just write it algebraically:
b = 3a
b = c-110
a+b+c=180
a + 3a + (110+3a) = 180
7a = 70
a=10
b=30
c=140
b = 3a
b = c-110
a+b+c=180
a + 3a + (110+3a) = 180
7a = 70
a=10
b=30
c=140
According to the problem, we're given three conditions:
1. "The measure of one angle is triple the measure of a second angle": This can be represented as x = 3y.
2. "The measure of one angle is 110 degrees less than the measure of the third angle": This can be represented as x = z - 110.
3. "The sum of the measures of three interior angles of a triangle is always 180 degrees": This can be represented as x + y + z = 180.
Now we can solve these equations to find the values of x, y, and z.
From condition 1, we can substitute the value of x in terms of y into equation 3:
3y + y + z = 180
4y + z = 180 ----(equation 1)
From condition 2, we can substitute the value of x in terms of z into equation 1:
z - 110 + y + z = 180
2z + y = 290 ----(equation 2)
Now we have a system of equations with two unknowns (y and z). We can solve these equations simultaneously to find the values of y and z.
Let's solve equations 1 and 2 using the method of substitution:
From equation 1:
4y + z = 180
z = 180 - 4y
Substitute this value of z into equation 2:
2(180 - 4y) + y = 290
360 - 8y + y = 290
-7y = -70
y = 10
Now that we have the value of y, substitute it back into equation 1 to find z:
4(10) + z = 180
40 + z = 180
z = 140
Finally, substitute the values of y and z back into equation 2 to find x:
2(140) + 10 = 290
280 + 10 = 290
x = 30
So, the measures of the angles in the triangle are:
Angle x = 30 degrees
Angle y = 10 degrees
Angle z = 140 degrees