Question

In Fig. 8.67, two tangents AB and AC are drawn to a circle with centre O such that $\angle BAC={120}^{\circ }$. Prove that OA =2AB.

Answer 1

Draw the figure. points B and C must both be on the circle. Now connect both points B and C to the circle center O.
Since both AB and AC are tangent to the circle then AB = AC

Now draw the line AO
$\angle$OAB and $\angle$OAC are equal.
Since BAC = 120
$\angle$OAB = 60 and $\angle$OBA = ${90}^{\circ }$

$\therefore$ $\angle$AOB is ${30}^{\circ }$.
So, sin$\angle$AOB = sin 30 = \frac{1}{2} = \frac{AB}{OA} and
OA = 2AB