In Figure 2, $PQ$ is tangent to the circle with centre at $O$, at the point $B$. If $\angle AOB = 100°$, then $\angle ABP$ is equal to

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In Figure 2, $PQ$ is tangent to the circle with centre at $O$, at the point $B$. If $\angle AOB = 100°$, then $\angle ABP$ is equal to

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Accepted Answer

Step 1: $OA = OB$ (radii), so $	riangle AOB$ is isosceles. $\angle OAB = \angle OBA = \frac{180° - 100°}{2} = 40°$. Since $PQ$ is tangent at $B$, $\angle OBP = 90°$. Therefore $\angle ABP = 90° - 40° = 50°$.

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