If A and B are positive acute angles satisfying 3 ^ 2 A+2 ^ 2 B=4 and (3 A)/( B)=(2 B)/( A) Then the value of A+2 B is equal to :

Correct answer: B. (π)/(2) Given, 3 ^ 2 A+2 ^ 2 B=4 2 ^ 2 B-1=4-3 ^ 2 A-1 2 B=3(1- ^ 2 A)=3 ^ 2 A (1) and 2 B B=3 A A 2 B=3 A A Now, (A+2 B)= A 2 B- A 2 B= A(3 ^ 2 A)- A(3 A A)=0 [using eqs. (1) and (2)] A+2 B=(π)/(2)

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

BITSAT 2016 · Mathematics · Trigonometric Ratios & Identities

If A and B are positive acute angles satisfying 3 ^ 2 A+2 ^ 2 B=4...

A

and

B

are positive acute angles satisfying

3 \cos ^{2} A+2 \cos ^{2} B=4

and

\frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A}

Then the value of

\mathrm{A}+2 \mathrm{ B}

is equal to :

Options

A $\frac{\pi}{6}$
B $\frac{\pi}{2}$
C $\frac{\pi}{3}$
D $\frac{\pi}{4}$

Correct answer

B. $\frac{\pi}{2}$

Step-by-step solution

Given,

3 \cos ^{2} \mathrm{ A}+2 \cos ^{2} \mathrm{ B}=4\Rightarrow 2 \cos ^{2} \mathrm{ B}-1=4-3 \cos ^{2} \mathrm{ A}-1\Rightarrow \cos 2 \mathrm{ B}=3\left(1-\cos ^{2} \mathrm{ A}\right)=3 \sin ^{2} \mathrm{ A} \ldots(1)

and

2 \cos \mathrm{B} \sin \mathrm{B}=3 \sin \mathrm{A} \cos \mathrm{A}\sin 2 \mathrm{ B}=3 \sin \mathrm{A} \cos \mathrm{A}

Now,

\cos (\mathrm{A}+2 \mathrm{ B})=\cos \mathrm{A} \cos 2 \mathrm{ B}-\sin \mathrm{A}\sin 2 \mathrm{ B}=\cos A\left(3 \sin ^{2} A\right)-\sin A(3 \sin A \cos A)=0

[using eqs. (1) and (2)]

\Rightarrow A+2 B=\frac{\pi}{2}

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