In the given ...

Question

In the given figure, O is the centre of a circle, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

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Solution

(c) 60°

OA = OB
⇒∠OAB = ∠OBA = 45°
$∠ ACB = \frac{1}{2} ∠ AOB = (\frac{1}{2} \times 90 °) = 45 °$
In Δ ABC, we have:
∠ACB + ∠CAB + ∠ABC = 180°
⇒ 45° + ∠CAB + 30° = 180°
⇒ ∠CAB = (180° - 75°) = 105°
Now, ∠CAO + ∠OAB = 105°
⇒ ∠CAO + 45° = 105°
⇒ ∠CAO = (105° - 45°) = 60°

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In the given figure, A, B and C are three points on the circle with centre O such that $∠ B O C = 30^{\circ}$ and $∠ A O B = 60^{\circ}$ . If D is a point on the circle that is not on the arc ABC, then find $∠ A D C .$

In the given figure, O is the centre of a circle, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ? (a) 30° (b) 45° (c) 60° (d) 90°

(c) 60° OA = OB ⇒∠OAB = ∠OBA = 45° ∠ACB=12∠AOB=12×90°=45° In Δ ABC, we have: ∠ACB + ∠CAB + ∠ABC = 180° ⇒ 45° + ∠CAB + 30° = 180° ⇒ ∠CAB = (180° - 75°) = 105° Now, ∠CAO + ∠OAB = 105° ⇒ ∠CAO + 45° = 105° ⇒ ∠CAO = (105° - 45°) = 60°

In the given figure, O is the centre of a circle, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°