Question:
In the given figure, two tangents AB and AC are drawn to a circle with centre O such that ∠BAC = 120°. Prove that OA = 2AB.
Solution:
Consider 
and
.
We have,
OB = OC (Since they are radii of the same circle)
AB = AC (Since length of two tangents drawn from an external point will be equal)
OA is the common side.
Therefore by SSS congruency, we can say that andare congruent triangles.
Therefore,
It is given that,
$\angle O A B+\angle O A C=120^{\circ}$
$2 \angle O A B=120^{\circ}$
$\angle O A B=60^{\circ}$
We know that,
$\cos \angle O A B=\frac{A B}{O A}$
$\cos 60^{\circ}=\frac{A B}{O A}$
We know that,
$\cos 60^{\circ}=\frac{1}{2}$
Therefore,
$\frac{1}{2}=\frac{A B}{O A}$
OA = 2AB