In an isosceles trapezoid, the larger base is 10. The lateral side is 4 and forms an angle of 60 degrees with the larger base.

In an isosceles trapezoid, the larger base is 10. The lateral side is 4 and forms an angle of 60 degrees with the larger base. find the tangent of the angle between the diagonal and the base of the trapezoid.

You are given an isosceles trapezoid ABCD, in which AD = 10, AB = CD = 4, angle BAD = 60 °. We need to find the tangent of the angle BDA.

Let’s draw the height from vertex B to AD and denote it BH.

In the resulting right-angled triangle BHA, the angle ABH = 90 ° – 60 ° = 30 °. Side AH lies opposite an angle of 30 ° and is equal to half the length of the hypotenuse:

AH = AB * 1/2 = 2.

In triangle ABH:

BH = AB * sinBAH = 4 * √3 / 2 = 2 * √3.

HD = AD – AH = 10 – 2 = 8.

In a right triangle BHD:

tgBDH = BH / HD = √3 / 4.

Thus, the tangent of the angle between the diagonal and the large base of the trapezoid is √3 / 4.

Answer: √3 / 4.