In the given figure, O is the centre of the circle and QPR is a tangent to it at P. Prove that $\angle QAP + \angle APR = 90°$.

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Step 1: $OA = OP$ (radii). $	herefore$ In $	riangle OAP$, $\angle OPA = \angle OAP$ ... (i)
Step 2: $\angle OPA + \angle APR = 90°$ (tangent is perpendicular to radius)
Step 3: $\Rightarrow \angle OAP + \angle APR = 90°$ (using (i))
Step 4: $\Rightarrow \angle QAP + \angle APR = 90°$

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