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Question
In Fig.3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠ OTS = ∠ OST = 30°.
Solution
In the given figure,
OP = 2r … (Given)
∠OTP = 90° … (radius drawn at the point of contact is perpendicular to the tangent)
In OTP,
sin OPT =`(OT)/(OP)=1/2`= sin30°
⇒ ∠OPT=30°
∴ ∠TOP=60°
∴ ΔOTP is a 30°-60°-90°, right triangle.
In ΔOTS,
OT = OS … (Radii of the same circle)
∴ ΔOTS is an isosceles triangle.
∴ ∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)
In ΔOTQ and ΔOSQ
OS = OT … (Radii of the same circle)
OQ = OQ ...(side common to both triangles)
∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are equal)
∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)
∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)
∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)
∴ ∠OTS + ∠OST = 180° – 120° = 60°
∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°
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