Originally Posted by

**sammy28** Hi all,

$\displaystyle \angle OAP = 90^{\circ}$, tangent is perpendicular to radius at point of contact

$\displaystyle \angle OAB = \frac{180-152}{2}=14^{\circ}$, angle sum of triangle is 180, triangle OAB is isosceles (equal angles opposite equal sides)

$\displaystyle \angle PAB = 90 -14 = 76^{\circ}$

$\displaystyle \angle PBA = 180 - 71 - 76 = 33^{\circ}$, angle sum triangle is 180

BUT....

Angle sum of quadrilateral is $\displaystyle 360^{\circ}$

In which case

$\displaystyle \angle PBA = 360 -152 - 90 - 71 = 47^{\circ}$

(Punch)

is this a poorly drawn up question or is the reasoning flawed?

thanks.