Answer:
m∠A=m∠D=60°, m∠B=m∠C=120°
Step-by-step explanation:
Consider isosceles trapezoid ABCD with midsegment MN. Draw two trapezoid heights BF and CE. Then quadrilateral FBCE is rectangle and EF=BC=17 units.
By the trapezoid midsegment theorem,
[tex]MN=\dfrac{AD+BC}{2},\\ \\30=\dfrac{AD+17}{2},\\ \\60=AD+17,\\ \\AD=60-17=43\ units.[/tex]
Segments FA and ED are equal to [tex]\dfrac{43-17}{2}=13\ units.[/tex]
In right triangle ABF, the hypotenuse AB is twice greater than the leg FA, then m∠FBA=30°.
m∠B=m∠FBA+m∠CBF=30°+90°=120°.
In isosceles trapezoid
m∠B=m∠C=120°,
m∠A=m∠D and m∠A=180°-m∠B=180°-120°=60°=m∠D.
