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If in a !ABC,a^(2)cos^(2)A=b^(2)+c^(2) ,...

If in a `!ABC,a^(2)cos^(2)A=b^(2)+c^(2)` , then

A

`0ltAltpi/4`

B

`pi/4ltAltpi/2`

C

`pi/2ltAltpi`

D

`A=pi/2`

Text Solution

Verified by Experts

We have,
`cosA=(b^(2)+c^(2)-a^(2))/(2bc)`
`rArrcosA=(a^(2)cos^(2)A-a^(2))/(2bc) [becauseb^(2)+c^(2)=a^(2)cos^(2)A]`
`rArrcosA=(-a^(2)sin^(2)A)/(2bc)lt0rArrpi/2ltAltpi`
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