Answer: The measure of <D is 130°.
Solution:
m<B=x+10
m<C=2x-30
If ABCD is an isosceles trapezoid with AD||BC, then:
(1) m<B=m<C
(2) m<A=m<D
Using equation (1) and replacing m<B by x+10 and m<C by 2x-30
(1) x+10=2x-30
Solving for x: Subtracting x and adding 30 both sides of the equation:
x+10-x+30=2x-30-x+30
Adding similar terms:
40=x
x=40
Replacing x in m<B=x+10:
m<B=40+10
m<B=50°
Replacing x in m<C:
m<C=2x-30
m<C=2(40)-30
m<C=80-30
m<C=50°
Now, we need to find m<D. Using that the sum of the interior angles of any quadrillateral is equal to 360°:
(3) m<A+m<B+m<C+m<D=360°
Replacing: m<A=m<D; m<B=50°; and m<C=50° in equation (3):
(3) m<D+50°+50°+m<D=360°
Solving for m<D: Adding smilar terms:
2 m<D+100°=360°
Subtracting 100° both sides of the equation:
2 m<D+100°-100°=360°-100°
2 m<D=260°
Dividing both sides of the equation by 2:
2 m<D / 2 =260°/2
m<D=130°
