In a triangle ABC ∠A=60∘,∠B=40∘ and ∠C=80∘ If P is the centre of the circumcircle of triangle ABC with radius unity, then the radius of the circumcircle of triangle BPC is
1
Let R and R1 be the radius of the circumcircle of ΔABC and ΔBPC respectively.
Using sine law in the ΔBPC
asin120∘=2R1
also asin60∘=2R( in ΔABC)
a=2Rsin60∘(It is given that R=1)a=√3
From (1),R1=2√3√3.12=1