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


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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