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If in triangle A B C ,sumsinA/2=6/5a n d...

If in triangle `A B C ,sumsinA/2=6/5a n dsumI I_1=9` (where `I_1,I_2a n dI_3` are excenters and `I` is incenter, then circumradius `R` is equal to `(15)/8` (b) `(15)/4` (c) `(15)/2` (d) `4/(12)`

A

`(15)/(8)`

B

`(15)/(4)`

C

`(15)/(2)`

D

`(4)/(12)`

Text Solution

Verified by Experts

The correct Answer is:
A
We know that `II_(1) = 4R sin A//2`
Now, `sum II_(1) = 9`
`rArr 4R sum sin.(A)/(2) = 9`
or `4R ((6)/(5)) = 9 " or " R = (15)/(8)`
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