in a triangle the sum of the angle is 180degrees. if angle A is three times angle B and Angle C is 10 degrees more than twice Angle A. Find the measures of the three anges to the newarest degree
A = 3B or B = A/3
C = 2A + 10 or
A + B + C = 180
A + A/3 + (2A+10) = 180
Solve for A, then B and C.
A = 3 B
C = 2 A + 10 °
C = 2 * 3 B + 10 °
C = 6 B + 10 °
A + B + C = 180 °
3 B + B + 6 B + 10 ° = 180 °
10 B + 10 ° = 180 ° Subtract 10 ° to both sides
10 B + 10 ° - 10 ° = 180 ° - 10 °
10 B = 170 ° Divide both sides by 10
B = 17 °
A = 3 * B = 3 * 17 = 51 °
C = 6 B + 10 °
C = 6 * 17 ° + 10 ° = 102 ° + 10 ° = 112 °
Proof :
A + B + C = 51 ° + 17 ° + 112 ° = 180 °
To solve this problem, we can assign variables to the angles and set up equations based on the given information.
Let's say angle B is represented by "x."
Since angle A is three times angle B, angle A can be represented by "3x."
And since angle C is 10 degrees more than twice angle A, angle C can be represented by "2(3x) + 10" or "6x + 10."
We know that the sum of the three angles in a triangle is 180 degrees. So, we can set up the equation:
angle A + angle B + angle C = 180
Substituting the values we found, we get:
3x + x + 6x + 10 = 180
Combining like terms:
10x + 10 = 180
Subtracting 10 from both sides:
10x = 170
Dividing both sides by 10:
x = 17
So, angle B = 17 degrees.
Substituting this value back into the equations, we can find the measures of angle A and angle C:
Angle A = 3x = 3 * 17 = 51 degrees
Angle C = 6x + 10 = 6 * 17 + 10 = 112 degrees
Therefore, the measures of the three angles are approximately: Angle A = 51 degrees, Angle B = 17 degrees, Angle C = 112 degrees.