Q. 14.4( 14 Votes )

From an external

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 = 50o


Sum of all angles of triangle is 180o


in triangle ABP, we have: PBA + PAB + APB = 180o


50o + 50o + APB = 180o


APB = 180o – 100o


= 80o


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


In cyclic quadrilateral OAPB, we have:


AOB + APB = 180o


AOB + 80o = 180o


∴∠ AOB = 100o

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