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

Given:
TP and TQ are tangents to a circle from an external point T.
O is the centre of the circle, and P and Q are the points of contact.
We are given: \( \angle OPQ = 15^\circ \)

Step 1: Use property of tangents
Tangents drawn from an external point to a circle are equal in length.
So, TP = TQ
Also, angles between the radius and tangent at point of contact are \(90^\circ\):
\[ \angle OPT = \angle OQT = 90^\circ \]
Triangles \( \triangle OTP \) and \( \triangle OTQ \) are right-angled and share side OT.
Since TP = TQ and OP = OQ (radii), we get:
\[ \triangle OTP \cong \triangle OTQ \quad \text{(by RHS criterion)} \]
Step 2: Use angle information
It is given that:
\[ \angle OPQ = 15^\circ \]
This is the angle between the radius and chord PQ at point P.
In triangle \( \triangle PTQ \), this angle contributes to the total angle at vertex T.
So, the full angle at vertex T is:
\[ \angle PTQ = 2 \cdot \angle OPQ = 2 \cdot 15^\circ = 30^\circ \]
Step 3: Let \( \theta = \angle PTO = \angle QTO \)
Because triangle \( \triangle PTQ \) is symmetric and point O lies on the angle bisector of \( \angle PTQ \), we can write:
\[ 2\theta = \angle PTQ = 30^\circ \Rightarrow \theta = 15^\circ \]
Step 4: Find value of \( \sin 2\theta \)
We are to find: \( \sin 2\theta \)
Since \( \theta = 15^\circ \),
\[ \sin 2\theta = \sin 30^\circ = \frac{1}{2} \]
Final Answer:
\[ \boxed{\sin 2\theta = \dfrac{1}{2}} \]