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If in a triangle ABC,4sin A = 4 sin B=3...

If in a triangle `ABC,4sin A = 4 sin B=3 sin C`, then `cos C=`

A

`1//3`

B

`1//9`

C

`1//27`

D

`1//18`

Text Solution

Verified by Experts

The correct Answer is:
B
Given, `(sinA)/(1//4)=(sin B)/(1//4)=(sin C)/(1//3)`
`:." "(sinA)/(3)=(sinB)/(3)=(sinC)/(4)`
Here, a = 3k, b = 3k and c = 4k, where k is a proportionality constant.
`:.cosC=(9k^(2)+9k^(2)-16k^(2))/(2xx3kxx3k)=(1)/(9)`
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