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If Aa n dB are acute positive angles sat...

If `Aa n dB` are acute positive angles satisfying the equations `3sin^2A+2sin^2B=1a n d3sin2A-2sin2B=0,t h e nA+2B` is equal to `pi` (b) `pi/2` (c) `pi/4` (d) `pi/6`

A

`pi//4`

B

`pi//3`

C

`pi//6`

D

`pi//2`

Text Solution

Verified by Experts

The correct Answer is:
D
From the given relation, we have
`2 cos^(2) B-1=3(1-cos^(2)A)`
or `cos 2B = 3 sin^(2)A` …(1)
`(3 sin A)/(sin B)=(2 cos B)/(cos A)`
`rArr (3)/(2) sin 2A = sin 2B`
`rArr 3 sin 2A = 2 sin 2B` …(2)
Now cos (A + 2B)
`= cos A cos 2B - sin A sin 2B`
`= cos A(3 sin^(2)A)-sin A.(3)/(2)sin 2A`
`= 3 cos A sin^(2)A-3 sin^(2)A cos A = 0`
`therefore A+2B=90^(@)`
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