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Let a,b,c be the sides of a triangle ABC...

Let a,b,c be the sides of a triangle ABC, a=2c,cos(A-C)+cos B=1. then the value of C is

A

`pi//6`

B

`pi//3`

C

`2pi//3`

D

`5pi//6`

Text Solution

Verified by Experts

The correct Answer is:
A, D
cos(A -C) + cos B = 1
`rArr cos(A-C)-cos(A+C)=1`
`rArr 2sin A sin C=1` …..(1)
Now a= 2c, so by sine rule sin A = 2sin C …..(2)
From (1) and (2), we get `4sin^(2)C=1`
`rArr sin C=(1)/(2)`
`rArr C={(pi)/(6),(5pi)/(6)}`
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