Question

$PQ$ is a tangent to a circle with centre $O$ at the point $P$. If $△OPQ$ is an isosceles triangle with P as vertex, then $\angle OQP$ is equal to

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (B)

${45}^o$

Given- O is the centre of a circle to which PQ is a tangent at P. $\Delta OPQ$ is isosceles whose vertex is P.
To find out- $\angle OQP=?$
Solution- OP is a radius through P, the point of contact of the tangent PQ with the given circle $\angle OPQ={90}^o$ since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. Now $\Delta OPQ$ is isosceles whose vertex is P.
$\therefore OP=PQ⟹\angle OQP=\angle QOP⟹\angle OQP+\angle QOP=2\angle OQP.$
$\therefore$ By angle sum property of triangles,
$\therefore \angle OPQ+2\angle OQP={180}^o⟹{90}^o+2\angle OQP={180}^o⟹\angle OQP={45}^o.$
Ans- Option- B.