Question

If PA and PB are tangents from an outside point P such that $PA=10$ cm and $\angle APB={{60}^{\circ }}$. Find the chord of length AB.

Answer 1

Hint: We find the characteristics of the tangents and use that to find the angles of the $\Delta PAB$. The relation between sides and angles gives an equilateral triangle along with the remaining side’s length.

Complete step-by-step answer:
We know that two tangents drawn from an outside point on a circle are of equal length.
In the given picture there are two tangents PA and PB drawn from an outside point P.
Therefore, $PA=PB=10$.
We know that in an isosceles triangle, the opposite angles of the equal sides are equal.
The sum of three angles of any triangle is also ${{180}^{\circ }}$.
In $\Delta PAB$, we have $PA=PB$ and $\angle APB={{60}^{\circ }}$. We get $\angle PAB+\angle PBA+\angle APB={{180}^{\circ }}$.
Therefore, $\angle PAB+\angle PBA={{180}^{\circ }}-\angle APB={{180}^{\circ }}-{{60}^{\circ }}={{120}^{\circ }}$.
As $PA=PB$, we can say $\angle PAB=\angle PBA=\dfrac{{{120}^{\circ }}}{2}={{60}^{\circ }}$.
All the angles of $\Delta PAB$ are equal. Therefore, $\Delta PAB$ is equilateral.
We get $PA=PB=AB=10$ as all sides of an equilateral triangle are equal.
The length of AB is 10 cm.

Note: Geometrically we can prove that the tangents are equal. The connecting line between centre and the outside point breaks the main triangle into two congruent triangles. The sides and the angles all become equal to their corresponding counters.