Question

To draw a pair of tangents to a circle which are inclined to each other at an angle of ${60}^o$, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be

• A. ${100}^o$
• B. ${140}^o$
• C. ${120}^o$
• D. ${135}^o$

Answer 1

Answer: (C)

${120}^o$

Given- $O$ is the centre of a circle to which a pair of tangents $PQ\&PR$ from a point $P$ touch the circle at $Q\&R$ respectively. $\angle RPQ={60}^o$.
To find out- $\angle ROQ=?$
Solution- $\angle OQP={90}^o=\angle ORP$ since the angle, between a tangent to a circle and the radius of the same circle passing through the point of contact, is ${90}^o$. $\therefore$ By angle sum property of quadrilaterals, we get $\angle OQP+\angle RPQ+\angle ORP+\angle ROQ={360}^o⟹{90}^o+{60}^o+{90}^o+\angle ROQ={360}^o⟹\angle ROQ={120}^o$.
Ans- Option C.