In the figure PQ is a daimeter and PT is a tangent of the circle with centre O if QOT=150°find PTO
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Answer:
∠QOT = 150°
∠POQ = 180° (LPA)
∠POT + 150° = 180°
∴∠POT = 30°
∠OPT = 90° (theorem...)
∴In ΔPOT,
∠POT + ∠OTP + ∠TPO = 180° (ASP of Δ)
30° +∠OTP + 90° = 180
∴∠OTP = 60°
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