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 °
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
2 answers
The answer is A + B = 62° + 28°
[=90°✓]
[=90°✓]