Question

12. If the angles of a triangle are in the ratio of 8 : 6:4, find the angles.
13. If the angles of a triangle are in the ratio of 8:8: 20. find the angles.​

Answer 1

Question :-

12. If the angles of a triangle are in the ratio of 8 : 6:4, find the angles.

13. If the angles of a triangle are in the ratio of 8:8: 20. find the angles.

Answer :-

8x + 6x + 4x = 180 [ Sum of all the angles of a triangle = 180 ]

18x = 180

x = 10

8x = 80

6x = 60

4x = 40

__________________________

8x + 8x + 20x = 180 [ Sum of all the angles of a triangle = 180 ]

36x = 180

x = 5

8x = 40

8x = 40

20x = 100

Step-by-step explanation:

Hope you understand ☆

Answer 2

Answer:

12. So , the angles are 80° , 60° and 40° .

13. So , the angles are 40° , 40° and 100° .

Step-by-step explanation:

Given :

12. Ratio of the angles of the triangle = 8:6:4

13. Ratio of the angles of the triangle = 8:8:20

To find :

The angles of the triangle

Taken :

Let the angles be x

To find the the angles use this formula -:

12. => $\boxed{\sf{8x+6x+4x={180}^{\circ}}}$

13. => $\boxed{\sf{8x+8x+20x={180}^{\circ}}}$

Here , 180° is the sum of all the angles of the triangle .

After that multiply the value of x with the ratio .

Solution :

12.

$:\implies\sf{8x+6x+4x={180}^{\circ}}$

$:\implies\sf{18x={180}^{\circ}}$

$:\implies\sf{x=\cancel{\dfrac{{180}^{\circ}}{18}}}$

$:\implies\sf{x={10}^{\circ}}$

First angle -:

$:\implies\sf{{10}^{\circ}\times8}$

$:\implies\sf{{80}^{\circ}}$

Second angle -:

$:\implies\sf{{10}^{\circ}\times6}$

$:\implies\sf{{60}^{\circ}}$

Third angle -:

$:\implies\sf{{10}^{\circ}\times4}$

$:\implies\sf{{40}^{\circ}}$

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13.

$:\implies\sf{8x+8x+20x={180}^{\circ}}$

$:\implies\sf{36x={180}^{\circ}}$

$:\implies\sf{x=\dfrac{{180}^{\circ}}{36}}$

$:\implies\sf{x={5}^{\circ}}$

First angle -:

$:\implies\sf{{5}^{\circ}\times8}$

$:\implies\sf{{40}^{\circ}}$

Second angle -:

$:\implies\sf{{5}^{\circ}\times8}$

$:\implies\sf{{40}^{\circ}}$

Third angle -:

$:\implies\sf{{5}^{\circ}\times20}$

$:\implies\sf{{100}^{\circ}}$