PA and PB are tangents from an outside point P to a circle with centre O, touching it at A and B. If $\angle APB = 50^\circ$ , what is $\angle OAB$ ?

25°Correct answerYour answer40°Correct answerYour answer65°Correct answerYour answer130°Correct answerYour answer

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CorrectNot quiteAnswer A

Correct answer: A

Tangents from one point are equal, $PA = PB$ , so $\angle PAB = \angle PBA = (180^\circ - 50^\circ) \div 2 = 65^\circ$ .
The radius OA is perpendicular to the tangent PA, so $\angle OAP = 90^\circ$ .
$\angle OAB = 90^\circ - 65^\circ = 25^\circ$ .

Worth noting

65° is $\angle PAB$ , and 130° is $\angle AOB$ . A neat check: $\angle OAB = \tfrac{1}{2}\angle APB$ .

Ten questions from Lines, Triangles and Circles, at your own pace.

PA and PB are tangents from an outside point P to a circle with centre O, touching it at A and B. If ∠APB=50∘\angle APB = 50^\circ∠APB=50∘, what is ∠OAB\angle OAB∠OAB?