Question 90

PQ is a diameter of a circle with centre O. RS is a chord parallel to PQ subtends an angle of 40° at the centre of the circle. If PR and QS are produced to meet at T, then what will be the measure (in degrees) of ∠PTQ?

Solution

∠POR + ∠ROS + ∠SOQ = 180

∠POR = ∠SOQ

2∠POR = 180 - 40 = 140

∠POR = ∠SOQ = 70

In triangle PRO,

∠OPR = ∠ORP

∠OPR + ∠ORP + ∠POR = 180

2∠OPR = 180 - 70 = 110

∠OPR = 55

In triangle OQS,

∠OQS = ∠OSQ

∠OQS + ∠OSQ + ∠SOQ = 180

2∠OQS = 180 - 70 = 110

∠OQS = 55

∠OPR = ∠TPQ and ∠OQS = ∠TQP

In triangle PQT,

∠TPQ + ∠TQP + ∠PTQ = 180

∠PTQ = 180 -55-55 = 70

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