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Question
PA and PB are tangents drawn to the circle with centre O as shown in the figure. Prove that ∠APB = 2∠OAB.
Solution
Let ∠APB = x
Now by theorem, the lengths of a tangents drawn from an external point to a circle are equal
So, PAB is an isosceles triangie
Therefore, ∠PAB = ∠PBA
= `1/2(180^circ - x)`
= `90^circ - x/2`
Also by theorem, the tangents at any point of a circle is perpendicular to the radius through the point of contact ∠OPT = 90°
Therefore, ∠OAB = ∠OAP – ∠PAB
= `90^circ - (90^circ - x/2)`
= `x/2`
= `1/2` ∠APB
Hence, ∠APB = 2∠OAB.
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