In the given figure, O is the centre of the circle. If ∠ABD = 30° and ∠BAC = 80° , find ∠ACB.

[2 Marks]

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Solution

Angle subtended by a diameter( in this case DOB) to any point on the periphery of the circle is always 90°.

Therefore ∠BAD = 90° [0.5 Marks]

In △BAD,

∠DBA + ∠BAD + ∠ADB = 180° [0.5 Marks]

⇒ 30° + 90° + ∠ADB = 180°

⇒ ∠ADB = 60° [0.5 Marks]

The angles subtended by the arc in the same segment are equal.

⇒ ∠ADB = ∠ACB = 60°. [0.5 Marks]

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Similar questions

In the given figure, O is the centre of the circle. If $∠$ ABD=35 $^{\circ}$ and $∠$ BAC=70 $^{\circ}$ , find $∠$ ACB.

In the given figure, the centre O of the small circle lies on the circumference of the bigger circle. If $∠$ APB = $75^{o}$ and $∠$ BCD = $40^{o}$ , find :

(i) $∠$ AOB,

(ii) $∠$ ACB,

(iii) $∠$ ABD,

(iv) $∠$ ADB.

Q. In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB = ?
(a) 30°
(b) 15°
(c) 60°
(d) 90°
Figure

In the given figure, O is the centre of a circle. If $∠ O A B = 40^{\circ}$ and C is a point on the circle then $∠ A C B = ?$

(a) $40^{\circ}$

(b) $50^{\circ}$

(d) $100^{\circ}$

In the given figure, O is the centre of the circle. If $∠ A C B = 50^{\circ}$ , find $∠$ OAB.

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In the given figure, O is the centre of the circle. If ∠ABD = 30° and ∠BAC = 80° , find ∠ACB. [2 Marks]

In the given figure, O is the centre of the circle. If ∠ABD = 30° and ∠BAC = 80° , find ∠ACB.

[2 Marks]