**Question:**

If *PA* and *PB* are tangents from an outside point *P* such that *PA* = 10 cm and ∠*APB* = 60°. Find the length of chord *AB*.

**Solution:**

Let us first put the given data in the form of a diagram.

From the property of tangents we know that the length of two tangents drawn to a circle from a common external point will always be equal. Therefore,

*PA=PB*

Consider the triangle *PAB*. Since we have *PA=PB*, it is an isosceles triangle. We know that in an isosceles triangle, the angles opposite to the equal sides will be equal. Therefore we have,

Also, sum of all angles of a triangle will be equal to . Therefore,

$\angle P A B+\angle P B A+\angle A P B=180^{\circ}$

$60^{\circ}+2 \angle P B A=180^{\circ}$

$2 \angle P B A=120^{\circ}$

$\angle P A B=60^{\circ}$

Since we know that $\angle P A B=\angle P B A$,

$\angle P A B=60^{\circ}$

Now if we see the values of all the angles of the triangle, all the angles measure. Therefore triangle *PAB* is an equilateral triangle.

We know that in an equilateral triangle all the sides will be equal.

It is given in the problem that side *PA* = 10 cm. Therefore, all the sides will measure 10 cm. Hence, *AB* = 10 cm.

Thus the length of the chord *AB* is 10 cm.

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