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`

Explanation

With the given details, we can create a diagram.

Please refer to video for the diagram.

Here, circumradius of `Delta ABC` = `sqrt16 = 4`

`:. a = 2*4sinA =>a = 8sinA->(1)`

Let `R_1` is the circumradius of the `Delta HBC`.

Then,

`a = 2R_1sin/_BHC`

`=2R_1sin(B+C)`

...