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
The answer is A + B = 62° + 28°
[=90°✓]
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 °
Let's assume that the first acute angle in the right triangle is represented by "x" degrees, and the second acute angle is represented by "y" degrees.
According to the problem, the sum of two acute angles in a right triangle is 90 degrees:
x + y = 90 ---(equation 1)
It is given that "twice the first angle is 40 degrees more than thrice the second angle":
2x = 3y + 40 ---(equation 2)
Now we have a system of two equations (equation 1 and equation 2) with two variables (x and y).
Let's solve the system of equations to find the values of x and y:
From equation 1, we have y = 90 - x.
Substituting this in equation 2, we get:
2x = 3(90 - x) + 40
Simplifying further:
2x = 270 - 3x + 40
2x + 3x = 310
5x = 310
x = 310/5
x = 62
Substituting the value of x in equation 1, we get:
62 + y = 90
y = 90 - 62
y = 28
Therefore, the first acute angle in the right triangle is 62 degrees, and the second acute angle is 28 degrees.
To solve this problem, let's assume that the two acute angles in the right triangle are x and y degrees.
According to the problem, the sum of the two acute angles is 90 degrees:
x + y = 90 ---(equation 1)
The problem also states that twice the first angle is 40 degrees more than thrice the second angle:
2x = 3y + 40 ---(equation 2)
Now, we have a system of two equations with two variables. We can solve these equations simultaneously to find the values of x and y.
Let's use the substitution method to solve the system of equations:
From equation 1, we can express x in terms of y as:
x = 90 - y
Substituting this value into equation 2, we get:
2(90 - y) = 3y + 40
Simplifying the equation:
180 - 2y = 3y + 40
180 - 40 = 3y + 2y
140 = 5y
y = 28
Now, we can substitute the value of y back into equation 1 to find x:
x + 28 = 90
x = 90 - 28
x = 62
Therefore, the acute angles in the right triangle are x = 62 degrees and y = 28 degrees.