Question

In the figure, AP is a tangent to the circle and $\Delta$APB is an isosceles triangle. Also $m\angle APB={50}^o$. Find the angle of $\Delta$ABC and $\Delta$ APB.

Answer 1

Since PB is not a tangent, PA and PB are not equal.
In $\Delta APB,m\angle BPA={50}^o$ (given)
$m\angle BAP={50}^o$ (Two angles in an isosceles triangle are equal)
So, now using angle sum property in $\Delta$APB, we have
$m\angle ABP={180}^o-({50}^o+{50}^o)={80}^o$
In $\Delta ABC,m\angle ABC={180}^o-{80}^o={100}^o$
$m\angle ACB=m\angle BAP={50}^o$
(Each of the two angles made by a tangent to a circle and a chord through the point of contact is equal to an angle in the segment on the other side of the chord.)
Now using angle sum property in $\Delta$ACB, we have
$m\angle BAC={180}^o-({100}^o+{50}^o)={30}^o$
Note: Given that $\Delta$APB is isosceles, but its equal sides are not given. The angles calculated above are based on the assumption that BA = BP. If we take AP = AB, it is not possible to find all the angles.