Question

In a triangle $ABC\angle A={60}^{\circ },\angle B={40}^{\circ }$ and $\angle C={80}^{\circ }$ If P is the centre of the circumcircle of triangle $ABC$ with radius unity, then the radius of the circumcircle of triangle $BPC$ is

• A. 1
• B. $\sqrt{3}$
• C. 2
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A

1

Let $R$ and $R_1$ be the radius of the circumcircle of $\Delta ABC$ and $\Delta BPC$ respectively.
Using sine law in the $\Delta BPC$
$\frac{a}{sin{120}^{\circ }}=2R_1$
also $\frac{a}{sin{60}^{\circ }}=2R\left(\text{in}\Delta ABC\right)$
$a=2Rsin{60}^{\circ }(\text{It is given that}R=1)a=\sqrt{3}$

From $\left(1\right),R_1=\frac{2\sqrt{3}}{\sqrt{3}}.\frac{1}{2}=1$