A phone company's circular exclusion zone, centre \(O\), has two straight maintenance tracks from an external junction \(P\), tangent to the zone at \(A\) a...
Question 1 Report
A phone company's circular exclusion zone, centre \(O\), has two straight maintenance tracks from an external junction \(P\), tangent to the zone at \(A\) and \(B\), with angle \(APB=50^{\circ}\). A drone \(C\) hovers on the major arc \(AB\), as shown.
- Give a reason why angle \(OAP=\) angle \(OBP=90^{\circ}\). (1)
- Work out the size of angle \(AOB\). (2)
- Work out the size of angle \(ACB\). (2)
Answer Details
(a) \(OA\) and \(OB\) are radii drawn to the points of contact \(A\) and \(B\), and a tangent always meets its radius at a right angle, so angle \(OAP =\) angle \(OBP = 90^{\circ}\). [1 mark]
(b) Quadrilateral \(OAPB\) has angles summing to \(360^{\circ}\): two right angles at \(A\) and \(B\), plus \(50^{\circ}\) at \(P\), leaves angle \(AOB = 360-90-90-50 = 130^{\circ}\). [2 marks]
(c) \(C\) lies on the major arc \(AB\) of the exclusion zone's circle, so the angle at the centre, angle \(AOB\), is twice the angle at the circumference, angle \(ACB\): angle \(ACB = 130 \div 2 = 65^{\circ}\). [2 marks]
This question chains the tangent-radius right angle, used twice to fix the quadrilateral's angle sum, with the centre-circumference angle theorem to reach \(C\).
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