If the radius of the circumcircle of an isosc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $R$ be the circumradius of $\Delta PQR$. Given $R = PQ = PR$.By the sine rule,$R = \frac{PQ}{2 \sin R} = \frac{PR}{2 \sin Q} = \frac{QR}{2 \si

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(D) Let $R$ be the circumradius of $\Delta PQR$. Given $R = PQ = PR$.By the sine rule,$R = \frac{PQ}{2 \sin R} = \frac{PR}{2 \sin Q} = \frac{QR}{2 \sin P}$.Since $R = PQ$,we have $PQ = \frac{PQ}{2 \sin R}$,which implies $\sin R = \frac{1}{2}$.Thus,$\angle R = \frac{\pi}{6}$.Since $PQ = PR$,the triangle is isosceles,so $\angle Q = \angle R = \frac{\pi}{6}$.The sum of angles in a triangle is $\pi$. Therefore,$\angle P = \pi - (\angle Q + \angle R) = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

If the radius of the circumcircle of an isosceles triangle $PQR$ is equal to $PQ$ (where $PQ = PR$),then the angle $P$ is

Similar Questions

If $s$ is the semi-perimeter of $\triangle ABC$ and if $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}$,then $\sum \sin^2\left(\frac{A}{2}\right) = $

In $\triangle ABC$,if $r_1=4, r_2=8, r_3=24$,then $a=$

In a $\Delta ABC,$ $2ac \sin \left( \frac{A - B + C}{2} \right)$ is equal to

The angles of a triangle are in the ratio $1:3:5$. Find the greatest angle. (in $\pi /9$)

In $\triangle ABC$,if $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$,then $2r_1=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module