In the early femoral trapezoid ABCD with bases BC and AD, the angle B is equal

In the early femoral trapezoid ABCD with bases BC and AD, the angle B is equal to 120 degrees, BC = 10, AD = 18. Find the side of the trapezoid.

Isosceles is a trapezoid in which the sides are equal:

AB = CD.

If we draw two heights from the vertices ∠В and ∠С, then the segments AH and KD will also be equal to each other.

Since the segment of the larger base, located between two heights, is equal to the length of the smaller base, then:

НK = BC;

AH = KD = (AD – НK) / 2;

AH = KD = (18 – 10) / 2 = 8/2 = 4 cm.

Consider the triangle ΔАВН. To calculate the length AB, we apply the theorem of sines. The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse:

sin B = AH / AB.

AB = AH / sin B.

To do this, we find the value of the angle ∠АВН. Since the height of the ВН intersects the bases at a right angle, then:

∠AВН= ∠ABС- 90º;

∠АВН = 120º – 90º = 30º;

sin 30º = 1/2 = 0.5;

AB = 4 / 0.5 = 8 cm.

Answer: the side of the trapezoid is 8 cm.