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 the point where the line segment AB touches the circle be point C. Since PQ, PR, and AB are all tangent lines to the circle, we can use the geometric property that tangents drawn from an external point to a circle subtend equal angles at the center of the circle.

For the external point A, the tangents are AQ and AC. Therefore, the line segment OA bisects the angle formed by the radii drawn to the points of tangency. This means:
$\angle AOQ=\angle AOC$

Similarly, for the external point B, the tangents are BC and BR. Therefore, the line segment OB bisects the angle formed by the radii to these respective points of tangency, giving us:
$\angle BOC=\angle BOR$

We are given in the problem statement that:
$\angle AOB={50}^{\circ }$

Notice that the angle AOB is composed of two adjacent smaller angles:
$\angle AOB=\angle AOC+\angle BOC$

Now, we want to find the total central angle QOR. This angle is the sum of all four individual small angles at the center:
$\angle QOR=\angle AOQ+\angle AOC+\angle BOC+\angle BOR$

By substituting the equal angles we established earlier into this equation, we get:
$\angle QOR=2(\angle AOC)+2(\angle BOC)$

Factoring out the 2 yields:
$\angle QOR=2(\angle AOC+\angle BOC)$

Since ∠AOC + ∠BOC is simply ∠AOB, we can substitute our known value:
$\angle QOR=2(\angle AOB)=2\times {50}^{\circ }={100}^{\circ }$

Next, let us analyze the quadrilateral PQOR. Another key circle property dictates that the radius of a circle is always perpendicular to a tangent line at the exact point of tangency. Therefore, the angles formed at Q and R are right angles:
$\angle OQP={90}^{\circ }$
and
$\angle ORP={90}^{\circ }$

Because the sum of the interior angles of any quadrilateral is always 360^°, the sum of the angles in quadrilateral PQOR is:
$\angle QPR+\angle OQP+\angle QOR+\angle ORP={360}^{\circ }$

Substituting the known angle values into our quadrilateral equation gives:
$\angle QPR+{90}^{\circ }+{100}^{\circ }+{90}^{\circ }={360}^{\circ }$

Simplifying the constant values together, we have:
$\angle QPR+{280}^{\circ }={360}^{\circ }$

Subtracting 280^° from both sides solves for angle QPR:
$\angle QPR={360}^{\circ }-{280}^{\circ }={80}^{\circ }$

Finally, because point A lies directly on line segment PQ, and point B lies directly on line segment PR, the angle denoted as ∠QPR spans the exact same vertex as ∠APB. Thus, ∠APB is strictly equal to 80^°.