Question 113

Two circles touch each other externally at a point P and a direct common tangent touches the circles at the points Q and R respectively. Then ∠QPR is

Solution

Since QS and SP are tangents from the same point S => QS = SP

=> In $$\triangle$$QSP, $$\angle$$SQP = QPS

=> $$\angle$$SQP + $$\angle$$QPS + $$\angle$$PSQ = 180°

=> $$\angle$$QPS = 90/2 = 45°

Similarly, $$\angle$$SPR = 45°

Adding above two equations, we get :

=> $$\angle$$QPS + $$\angle$$SPR = 45° + 45°

=> $$\angle$$QPR = 90°

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