Two tangents drawn from a point p and a circle with center O at point Q and R. Point A and B lie on PQ and PR, repectively, Such that AB is also a tangent to the same circle. Ir ∠A0B = 50⁰ , then ∠APB, in degrees equals

The correct answer is 80.

Let us solve this step-by-step using geometric properties of tangents to a circle from an external point.

1. Understand the Setup:
We have a circle with center $O$.
From an external point $P$, two tangents $PQ$ and $PR$ are drawn to touch the circle at points $Q$ and $R$ respectively.
A third tangent touches the circle at some point, say $C$, and intersects segment $PQ$ at point $A$ and segment $PR$ at point $B$.

2. Angle Subtended by Tangents at the Center:
The line joining an external point to the center of the circle bisects the angle subtended by the contact points at the center.
Specifically:
- From point $A$, tangents $AQ$ and $AC$ touch the circle. Therefore, line $OA$ bisects $\angle QOC$.
Hence, $\angle AOC=\frac{1}{2}\angle QOC$.
- From point $B$, tangents $BR$ and $BC$ touch the circle. Therefore, line $OB$ bisects $\angle ROC$.
Hence, $\angle BOC=\frac{1}{2}\angle ROC$.

3. Relating $\angle AOB$ to $\angle QOR$:
We know that $\angle AOB=\angle AOC+\angle BOC$.
Substituting the expressions from above:

$\angle AOB=\frac{1}{2}\angle QOC+\frac{1}{2}\angle ROC=\frac{1}{2}(\angle QOC+\angle ROC)=\frac{1}{2}\angle QOR$

Given that $\angle AOB={50}^{°}$:

${50}^{°}=\frac{1}{2}\angle QOR\Rightarrow \angle QOR={100}^{°}$

4. Finding $\angle APB$ (which is $\angle QPR$):
In quadrilateral $OQPR$:
- The radii are perpendicular to the tangents at the points of contact, so $\angle OQP={90}^{°}$ and $\angle ORP={90}^{°}$.
- The sum of interior angles of quadrilateral $OQPR$ is ${360}^{°}$.

Therefore, the opposite angles $\angle QOR$ and $\angle QPR$ are supplementary:

$\angle QPR+\angle QOR={180}^{°}$

$\angle APB={180}^{°}-\angle QOR$

$\angle APB={180}^{°}-{100}^{°}={80}^{°}$

Thus, $\angle APB$ equals 80 degrees.