Given a triangle ABC of side AC and BC = 5, AB = 2 √21?

Given a triangle ABC:

It is known:

Sides AC = BC = 5;
AB = 2√21.
Find sin a.

Decision.

1) The height CH, drawn from point C to the base of AB, divides the side AB in half.

Hence, ВН = AC / = 2√21 / 2 = √21;

2) Consider a triangle AСН.

In this triangle, the angle H is 90 °.

The hypotenuse of this triangle is AC = 5 and the leg AH = √21.

If the values of the adjacent leg and hypotenuse are known, then we can find the cosine of the angle between the sides.

cos a = AH / AC = √21 / 5;

3) Find sin a.

sin a = √ (1 – 21/25) = √4 / 25 = 2/5 = 0.4;

Answer: sin a = 0.4.