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In the figure (ii) given below, O is the centre of the circle. AB is a diameter, TPT' is a tangent to the circle at P. If ∠BPT' = 30°, calculate

(i) ∠APT

(ii) ∠BOP

In the figure (ii) given below, O is the centre of the circle. AB is a diameter, TPT' is a tangent to the circle at P. If ∠BPT' = 30°, calculate (i) ∠APT (ii) ∠BOP. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Circles

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Answer

From figure,

∠APB = 90° (∵ angles in semicircle is equal to 90)

From figure,

∠APT + ∠APT' = 180° (∵ they form linear pair)

⇒ ∠APT + ∠APB + ∠BPT' = 180°
⇒ ∠APT + 90° + 30° = 180°
⇒ ∠APT + 120° = 180°
⇒ ∠APT = 180° - 120°
⇒ ∠APT = 60°.

Hence, the value of ∠APT = 60°.

(ii) From figure,

BAP = BPT' = 30. (∵ angles in alternate segment are equal.)

Arc BP subtends ∠BOP at the centre and ∠BAP at the remaining part of the circle.

∴ ∠BOP = 2∠BAP (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle.)

∠BOP = 2 × 30° = 60°.

Hence, the value of ∠BOP = 60°.

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