Mathematics
In the given figure, O is the center of the circle, AB is diameter and CPD is a tangent to the circle at P. If ∠BPD = 30°, find :
(i) ∠APC
(ii) ∠BOP
(iii) ∠OAP
Circles
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Answer
(i) From figure,
∠APB = 90° [∵ angle in semicircle is equal to 90]
From figure,
⇒ ∠APC + ∠APD = 180° [∵ they form linear pair]
⇒ ∠APC + ∠APB + ∠BPD = 180°
⇒ ∠APC + 90° + 30° = 180°
⇒ ∠APC + 120° = 180°
⇒ ∠APC = 180° - 120°
⇒ ∠APC = 60°.
Hence, the value of ∠APC = 60°.
(ii) From figure,
⇒ ∠BAP = ∠BPD = 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°.
(iii) From figure,
⇒ ∠AOP + ∠BOP = 180° [∵ they form linear pair]
⇒ ∠AOP + 60° = 180°
⇒ ∠AOP = 180° - 60° = 120°.
From figure,
⇒ OA = OP [Radius of the same circle]
⇒ ∠OAP = ∠OPA = a(let) (Angles opposite to equal sides are equal)
In △ OAP,
⇒ ∠OAP + ∠OPA + ∠AOP = 180°
⇒ a + a + 120° = 180°
⇒ 2a = 180° - 120°
⇒ 2a = 60°
⇒ a = = 30°
Hence, the value of ∠OAP = 30°.
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