The given question is a hard 2-mark question from the chapter Circles. It combines multiple concepts. Concept tested: Properties of tangents to a circle, values of trigonometric ratios, and properties of isosceles triangles. It is an important question and appeared in Section B of the 2016 CBSE class 10 board paper.

Question 10: 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°

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∠OTP = 90° because the angle between a tangent and a radius where the tangent meets the circle is a right angle.

∴ ΔOTP is a right triangle

Sin ∠OPT = \\frac{\text{Opposite side}}{\text{Hypotenuse}}) = \\frac{\text{OT}}{\text{OP}})

OT is a radius to the circle. So, OT = r.

**Given Data**: OP = 2r.

∴ Sin ∠OPT = \\frac{\text{r}}{\text{2r}}) = \\frac{\text{1}}{\text{2}})

Sin 30° = \\frac{\text{1}}{\text{2}})

∴ ∠OPT = 30°

Sum of the interior angles of a triangle = 180°

In Δ OPT, ∠OPT + ∠OTP + ∠POT = 180°

30° + 90° + ∠POT = 180°

Or ∠POT = 60°

By RHS rule Δ OPT and Δ OPS are congruent

∴ ∠QOT = ∠QOS = 60° (∠QOT is the same as ∠POT = 60°)

Hence, ∠SOT = ∠QOT + ∠QOS = 60 + 60 = 120°

Δ SOT is an isosceles triangle because OS and OT are radii to the circle.

∴ ∠OTS = ∠OST = x

In Δ SOT, ∠SOT + ∠OTS + ∠OST = 180°

120° + x + x = 180°

Or x = 30°

i.e., ∠OTS = ∠OST = 30°

Class 10 Maths

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CBSE Class 10 Maths - 2022

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