In a trapezoid, three sides are 5 cm, and the larger base is 10 cm. Find a larger angle.

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

НC = BC.

Since the trapezoid is isosceles, then:

AH = KD = (AD – BC) / 2;

AH = KD = (10 – 5) / 2 = 5/2 = 2.5 cm.

The angle АВС is equal to the sum of the angles ∠АВН and ∠НВС, which is straight, since it is formed by the height of the ВН:

∠ABС = ∠ABН + ∠НВС = ∠АВН + 90º.

Therefore, to calculate the angle ABC, it is necessary to calculate the value of the angle ABH. To do this, consider the triangle ΔАВН.

Let’s apply the theorem of sines, since we know the length of the opposite leg and the hypotenuse of the given triangle:

sin ABH = AH / AB;

sin ABH = 2.5 / 5 = 1/2.

Since 1/2 corresponds to a sine of 30º, the degree measure of the angle ∠ABH is equal to 30º.

∠ABC = 30º + 90º = 120º

Answer: The degree measure of the larger trapezoid angle is 120º.