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, $$\angle PAQ=2\angle OPQ$$.Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More