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If sinA=sin^2Ba n d2cos^2A=3cos^2B then ...

If `sinA=sin^2Ba n d2cos^2A=3cos^2B` then the triangle `A B C` is right angled (b) obtuse angled isosceles (d) equilateral

A

right angled

B

obtuse angled

C

ospsceles

D

equilateral

Text Solution

Verified by Experts

The correct Answer is:
B
`sinA=sin^2Band2cos^2A=3cos^2B`
`rArr2-2sin^2A=3-3sin^2B`
`rArr2sin^2A-3sinA+1=0`
`rArr(2sinA-1)(sinA-1)=0`
`rArrA=30^@rArrA=90^@`
If `A=30^@rArrB=45^@rArrC=105^@`
If `A=90^@rArrB=90^@`, which is not possible
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