Angles Tucked Inside A Circle

Given that $PA$ and $PB$ are tangents to the circle, then triangle $ABP$ is an isosceles triangle with $AP = BP$ , then means that the angles $\angle ABP = \angle BAP$ .

Recall that the sum of angles of a triangle is $180^\circ$ , then for triangle $ABP$ ,

$\begin{aligned} \angle ABP + \angle BAP + \angle APB &=& 180^\circ \\ 2 \angle BAP + 70^\circ &=& 180^\circ \\ \angle BAP&=& \dfrac{180^\circ - 70^\circ }2 = 55^\circ \end{aligned}$

And since $AP$ is a straight line that is tangent to the circle, then $\angle OAP = 90^\circ$ , and so

$\begin{aligned} \angle OAP &= & \angle OAB + \angle BAP \\ 90^\circ &= & x^\circ + 55^\circ \\ x^\circ&= & 90^\circ - 55^\circ \\ &= & 35^\circ \end{aligned}$

Hence, $x = \boxed{35}$ .