Question

$AB$ and $AC$ are $2$ tangent to a circle with centre $O$, if $\angle BAC={120}^o$ then Prove that $OA=2AB$

Answer 1

AB and AC are two tangents to circle with centre O, if $\angle BAC={120}^o$ then prove that $OA=2AB$.

Solution:-

Given: $1)$ AB and AC are tangents to circle

$2)$ $\angle BAC={120}^o$

To prove: $OA=2AB$

Construction: Join OA, OB & OC

Proof: $1)$ We know the property of tangents that radius drawn to the point at which a line is tangent to circle is perpendicular to tangent.

$\therefore OB\perp AB$ and $OC\perp AC$

$2)$ Also, clearly OA bisects $\angle BAC$

$\therefore \angle BAO=\angle CAO=6{-}^o$

$3)$ Now in $\Delta BAO,\angle BAO={60}^o,\angle B={90}^o$

$\therefore \angle AOB={30}^o$

$\sin \angle AOB=\frac{AB}{OA}$

$\sin {30}^o=\frac{AB}{OA}$

$\frac{1}{2}=\frac{AB}{OA}$

$OA=2AB$

Hence proved.