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°

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Solution

(c) 80°

We have :
$∠ A O C + ∠ B O C = 180 ° [Since AOB is a straight line] \Rightarrow 4 x + 5 x = 180 ° \Rightarrow 9 x = 180 ° \Rightarrow x = 20 ° ∴ ∠ A O C = 4 \times 20 ° = 80 °$

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Similar questions

In the given figure, AOB is a straight line. If $∠ A O C = 4 x^{\circ}$ and $∠ B O C = 5 x^{\circ}$ then $∠ A O C = ?$

(a) $40^{\circ}$

(b) $60^{\circ}$

(d) $100^{\circ}$

Q. In the given figure, AOB is a straight line. If ∠AOC = (3x − 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 50°

In the given figure, $A O B$ is a straight line. If $∠ A O C = (3 x + 10)^{\circ}$ and $∠ B O C = (4 x - 26)^{\circ}$ , then $∠ B O C = ?$

(a) $96^{\circ}$

(b) $86^{\circ}$

(d) $106^{\circ}$ .

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 $∠ A O C = ?$

(a) $40^{\circ}$

(b) $60^{\circ}$

(d) $50^{\circ}$

Q. In the given figure, AOB is a straight line. If ∠AOC = (3x + 10)° and ∠BOC (4x − 26)°, then ∠BOC = ?
(a) 96°
(b) 86°
(c) 76°
(d) 106°

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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°

(c) 80° We have : ∠AOC+∠BOC=180° Since AOB is a straight line⇒4x+5x=180°⇒9x=180°⇒x=20°∴∠AOC=4×20°=80°

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°

(c) 80°

We have :
$∠ A O C + ∠ B O C = 180 ° [Since AOB is a straight line] \Rightarrow 4 x + 5 x = 180 ° \Rightarrow 9 x = 180 ° \Rightarrow x = 20 ° ∴ ∠ A O C = 4 \times 20 ° = 80 °$