In the given figure, ABCD is a cyclic quadrilateral, OB is the radius, PB is the tangent at point B and ∠OBC=30∘. AOC is a straight line. Then, the value of x is
60∘
In the given figrue,
∠OBP=90∘
and ∠OBC=30∘
∴∠CBP=90∘−30∘=60∘⇒∠CAB=60∘
[∵∠CAB=∠CBP, alternate segment of chord]
In ΔOAB,
OA =OB
⇒∠OAB=∠OBA=60∘
In ΔOAB,
∠OAB+∠OBA+∠AOB=180∘⇒60∘+60∘+∠AOB=180∘∴∠AOB=180∘−120∘=60∘