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If in a triangle ABC, theta is the angle...

If in a triangle ABC, `theta` is the angle determined by `cos theta=(a-b)//c`, then

A

`((a+b)sin theta)/(2sqrt(ab))=cos(A-B)/(2)`

B

`((a+b)sin theta)/(2sqrt(ab))=cos(A+B)/(2)`

C

`(c sin theta)/(2sqrt(ab))=cos(A-B)/(2)`

D

`(c sin theta)/(2sqrt(ab))=cos(A+B)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, D
`cos theta=(a-b)/(c )=(sin A- sin B)/(sin C)`
`=(2 cos.(A+B)/(2)sin.(A-B)/(2))/(2sin.(C )/(2)cos.(C )/(2))`
`=(sin.(A-B)/(2))/(sin.(A+B)/(2))`
`rArr sin theta =(sqrt("sin"^(2)(A+B)/(2)-"sin"^(2)(A-B)/(2)))/(sin.(A+B)/(2))`
`=(sqrt(sin A sin B))/(sin.(A+B)/(2))`
`rArr ((a+b)sin theta)/(2sqrt(ab))`
`=(sin A+sin B)/(2sqrt(sin A sin B))(sqrt(sin A sin B))/(sin.(A+B)/(2))`
`=(2sin.(A+B)/(2)cos.(A-B)/(2))/(2sin.(A+B)/(2))=cos.(A-B)/(2)`
`(c sin theta)/(2sqrt(ab))=(sin C)/(2sqrt(sin A sin B))(sqrt(sin A sin B))/(sin.(A+B)/(2))`
`=(2sin(C )/(2)cos(C )/(2))/(2sin(A+B)/(2))`
`=cos.(A+B)/(2)`
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