Question

9. The larger of the two
supplementary angles exceeds
thrice the smaller by 20
degrees. The two angles in
degrees are
30,90
O 40, 140
50, 100
O 10, 110
​

Answer 1

$\huge \underline \mathbb \purple {ANSWER :-}$

According to the question :-

Let one angle be x

The other angle be 3x+20

$x + 3x + 20 = 180 \\ \\ 4x + 20 = 180 \\ \\ 4x = 180 - 20 \\ \\ 4x = 160 \\ \\ x = \frac{160}{4} \\ \\ x = 40 \\ \\ \\ 3x + 20 = 3 \times 40 + 20 \\ = 140$

Therefore the angles are = 40°,140 °

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Answer 2

Answer:

The angles are 140 ° and 40 °.

Solution:

Let the smaller angle be xº

Given

Larger angle = Exceeds thrice the smaller by 20 °

= 3 (x) + 20

= (3x + 20 °)

= (3x + 20) + x

= 180 °

(Because given angles are supplementary)

3x + 20 + x = 180°

4x + 20 = 180°

4x = 180° - 20°

4x = 160

x = 160/4

x = 40

Smaller angle = x = 40 °

Larger angle = (3x + 20)

= 3 ( 40) + 20)

= 120 + 20

= 140°