KVPY SB SX Exam 2011 Question Paper

Question : 10 of 120

Marks: +1, -0

In triangle ABC, we are given that

3 sin A + 4cos B = 6

and

4 sin B + 3cos A = 1

. Then the measure of the angle C is

Solution:

3 \sin {A}+4 \cos {B}=6

4 \sin {B}+3 \cos {A}=1 \;\;

....(ii)
squaring and adding (i) and (ii)

(9 \sin ^{2} {A}+16 \cos ^{2} {B}+24 \sin {A} \cos {B})+(16 \sin ^{2} {B}+9 \cos ^{2} {A}+24 \sin {B} \cos {A})=37

9(\sin ^{2} {A}+\cos ^{2} {A})+16(\sin ^{2} {B}+\cos ^{2} {B})+24(\sin {A} \cos {B}+\sin {B} \cos {A})=37

25+24 \sin ( {A}+ {B})=37

24 \sin ( {A}+ {B})=12

\sin ( {A}+ {B})=\;{12}/{24}=\;{1}/{2}

{As} {A}+ {B}+ {C}=π

{A}+ {B}=π- {C}

\sin (π- {C})=\;{1}/{2}

π- {C}=\;{π}/{6}

⇒ {C}=\;{5 π}/{6}=150^{∘}

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