In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then x = ?
(a) 20
(b) 25
(c) 30
(d) 35

Open in App

Solution

(a) 20

We have:
$∠ A O C + ∠ C O D + ∠ B O D = 180 ° [∵ A O B is a straight line] \Rightarrow 40 + 4 x + 3 x = 180 ° \Rightarrow 7 x = 140 ° \Rightarrow x = 20 °$

Suggest Corrections

Similar questions

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°

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 $∠ A O C = ?$

(a) $40^{\circ}$

(b) $60^{\circ}$

(d) $50^{\circ}$

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

Q. In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 100°

Q. In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC, ∠COD and ∠BOD.

Join BYJU'S Learning Program

In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then x = ? (a) 20 (b) 25 (c) 30 (d) 35

(a) 20 We have: ∠AOC+∠COD+∠BOD=180° ∵AOB is a straight line⇒40+4x+3x=180°⇒7x=140°⇒x=20°

In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then x = ?
(a) 20
(b) 25
(c) 30
(d) 35

(a) 20

We have:
$∠ A O C + ∠ C O D + ∠ B O D = 180 ° [∵ A O B is a straight line] \Rightarrow 40 + 4 x + 3 x = 180 ° \Rightarrow 7 x = 140 ° \Rightarrow x = 20 °$