Asked by Rin

The sum of two acute angles in a right triangle is 90 degrees. Find the acute angles on the right triangel wherein twice the first is 40 degrees more than thrice the second
Answered by Bosnian
A = first angle

B = second angle

In the right triangel summ of angles = 180 °

If the sum of two acute angles in a right triangle is 90 degrees this mean :

A + B = 180 ° - 90 ° = 90 °

A + B = 90 °


Twice the first is 40 degrees more than thrice the second.

This mean :

2 A = 40 ° + 3 B Divide both sides by 2

2 A / 2 = 40 ° / 2 + 3 B / 2

A = 20 ° + ( 3 / 2 ) B


A + B = 90 °

20 ° + ( 3 / 2 ) B + B = 90 ° Subtract 20 ° to both sides

20 ° + ( 3 / 2 ) B + B - 20 ° = 90 ° - 20 °

( 3 / 2 ) B + B = 70 °

( 3 / 2 ) B + ( 2 / 2 ) B = 70 °

( 5 / 2 ) B = 70 ° Multiply both sides by 2

5 B = 2 * 70 °

5 B = 140 ° Divide both sides by 5

B = 140 ° / 5

B = 28 °



A = 20 ° + ( 3 / 2 ) B


A = 20 ° + ( 3 / 2 ) * 28 °

A = 20 ° + 3 * 28 ° / 2

A = 20 ° + 84 ° / 2

A = 20 ° + 42 °

A = 62 °


Proof A + B = 62 ° + 28 ° = 90 °










Answered by Julian
The answer is A + B = 62° + 28°

[=90°✓]
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