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

Let \(P\) be an external point from which tangents \(PQ\) and \(PR\) touch the circle with centre \(O\) at points \(Q\) and \(R\). Points \(A\) and \(B\) lie on \(PQ\) and \(PR\) respectively such that \(AB\) is also a tangent to the circle at \(T\). Given \(\angle AOB = 50^\circ\), we are required to find \(\angle APB\). Step 1: Use the property of equal tangents from an external point. From point \(A\), the tangents \(AQ\) and \(AT\) are drawn to the circle. Since tangents drawn from the same external point are equal, \[ AQ = AT. \] Hence, triangles \(AOQ\) and \(AOT\) are congruent, which gives \[ \angle AOQ = \angle AOT = x \quad (\text{say}). \] Similarly, from point \(B\), tangents \(BR\) and \(BT\) are drawn. Thus, triangles \(BOR\) and \(BOT\) are congruent, giving \[ \angle BOR = \angle BOT = y \quad (\text{say}). \] Step 2: Use the given central angle. At the centre \(O\), \[ \angle AOB = \angle AOT + \angle TOB = x + y = 50^\circ. \tag{1} \] Step 3: Find the central angle \(\angle QOR\). The angle subtended at the centre by arc \(QR\) is \[ \angle QOR = \angle QOA + \angle AOT + \angle TOB + \angle BOR = x + x + y + y = 2(x+y). \] Using equation (1), \[ \angle QOR = 2 \times 50^\circ = 100^\circ. \] Step 4: Apply angle sum in quadrilateral \(PQOR\). The radius is perpendicular to the tangent at the point of contact, hence \[ \angle OQP = 90^\circ, \qquad \angle ORP = 90^\circ. \] In quadrilateral \(PQOR\), \[ \angle QPR + \angle OQP + \angle QOR + \angle ORP = 360^\circ. \] Since \(\angle QPR = \angle APB\), we get \[ \angle APB + 90^\circ + 100^\circ + 90^\circ = 360^\circ, \] \[ \angle APB + 280^\circ = 360^\circ, \] \[ \angle APB = 80^\circ. \] Therefore, the value of \(\angle APB\) is \[ 80^\circ. \]

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, $\angle APB = 80^{\circ}$.