What is \angle BAP equal to?
Consider the following for the next two (02) items that follow: AB is a diameter of a circle with centre O. Radius OP is perpendicular to AB. Let Q be any point on arc PB.
- A. 30^{\circ}
- B. 40^{\circ}
- C. 45^{\circ} ✓
- D. 60^{\circ}
Correct Answer: C. 45^{\circ}
Explanation
In \triangle AOP, \angle AOP = 90^{\circ} (since OP \perp AB) and OA = OP (radii of the same circle). Therefore, \triangle AOP is an isosceles right-angled triangle. This gives \angle OAP = \angle OPA = \frac{180^{\circ} - 90^{\circ}}{2} = 45^{\circ}. Since A, O, B are collinear, \angle BAP is 45^{\circ}.
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