A and B are positive acute angles such that 2 B=3 ^2 A and 3 2 A=2 2 B . What is the value of (A+2 B) ?

Correct answer: D. (π)/(2) From question, we have 2 B=3 ^2 A and 3 2 A=2 2 B Now, aligned & (A+2 B)= A · 2 B- A · 2 B & = A · 3 ^2 A- A · (3)/(2) 2 A aligned aligned & =3 A · ^2 A-(3)/(2) · A ·(2 A- A) & =3 A · ^2 A-3 A · ^2 A=0 & (A+2 B)= ((π)/(2)) & A+2 B=(π)/(2) aligned

NDA 2020

Exam: NDA
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

A and B are positive acute angles such that 2 B=3 ^2 A and 3 2 A=2...

A

and

B

are positive acute angles such that

\cos 2 B=3 \sin ^2 A

and

3 \sin 2 A=2 \sin 2 B

. What is the value of

(A+2 B)

Options

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

Correct answer

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

Step-by-step solution

From question, we have

\cos 2 B=3 \sin ^2 A \text{\text{ and } } 3 \sin 2 A=2 \sin 2 B

Now,

\begin{aligned} & \cos (A+2 B)=\cos A \cdot \cos 2 B-\sin A \cdot \sin 2 B \\ & =\cos A \cdot 3 \sin ^2 A-\sin A \cdot \frac{3}{2} \sin 2 A \end{aligned}

\begin{aligned} & =3 \cos A \cdot \sin ^2 A-\frac{3}{2} \cdot \sin A \cdot(2 \sin A-\cos A) \\ & =3 \cos A \cdot \sin ^2 A-3 \cos A \cdot \sin ^2 A=0 \\ & \therefore \cos (A+2 B)=\cos \left(\frac{\pi}{2}\right) \\ & \therefore A+2 B=\frac{\pi}{2}\end{aligned}

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