Answer:
[tex]m\angle A=m\angle D=60^{\circ},\ m\angle B=m\angle C=120^{\circ}[/tex]
Step-by-step explanation:
ABCD is isosceles trapezoid, because AB=CD.
1. The midsegment MN=30, the smaller base BC=17, thenyou can find the greater base:
[tex]MN=\dfrac{BC+AD}{2},\\ \\30=\dfrac{17+AD}{2},\\ \\17+AD=60,\\ \\AD=43.[/tex]
2. Draw two trapezoid's heights BH and CK. Triangles ABH and CDK are congruent and
[tex]AH=DK=\dfrac{AD-BC}{2}=\dfrac{43-17}{2}=13.[/tex]
3. In right triangle ABH, the hypotenuse AB=26 and the leg AH=13, then angle opposite to the leg AH is 30°. This means that
[tex]m\angle ABH=30^{\circ}.[/tex]
Thus,
[tex]m\angle ABC=30^{\circ}+90^{\circ}=120^{\circ}=m\angle BCD.[/tex]
[tex]m\angle DAB=180^{\circ}-120^{\circ}=60^{\circ}=m\angle CDA.[/tex]
The measure of angle are [tex]\angle A=60^0,\angle D =60^0and \angle B=120^0,\angle,C=120^0[/tex]
We have given that the ABCD is isosceles trapezoid, because AB=CD.
Also the mid segment MN=30, the smaller base BC=17
we have to find [tex]m\angle A[/tex][tex],m\angle B,m\angle C ,and m\angle D[/tex]
Suppose that a and b are two points we have to find the midpoint [tex]midpoint=\frac{a+b}{2}[/tex][tex]
MN=\frac{BC+AD}{2}[/tex][tex]30=\frac{17+AD{2}\\
17+AD=60\\
AD=43[/tex]
Now next we have ,In right triangle ABH,
The hypotenuse AB=26 and The length AH=13,
Then angle opposite to the length AH is 30°.
Therefore we get,[tex]m\angle ABH=30^0[/tex]
Therefore we get,[tex]m\angle ABC=30^0+90^0=120^0=m\angle BCD\\\\
m\angle DAB=180^0-120^0=60^0=m\angle CDA[/tex]
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