In triangle ABC, angle C is 90 degrees, CH is height, BH = 12, sin A = 2/3. Find AB.

In triangle ABC it is known:

Angle C = 90 °;
CH – height;
BH = 12;
sin A = 2/3.
Find the hypotenuse AB.

Decision:

1) In a right-angled triangle sin a = cos b, so cos b = 2/3;

sin b = √ (1 – cos ^ 2 b) = √ (1 – (2/3) ^ 2) = √ (1 – 4/9) = √ (9/9 – 4/9) = √ (9 – 4) / √9 = √5 / 3;

2) tan b = sin b / cos b = (√5 / 3) / (2/3) = √5 / 3 * 3/2 = √5 / 2;

3) cos b = BH / BC;

BC = BH / cos b = 12 / (2/3) = 12/1 * 3/2 = 12/2 * 3 = 6 * 3 = 18;

4) sin A = BC / AB;

AB = BC / sin a;

Substitute the known values and calculate the hypotenuse.

AB = 18 / (2/3) = 18 * 3/2 = 18/2 * 3 = 9 * 3 = 27;

Answer: AB = 27.