Question

Two tangents drawn from a point $P$ touch a circle with center $O$ at points $Q$ and $R$. Points $A$ and $B$ lie on $PQ$ and $PR$, respectively, such that $AB$ is also a tangent to the same circle. If $\angle AOB = 50^{\circ}$, then $\angle APB$, in degrees, equals:

Hint:

Tangents from an external point to a circle are equal, and the line from the center to the point of tangency is perpendicular to the tangent. These facts often allow you to form congruent triangles and use angle sums in quadrilaterals involving the center.

Answer 1

Correct Answer: 80

Step 1: Since \(AQ\) and \(AS\) are tangents from \(A\), \(OA\) bisects the angle between them, so \(\angle OAB = \dfrac{180^{\circ} - \angle PAB}{2}\) (as \(\angle QAS = 180^{\circ} - \angle PAB\), being the supplement along the straight line through \(P, A, Q\)). Similarly \(\angle OBA = \dfrac{180^{\circ} - \angle PBA}{2}\).
Step 2: In triangle \(OAB\): \(\angle AOB = 180^{\circ} - \angle OAB - \angle OBA = \dfrac{\angle PAB + \angle PBA}{2}\).
Step 3: In triangle \(PAB\), \(\angle PAB + \angle PBA = 180^{\circ} - \angle APB\). Substituting, \(\angle AOB = \dfrac{180^{\circ} - \angle APB}{2} = 90^{\circ} - \dfrac{\angle APB}{2}\).
Step 4: Given \(\angle AOB = 50^{\circ}\): \(50^{\circ} = 90^{\circ} - \dfrac{\angle APB}{2}\), so \(\angle APB = 2(90^{\circ}-50^{\circ})\).
\[ \boxed{\angle APB = 80^{\circ}} \]