In triangle `A B C ,/_A=60^0,/_B=40^0,a n d/_C=80^0dot` If `P` is the center of the circumcircle of triangle `A B C` with radius unity, then the radius of the circumcircle of triangle `B P C` is 1 (b) `sqrt(3)` (c) 2 (d) `sqrt(3)` `2`

Options

A. 1
B. `sqrt3`
C. 2
D. `sqrt3//2`

Answer

A

Explanation

Let R' be the radius of the circumcircle of `DeltaABC` using the sine rule in triangle BPC. Then

`(a)/(sin 120^(@)) = 2R_(1)`...(i)
Aslo, `(a)/(sin 60^(@)) = 2R " " ("in " DeltaABC)`
`:. A = 2R sin 60^(@)`
or `a = sqrt3 " " [R = 1, "give"]`
From Eq. (i), we get `R_(1) = (2sqrt3)/(sqrt3) xx (1)/(2)`