If $ P A $ and $ P B $ are tangents from an outside point $ P $. such that $ P A=10 \mathrm{~cm} $ and $ \angle A P B=60^{\circ} $. Find the length of chord $ A B $.

Given:

\( P A \) and \( P B \) are tangents from an outside point \( P \). such that \( P A=10 \mathrm{~cm} \) and \( \angle A P B=60^{\circ} \).

To do:

We have to find the length of chord \( A B \). 

Solution:

Tangents drawn from a point outside the circle are equal

This implies,

$PA = PB = 10\ cm, \angle PAB = \angle PBA$    (Angles opposite to equal sides are equal)

In $\triangle APB$,

$\angle APB + \angle PAB + \angle PBA = 180^o$

$60^o + \angle PAB + \angle PAB = 180^o$

$2 \angle PAB = 180^o - 60^o$

$\angle PAB= \frac{120^o}{2}$

$\angle PAB = 60^o$

$\angle PBA = \angle PAB = 60^o$

$PA = PB = AB = 10\ cm$    (Equilateral triangle)

The length of the chord AB is 10 cm.

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Updated on: 10-Oct-2022

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