question_answer

From an external point P, tangents PA and PB are drawn to a circle with centre O. If \[\angle PAB=50{}^\circ \], then find \[\angle AOB.\]

Answer:

Since, tangents from an external point are equal.

i.e.,                   \[AP=BP\]

Given,           \[\angle PAB=50{}^\circ \]

\[\therefore \angle PBA=50{}^\circ \]

In \[\Delta \text{ }APB\]

\[\angle APB=180{}^\circ -(50{}^\circ +50{}^\circ )=80{}^\circ \]

\[\therefore \angle AOB=180{}^\circ -80{}^\circ =100\]