In the given figure, O is the centre of the circle. If ∠AOB = 100° and ∠AOC = 90°, then find ∠BAC.

80°

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90°

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75°

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85°

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Solution

The correct option is D 85°
$∠ A O B + ∠ A O C + ∠ B O C = 360 °$ (sum of all the angles at a point)
$100 ° + 90 ° + ∠ B O C = 360 °$
$∠ B O C = 170 °$

$∠ B A C = \frac{1}{2} ∠ B O C = \frac{1}{2} 170 ° = 85 °$

(The angle subtended by an arc of a circle at its centre is twice the angle it subtends anywhere on the circle’s circumference.)

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Similar questions

Q. In the given figure, O is the centre of the circle. If

∠ A O B = 100^{\circ}

and

∠ A O C = 90^{\circ}

, then find

∠ B A C

In the given figure, $O$ is the centre of a circle. If $∠ A O B = 100^{\circ}$ and $∠ A O C = 90^{\circ}$ , then $∠ B A C = ?$

(a) $85^{\circ}$

(b) $80^{\circ}$

(d) $75^{\circ}$

In the given figure, O is the centre of a circle. If $∠ A O B = 100^{\circ} a n d ∠ A O C = 90^{\circ} t h e n ∠ B A C = ?$

(a) $85^{\circ}$

(b) $80^{\circ}$

(d) $75^{\circ}$

Q. In the figure, O is the centre of the circle with

∠ A O B = 85^{\circ} a n d ∠ A O C = 115^{\circ}, t h e n ∠ B A C

is :

(i) In figure (1), $O$ is the centre of the circle. If $∠ O A B = 40^{\circ}$ and $∠ O C B = 30^{\circ}$ . Find $∠ A O C$ .

(ii) In figure (2), $A, B$ and $C$ are three points on the circle with centre $O$ such that $∠ A O B = 90^{\circ}$ and $∠ A O C = 110^{\circ}$ . Find $∠ B A C$ .

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In the given figure, O is the centre of the circle. If ∠AOB = 100° and ∠AOC = 90°, then find ∠BAC.

The correct option is D 85°∠AOB+∠AOC+∠BOC=360° (sum of all the angles at a point) 100°+90°+∠BOC=360° ∠BOC=170° ∠BAC=12∠BOC=12170°=85° (The angle subtended by an arc of a circle at its centre is twice the angle it subtends anywhere on the circle’s circumference.)