In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ?
(a) 95°
(b) 85°
(c) 90°
(d) 55°

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Solution

(b) 85°

We have :
$∠ A O C + ∠ B O D + ∠ C O D = 180 ° [Since AOB is a straight line] \Rightarrow (∠ A O C + ∠ B O D) + ∠ C O D = 180 ° \Rightarrow 95 ° + ∠ C O D = 180 ° \Rightarrow ∠ C O D = 85 °$

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Q. In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then ∠COD = ?
(a) 80°
(b) 100°
(c) 120°
(d) 140°

Q. In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then x = ?
(a) 20
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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, AOB is a straight line. If $∠ A O C = (3 x - 10)^{\circ}$ , \angle COD = 50^{\circ}\) and $∠ B O D = (x + 20)^{\circ}$ then $∠ A O C = ?$

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In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ? (a) 95° (b) 85° (c) 90° (d) 55°

(b) 85° We have : ∠AOC+∠BOD+∠COD=180° Since AOBis a straight line⇒∠AOC+∠BOD+∠COD=180°⇒95°+∠COD=180°⇒∠COD=85°

In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ?
(a) 95°
(b) 85°
(c) 90°
(d) 55°

(b) 85°

We have :
$∠ A O C + ∠ B O D + ∠ C O D = 180 ° [Since AOB is a straight line] \Rightarrow (∠ A O C + ∠ B O D) + ∠ C O D = 180 ° \Rightarrow 95 ° + ∠ C O D = 180 ° \Rightarrow ∠ C O D = 85 °$