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Question

Tangent AP and AQ are drawn to circle with centre O, from an external point A. Prove that $$\angle PAQ = 2. \angle OPQ$$.


Solution

Consider the above drawn figure:

$$OP=OQ$$    (Radii of same circle)

Thus, $$\angle OPQ=\angle OQP$$       (Angle opposite to equal sides are equal)

In $$\triangle OPQ$$, the sum of the angles is $$180^0$$ that is:

$$\angle OPQ+\angle OQP+\angle POQ=180^{ 0 }$$

But $$\angle OPQ=\angle OQP$$, therefore,

$$\angle OPQ+\angle OQP+\angle POQ=180^{ 0 }\\ \Rightarrow \angle OPQ+\angle OPQ+\angle POQ=180^{ 0 }\\ \Rightarrow 2\angle OPQ+\angle POQ=180^{ 0 }\\ \Rightarrow 2\angle OPQ=180^{ 0 }-\angle POQ\quad ........(1)$$

We also know that 

$$\angle POQ+\angle PAQ=180^{ 0 }\\ \Rightarrow \angle PAQ=180^{ 0 }-\angle POQ\quad ........(2)$$

From equations 1 and 2, we get

$$2\angle OPQ=\angle PAQ$$

Hence, $$\angle PAQ=2\angle OPQ$$.


639293_564027_ans.png

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