In the given figure, AOB is a straight line and OC is ray such that∠AOC = (3x + 20)° and ∠BOC = (2x − 10)°. Find the value of x and hence find (i) ∠AOC and ∠BOC.

Open in App

Solution

$\begin{array}{l} Here, 3 x + 20 + 2 x - 10 = 180 \\ = > 5 x + 10 = 180 \\ =>5 x = 170 \\ = > x = 34 \\ ∠ AOC = (3 \times 34 + 20) ° \\ = (102 + 20) ° \\ =122 ° \\ ∠ BOC =(2 \times 34 - 10) ° \\ =(68 - 10) ° \\ = 58 ° \end{array}$

Suggest Corrections

Similar questions

Q. In the given figure, AOB is a straight line and the ray OC stands on it.
If ∠AOC = (2x − 10)° and ∠BOC = (3x + 20)°, find the value of x.
Also, find ∠AOC and ∠BOC

In the given figure, $A O B$ is a straight line and the ray $O C$ stands on it.
If $∠ A O C = 64^{\circ}$ and $∠ B O C = x^{\circ},$ find the value of $x .$

Q. In the given figure AOB is a straight line.

∠ A O C = 3 x, ∠ B O C = 2 x

. The value of x is

Q. A ray

O C

stands on a line

A O B

. Find

x

∠ A O C = x + 20^{o}

and

∠ B O C = 3 x - 8^{o}

O C

and

O D

are two rays from point

O

on straight line

A O B

∠ A O C = 2 x, ∠ C O D = 3 x, ∠ B O C = 5 x

.
Find each angle :

∠ A O C, ∠ C O D, ∠ D O B

Join BYJU'S Learning Program

In the given figure, AOB is a straight line and OC is ray such that∠AOC = (3x + 20)° and ∠BOC = (2x − 10)°. Find the value of x and hence find (i) ∠AOC and ∠BOC.

Here, 3x+20+2x−10=180 =>5x+10=180 =>5x=170 =>x=34 ∠AOC=(3×34+20)° =102+20° =122° ∠BOC =(2×34−10)° =(68−10)° =58°

$\begin{array}{l} Here, 3 x + 20 + 2 x - 10 = 180 \\ = > 5 x + 10 = 180 \\ =>5 x = 170 \\ = > x = 34 \\ ∠ AOC = (3 \times 34 + 20) ° \\ = (102 + 20) ° \\ =122 ° \\ ∠ BOC =(2 \times 34 - 10) ° \\ =(68 - 10) ° \\ = 58 ° \end{array}$