Angle A is three times the measure of angle B. The sum of their measures is 128 degrees. Angle B is (x + 2) degrees. Find the value of x.
if <A is degrees, how big is <B
A=3B
B=(x+2)
A+B= 128
Now substitute
3B+B=128
3(x+2)+(x+2)=128
3x+6+x+2=128
4x+8=128
4x=120
x=30
Let's set up the equations based on the given information:
1. Angle A is three times the measure of angle B:
A = 3B
2. The sum of their measures is 128 degrees:
A + B = 128
3. Angle B is (x + 2) degrees:
B = x + 2
Now we can solve for x by substituting the values into equation 2 and 3:
A + B = 128
3B + B = 128 [since A = 3B]
4B = 128
B = 32
Substituting the value of B back into equation 3, we have:
B = x + 2
32 = x + 2
x = 32 - 2
x = 30
Therefore, the value of x is 30.
To solve this problem, we need to set up an equation based on the given information and solve for x.
Let's denote the measure of angle B as B degrees. We are given that angle A is three times the measure of angle B, so angle A can be written as 3B degrees.
The sum of the measures of angle A and angle B is given as 128 degrees, so we can set up the equation:
A + B = 128
Substituting the values for angle A and angle B:
3B + B = 128
Combining the like terms:
4B = 128
To find the value of x, we need to first find the measure of angle B.
Dividing both sides of the equation by 4:
B = 128 / 4
B = 32
Since we are given that angle B is (x + 2) degrees, we can now solve for x:
32 = x + 2
Subtracting 2 from both sides of the equation:
30 = x
Therefore, the value of x is 30.