Question

In triangle $ABC$, $\angle A={60}^o,\angle B={40}^o,and\angle C={80}^o$ and if $P$ is the centre of the circumcircle of triangle $ABC$ with circumradius unity, then the radios of circumcircle of triangle $BPC$ is

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

Answer 1

Answer: (A)

$1$

we have given,

$\begin{array}{c}\angle A={60}^o,\angle B={40}^o,and\angle C={80}^{o}then\\ R^{{}^{\prime }}betheradiusofcircumcircleofABC:[radius\left(r\right)=1\\ a=2R^{{}^{\prime }}\sin A\\ a=2R{}^{{}^{\prime }}\sin 60------\left(i\right)\\ and,R_{1}isthecircumcircleofPBC:\\ a=2R_1\sin {120}^0-----\left(ii\right)\\ Now,comparebothequ..............\\ \Rightarrow 2R^{{}^{\prime }}\sin 60=2R_1\sin {120}^0\\ \Rightarrow R^{{}^{\prime }}\frac{\sqrt{3}}{2}=R_1\frac{\sqrt{3}}{2}\\ \therefore R^{{}^{\prime }}=R_1[\therefore radius\left(r\right)=1\\ weknowthatcircumcircleofABCisequaltocircumcircleofPBC\\ sovalueof{R}_1=1units,andthecorrectoptionisA.\\ \end{array}$