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

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

Using Sine Rule,

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

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

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

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

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

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

$\displaystyle \\$

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

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

$\displaystyle \therefore \angle A = 30^o$

Using sine rule,

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

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

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

$\displaystyle \\$

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

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

Using sine rule,

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

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

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

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

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In any triangle ABC, prove the following:

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \\$

$\displaystyle \text{Question 11: } b \sin B - c \sin C = a \sin (B-C)$

Using cosine rule

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

$\displaystyle \text{RHS } = a \sin (B-C)$

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{LHS } = a^2 \sin (B - C)$

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{LHS } = a( \sin A - \sin C) + b ( \sin C - \sin A) + c ( \sin A - \sin B)$

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

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

$\displaystyle = k ( 0) = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle = k^2(0) = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

$\displaystyle \text{LHS } = b \cos B + c \cos C$

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

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

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

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

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

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

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

$\displaystyle = a \cos (B-C) = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle = k^2 ( c-b+a-c+b-a) = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle = k ( ab - ac + bc - ba + ca - cb) = k(0) = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

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

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

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

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

$\displaystyle = k$

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

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

Hence proved.

$\displaystyle \\$

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

$\displaystyle \text{LHS } = a \cos A + b \cos B + c \cos C$

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

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

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

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

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

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

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

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

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

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

$\displaystyle = 2a \sin B \sin C = \text{ RHS. Hence proved. }$

$\displaystyle \text{Similarly, } a \cos A + b \cos B + c \cos C = 2b \sin A \sin C = 2 c \sin A \sin B$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle = a\sin B \sin C$

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

$\displaystyle \text{Similarly, } b( \cos A \cos C + \cos B) = k \sin A \sin B \sin C$

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

$\displaystyle \\$

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

$\displaystyle \text{LHS } = a( \cos C - \cos B)$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \text{RHS } = c \cos ( A - \theta) + a \cos (C + \theta)$

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

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

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

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

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

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

$\displaystyle = k \cos \theta \sin B$

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

$\displaystyle \\$

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

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

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

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

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

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

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

Given $\displaystyle a^2 , b^2, c^2 \text{ are in AP }$

$\displaystyle \Rightarrow -2a^2 , -2b^2, -2c^2 \text{ are in AP }$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow \cot A , \cot B, \cot C \text{ are in AP }$

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

Using sine rule,

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

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

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

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

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

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

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

Let $\displaystyle AF = x m$

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

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

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

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

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

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

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

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

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

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

The person is observing the peak from point $\displaystyle Q$

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

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

Now consider $\displaystyle \triangle QUT$

$\displaystyle \angle TQU = \beta - \alpha$

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

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

Now consider $\displaystyle \triangle PQR$

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

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

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

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

Now consider $\displaystyle \triangle TPU$,

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

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

Substituting the value of $\displaystyle a$ from i)

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

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

$\displaystyle PR = PS + SR$ … … … … … iii)

From $\displaystyle \triangle PST$,

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

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

From $\displaystyle \triangle QTW$

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

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

$\displaystyle \therefore PR = PS + SR$

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

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

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

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

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

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

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

Hence proved.

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

If sides $\displaystyle a, b, c \text{ are in HP }$

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

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

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

Using sine rule,

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

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

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

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

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

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

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

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

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

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

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

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

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