In an isosceles trapezoid, the larger base is 3.7 in, the lateral side is 1.5 in, and the angle between them is 60

In an isosceles trapezoid, the larger base is 3.7 in, the lateral side is 1.5 in, and the angle between them is 60 degrees. Calculate the midline of the trapezoid.

It is known that the length of the midline of a trapezoid is equal to the half-sum of its bases: EF = (AB + CD) / 2.
We have AB = 3.7 dm. Let’s find the CD.
Let’s draw two lines DG ┴ AB and CH ┴ AB.
It is clear that ΔAGD and ΔBHC are right-angled triangles and CB = AD = 1.5 dm and ∠CBH = ∠DAG = 60 °.
We have AG / AD = cos60 ° = ½, whence AG = ½ * 1.5 dm. Similarly, HB = ½ * 1.5 dm.
It is clear that CD = GH = AB – AG – HB = 3.7 dm – 1.5 dm = 2.2 dm.
So, EF = (AB + CD) / 2 = (3.7 dm + 2.2 dm) / 2 = 2.95 dm.
Answer: 2.95 dm.