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In the given figure, two tangents *AB* and *AC* are drawn to a circle with centre *O* such that ∠*BAC* = 120°. Prove that *OA* = 2*AB*.

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#### Solution

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*

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