In the given figure, arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. Find :
(i) ∠AOB
(ii) ∠BOC
(iii) ∠OBA
(iv) ∠OCB
(v) ∠ABC

Question

In the given figure, arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. Find : (i) ∠AOB (ii) ∠BOC (iii) ∠OBA (iv) ∠OCB (v) ∠ABC

KnowledgeBoat · Accepted Answer

Given: arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. If two arcs of a circle is divided into a certain ratio, the angles subtended by the parts of the arc at the center of the circle will also be in the same ratio. ∠AOB : ∠BOC = 2 : 3 Let ∠AOB be 2x and ∠BOC be 3x. As we know, ∠AOB + ∠BOC = ∠AOC ⇒ ∠AOB + ∠BOC = 150° ⇒ 2x + 3x = 150° ⇒ 5x = 150° ⇒ x = ⇒ x = 30° (i) ∠AOB = 2x = 2 30° = 60° Hence, ∠AOB = 60°. (ii) ∠BOC = 3x = 3 30° = 90° Hence, ∠BOC = 90°. (iii) OA = OB (Radii of same circle), triangle OBA is an isosceles triangle and thus: ∠OBA = ∠OAB The sum of the angles in triangle OBA is 180°. ⇒ ∠OAB + ∠OBA + ∠AOB = 180° ⇒ ∠OBA + ∠OBA + 60° = 180° ⇒ 2∠OBA + 60° = 180° ⇒ 2∠OBA = 180° - 60° ⇒ ∠OBA = ⇒ ∠OBA = 60° Hence, ∠OBA = 60°. (iv) OB = OC (Radii of same circle), triangle OCA is an isosceles triangle and thus: ∠OBC = ∠OCB The sum of the angles in triangle OCB is 180°. ⇒ ∠OCB + ∠OBC + ∠COB = 180° ⇒ ∠OCB + ∠OCB + 90° = 180° ⇒ 2∠OCB + 90° = 180° ⇒ 2∠OCB = 180° - 90° ⇒ 2∠OCB = 90° ⇒ ∠OCB = ⇒ ∠OCB = 45° Hence, ∠OCB = 45°. (v) ∠ABO = 60° and ∠OCB = ∠OBC = 45° ∠ABC = ∠ABO + ∠OBC = 60° + 45° = 105° Hence, ∠ABC = 105°.

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Mathematics

In the given figure, arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. Find :
(i) ∠AOB
(ii) ∠BOC
(iii) ∠OBA
(iv) ∠OCB
(v) ∠ABC

Circles

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Mathematics

In the given figure, arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. Find :(i) ∠AOB(ii) ∠BOC(iii) ∠OBA(iv) ∠OCB(v) ∠ABC

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Two chords AB and CD of lengths 24 cm and 10 cm respectively of a circle are parallel. If the chords lie on the same side of the centre and distance between them is 7 cm, find the length of a diameter of the circle.

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In the given figure, arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. Find :
(i) ∠AOB
(ii) ∠BOC
(iii) ∠OBA
(iv) ∠OCB
(v) ∠ABC

In given figure, AB is a side of a regular pentagon and BC is the side of a regular hexagon. Find :
(i) ∠AOB
(ii) ∠OBC