In triangle ABC, angle C is 90 °, height CH is 20, BC = 25. Find sinA.

In triangle ABC it is known:

Angle C is 90 °;
Height CH = 20;
BC = 25.
Find sin A.

1) sin a = BC / AB;

2) Consider the СНВ triangle, where the angle H = 90 °.

If the BC is known the hypotenuse and the СН of the leg, then we find the ВН.

BH = √ (CB ^ 2 – CH ^ 2) = √ (25 ^ 2 – 20 ^ 2) = √ ((25 – 20) * (25 + 20)) = √ (5 * 45) = √ (25 * 9) = 5 * 3 = 15;

3) Find AH from the formula CH ^ 2 = AH * BH.

AH = CH ^ 2 / BH;

AH = 20 ^ 2/15 = 20 * 20/15 = 4 * 20/3 = 80/3;

4) Find AB from the triangle ABC.

AB = AH + BH = 80/3 + 15 = (80 + 45) / 3 = 125/3;

5) Find sin a from triangle ABC.

sin a = BC / AB;

sin a = 25 / (125/3) = 25 * 3/125 = 25 * 3 / (25 * 5) = 3/5 = 0.6.

Answer: sin a = 0.6.