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.

Answer 1

Consider Δ OAB and Δ OAC.

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 Δ OAB and Δ OAC are congruent triangles.

Therefore,

∠OAC = ∠OAC

It is given that,

`∠OAB +∠OAC=120^o`

`2∠OAB=120^o`

`∠OAB=60^o`

We know that,

`cos∠OAB =(AB)/(OA) `

`cos 60^o =(AB)/(OA) `

We know that,

`cos 60^o =1/2`

Therefore,

`1/2=(AB)/(OA)`

OA = 2AB