2} \sin^2 \frac{A}{2} - \sin^2 \frac{C}{2} \sin^2 \frac{B}{2} = \sin^2 \frac{A}{2} \sin^2 \frac{B}{2} - \sin^2 \frac{A}{2} \sin^2 \frac{C}{2}$

$\Rightarrow$ $\frac{1}{ \sin^2 \frac{B}{2} } - \frac{1}{ \sin^2 \frac{A}{2} } = \frac{1}{ \sin^2 \frac{C}{2} } - \frac{1}{ \sin^2 \frac{B}{2} }$

$\Rightarrow$ $\frac{1}{ \sin^2 \frac{A}{2} }$ $,$ $\frac{1}{ \sin^2 \frac{B}{2} }$ $,$ $\frac{1}{ \sin^2 \frac{C}{2} }$ are in AP

$\Rightarrow$ $\therefore \sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in HP