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

Approach: The picture is a triangle \(PAB\) with the circle as its incircle (it touches all three sides \(PQ\), \(PR\) and \(AB\)). For any incircle, the angle a side subtends at the centre is fixed relative to the opposite vertex angle, which links \(\angle AOB\) to \(\angle APB\) immediately.

Step 1: Note that \(PQ\), \(PR\), and \(AB\) are all tangents to the same circle, and they form triangle \(PAB\) with the circle sitting inside it. So \(O\) is the incentre of triangle \(PAB\) and the circle is its incircle, touching \(AB\) at a point \(T\).

Step 2: Look at the angle \(\angle AOB\) at the incentre, made by the two cevians \(OA\) and \(OB\). A standard incentre result says the angle subtended at the incentre by a side equals \(90^\circ\) plus half the opposite angle: \[ \angle AOB = 90^\circ + \tfrac{1}{2}\angle APB. \]

Step 3: Substitute \(\angle AOB = 50^\circ\): \[ 50^\circ = 90^\circ + \tfrac{1}{2}\angle APB \;\Rightarrow\; \tfrac{1}{2}\angle APB = -40^\circ. \] A negative value signals the circle is the excircle opposite \(P\), not the incircle (it touches side \(AB\) and the extensions of \(PA\), \(PB\) beyond \(A,B\) — exactly the "tangent line cutting across" configuration here). For the \(P\)-excircle the relation is \[ \angle AOB = 90^\circ - \tfrac{1}{2}\angle APB. \]

Step 4: Now \[ 50^\circ = 90^\circ - \tfrac{1}{2}\angle APB \;\Rightarrow\; \tfrac{1}{2}\angle APB = 40^\circ \;\Rightarrow\; \angle APB = 80^\circ. \]

Step 5 (clean cross-check): Split the central angles by the tangent contact points. Tangents from \(A\) give \(\angle AOT=\angle AOQ=x\); tangents from \(B\) give \(\angle BOT=\angle BOR=y\). Then \(\angle AOB = x+y = 50^\circ\), and the full angle \(\angle QOR = 2x+2y = 100^\circ\). In quadrilateral \(PQOR\), with \(\angle OQP=\angle ORP=90^\circ\), \[ \angle APB = 360^\circ - 90^\circ - 90^\circ - 100^\circ = 80^\circ. \] Final answer: \(\angle APB = 80^\circ\).

Answer 2

Let $P$ be an external point from which tangents $PQ$ and $PR$ are drawn to the circle with center $O$, touching it at $Q$ and $R$ respectively. Points $A$ and $B$ lie on $PQ$ and $PR$ such that $AB$ is also tangent to the circle and touches it at point $T$. We are given $\angle AOB = 50^{\circ}$ and we need to find $\angle APB$. Step 1: Use tangent–radius property. From point $A$, two tangents are drawn: $AQ$ and $AT$. Since tangents from an external point are equal, we have: \[ AQ = AT, \quad \Rightarrow\quad \triangle AOQ \cong \triangle AOT. \] Hence, \[ \angle AOQ = \angle AOT = x \quad\text{(say)}. \] Similarly, from point $B$, tangents $BR$ and $BT$ are drawn, so: \[ \triangle BOR \cong \triangle BOT \Rightarrow \angle BOR = \angle BOT = y \quad\text{(say)}. \]
Step 2: Relate $x$ and $y$ using $\angle AOB$. At the center, \[ \angle AOB = \angle AOT + \angle TOB = x + y = 50^{\circ}. \tag{1} \]
Step 3: Find $\angle QOR$. Angle between radii $OQ$ and $OR$ is: \[ \angle QOR = \angle QOA + \angle AOT + \angle TOB + \angle BOR = x + x + y + y = 2(x + y). \] Using (1), \[ \angle QOR = 2 \times 50^{\circ} = 100^{\circ}. \]
Step 4: Use quadrilateral $PQOR$. Since the radius is perpendicular to the tangent at the point of contact, \[ \angle OQP = 90^{\circ}, \qquad \angle ORP = 90^{\circ}. \] In quadrilateral $PQOR$: \[ \angle QPR + \angle OQP + \angle QOR + \angle ORP = 360^{\circ}. \] But $\angle QPR = \angle APB$ (same angle at $P$), hence \[ \angle APB + 90^{\circ} + 100^{\circ} + 90^{\circ} = 360^{\circ} \] \[ \Rightarrow\; \angle APB + 280^{\circ} = 360^{\circ} \Rightarrow\; \angle APB = 80^{\circ}. \] Therefore, $\angle APB = 80^{\circ}$.