From an external point P, tangents PA and PB are drawn to a circle with center O. If ∠PAB = 50°, then find ∠AOB.

Question

Philoid · Accepted Answer

As per the given condition the figure can be drawn as:

We know that, tangents drawn from external point to a circle are equal

∴ PA = PB

Also, angles opposite to equal sides are equal to each other

∴∠ PBA = ∠ PAB = 50^o

_{Sum of all angles of triangle is 180}^o

_∴_{in triangle ABP, we have:} _∠ _{PBA +} _∠ _{PAB +} _∠ _{APB = 180}^o

₅₀^o + 50^o + _∠ _{APB = 180}^o

_∴ _∠ _{APB = 180}^o – 100^o

_{= 80}^o

_{Now, we know that in a cyclic quadrilateral sum of opposite angles is equal to 180}^o

_∴_{In cyclic quadrilateral OAPB, we have:}

∠ AOB + ∠ APB = 180^o

∠ AOB + 80^o = 180^o

∴∠ AOB = 100^o