Question

Two tangents PA and PB are drawn to a circle with centre O from an external point P. If ∠OAB is 30°.

Find angle APB.​

Answer 1

Answer:

⇒ ∠APO = ∠OPB = 30° …(i)

∠OAP = ∠OBP = 90° …radius is perpendicular to tangent …(ii)

Consider quadrilateral OAPB

⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°…sum of angles of quadrilateral From figure ∠APB = ∠APO + ∠OPB

⇒ ∠OAP + ∠APO + ∠OPB + ∠PBO + ∠AOB = 360°

Using (i) and (ii)

⇒ 90° + 30° + 30° + 90° + ∠AOB = 360°

⇒ 240° + ∠AOB = 360°

⇒ ∠AOB = 120°

Hence ∠AOB is 120°