Answer:
Since, tangents from an external point are equal. i.e., \[AP=BP\] Given, \[\angle PAB=50{}^\circ \] \[\therefore \angle PBA=50{}^\circ \] In \[\Delta \text{ }APB\] \[\angle APB=180{}^\circ -(50{}^\circ +50{}^\circ )=80{}^\circ \] \[\therefore \angle AOB=180{}^\circ -80{}^\circ =100\]
You need to login to perform this action.
You will be redirected in
3 sec