Question

In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$.

Hint:

Use tangent-radius property and basic triangle geometry to relate angles in circle problems.

Answer 1

TP and TQ are tangents ⇒ triangle OTP and OTQ are congruent.
$\angle OPQ = 15^\circ$ is the angle between radius and tangent. Since triangle OPT is isosceles right at $O$, $\angle PTO = \angle QTO = \theta$.
Angle subtended at centre by chord PQ = $2\theta$ (since triangle PTQ is isosceles and split by the radius).
From figure: $\angle PTQ = 2 \cdot \angle OPQ = 2 \cdot 15^\circ = 30^\circ$
So $\theta = 15^\circ \Rightarrow 2\theta = 30^\circ$
Hence, \[ \sin 2\theta = \sin 30^\circ = \dfrac{1}{2} \] Answer: $\sin 2\theta = \dfrac{1}{2}$