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.

Answer:

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 .

Answer:

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

Answer:

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} )}

Answer:

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)

Answer:

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}}

Answer:

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}}

Answer:

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} }

Answer:

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}

Answer:

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)}

Answer:

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)

Answer:

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

Answer:

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}

Answer:

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

Answer:

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

Answer:

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.

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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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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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

Answer:

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.

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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

Answer:

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.

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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}

Answer:

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.

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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

Answer:

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

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

Answer:

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 

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

Answer:

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

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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. 

Answer:2020-05-20_21-28-23

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

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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.

Answer:

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 m2020-05-20_21-34-08

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

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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)}

Answer:

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 2020-05-20_21-40-57

\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.

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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.

Answer:

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