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प्रश्न
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.
उत्तर
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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