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A circle is inscribed in a triangle `A B C` touching the side `A B` at `D` such that `A D=5,B D=3,if/_A=60^0` then length `B C` equals. 9 (b)`(120)/(13)`(c) `13`(d) `12`

A

9

B

`(120)/(13)`

C

13

D

12

Text Solution

Verified by Experts

The correct Answer is:
C

`tan 30^(@) = (r)/(5)`
or `r = (5)/(sqrt3)`
We have `tan.(B)/(2) = (r)/(3) = (5)/(3sqrt3)`...(i)
`:. cos B (1-tan^(2).(B)/(2))/(1+tan^(2).(B)/(2)) = (1)/(26)`
`:. sin B = (sqrt(675))/(26) = (15 sqrt3)/(26)`
`:. sin C = sin (A +B) = sin A cos B + cos A sin B = (4sqrt3)/(13)`
Now `(a)/(sinA) = (c)/(sinC) rArr a = 13`
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