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In a triangle ABC cos A = 7/8 , cos B = ...

In a triangle ABC cos A = `7/8` , cos B = `11/16`.then , cos C is equal to

A

`-1/4`

B

`-1/2`

C

0

D

`1/4`

Text Solution

Verified by Experts

We have,
`cosA=7/8,cosB=11/16rArrtanA=(sqrt(15))/7,tanB=(3sqrt(15))/11`
Now,
tan A+ tan B + tan C = tan A tan B tan C
`rArrtanC=(tanA+tanB)/(tanAtanB-1)`
`rArrtanC=-sqrt(15)rArrcosC=-1/4`
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