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In the given figure, arc AB = twice arc BC and ∠AOB = 80°. Find:

(i) ∠BOC

(ii) ∠OAC

In the given figure, arc AB = twice arc BC and ∠AOB = 80°. Find: Chord Properties of a Circle, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Circles

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Answer

(i) We know that,

Ratio of the angles subtended by the chords on the center is equal to the ratio of the chords.

⇒∠AOB∠BOC=ABBC⇒80°∠BOC=21⇒∠BOC=80°×12⇒∠BOC=40°.\Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{80°}{∠BOC} = \dfrac{2}{1} \\[1em] \Rightarrow ∠BOC = 80° \times \dfrac{1}{2} \\[1em] \Rightarrow ∠BOC = 40°.

Hence, ∠BOC = 40°.

(ii) Join AC.

From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 80° + 40° = 120°.

In △ OAC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OCA = ∠OAC = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 120° = 180°

⇒ 2x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = 60°2\dfrac{60°}{2}

⇒ x = 30°

⇒ ∠OAC = 30°.

Hence, ∠OAC = 30°.

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