Mathematics
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
ICSE
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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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