PA and PB are the tangents drawn to a circle from an external point P. If PA = AB, find APB

Given:

PA and PB are the tangents drawn to a circle from an external point P. If PA = AB, find ∠APB.

Formula used:

In a circle, the tangents drawn from an external point to a circle are equal in length and make equal angles with the line segment joining the center of the circle to the point of tangency.

Calculation:

Let O be the center of the circle, and A and B be the points of tangency on the circle.

Given PA = PB (since both are tangents from point P) and PA = AB.

Since PA = PB, △OPA and △OPB are congruent by the RHS (Right angle-Hypotenuse-Side) criterion.

Therefore, ∠OPA = ∠OPB.

Also, since PA = AB, △PAB is an isosceles triangle with PA = AB.

In △PAB, since PA = AB, ∠PAB = ∠PBA.

Now, let ∠PAB = ∠PBA = θ.

Since the sum of angles in a triangle is 180°, we have:

∠APB + 2θ = 180°

As PA and PB are tangents to the circle, ∠OAP and ∠OBP are right angles (90°).

Thus, ∠APB = 180° - 2θ.

But, since PA = AB, we have θ = 45°.

Therefore, ∠APB = 180° - 2 * 45° = 180° - 90° = 90°.

∴ ∠APB = 90°.

Answer: 1) 90°