WAEC SSCE - General Mathematics - 2011 (Objective)

Question 1 Report

In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240^o. Calculate the value x

60^o Correct answer

55^o

50^o

45^o

Answer Details

In the given diagram, we have a circle with center O and radii OP and OR. The reflex angle PQR is 240° and |PQ| = |QR|. We need to find the value of x. Since |PQ| = |QR|, we know that triangle PQR is an isosceles triangle. Therefore, the angles opposite to PQ and QR are equal. Let's denote the angle PQR by y. Then we have: 2y + 60° = 360° (sum of angles in a triangle) 2y = 300° y = 150° Therefore, each of the angles opposite to PQ and QR is equal to (180° - 150°)/2 = 15°. Now, consider the triangle OQP. We know that the sum of angles in a triangle is 180°. Therefore: ∠OQP + ∠QOP + ∠OPQ = 180° Since OP and OQ are radii, ∠QOP = ∠OPQ. Let's denote this angle by z. Then we have: z + z + 15° = 180° 2z = 165° z = 82.5° Finally, consider the triangle OXR. We know that the sum of angles in a triangle is 180°. Therefore: ∠OXR + ∠ORX + ∠ROX = 180° Since OR and OX are radii, ∠ORX = ∠ROX. Let's denote this angle by x. Then we have: x + x + 60° = 180° 2x = 120° x = 60° Therefore, the value of x is 60°. Answer: 60°.






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