In triangle ABC, angle A is 90 °, angle B is 30 °, AB is 6 cm. Find the other two sides of the triangle.

Given ∆ABS with legs AB and AC, angle A = 90 ° and hypotenuse BC. Length AB is 6:

| AB | = 6;

and the angle B is 30 °. Then:

sin (∠В) = 1/2;

cos (∠В) = √3 / 2;

We will use the fact that the cosine of the angle is equal to the ratio of the lengths of the adjacent leg and the hypotenuse, and the sine of the angle is equal to the ratio of the lengths of the opposite leg and the hypotenuse:

cos (∠В) = | AB | / | ВС |;

sin (∠В) = | AC | / | ВС |;

We get:

cos (∠В) = 6 / | ВС | ⟹ | Sun | = 6 * 2 / √3 = 4 * √3;

sin (∠В) = | AC | / | Sun | ⟹ | AC | = ½ * (4 * √3) = 2 * √3;

Answer: other sides are 2√3 and 4√3