In triangle ABC, angle C is 90 °, CH is height, BC = 5, sin A = 0.2. Find BH.

Given: triangle ABC (C = 90); CH – height; sinA = 0.2; BC = 5
1. Let us express sinA = BC / AB
Find AB. AB = BC / sinA = 5 / 0.2 = 25
2. By the Pythagorean theorem: AB2 = AC2 + BC2
From here we find the AC.
AC ^ 2 = 25 ^ 2 – 5 ^ 2 = 62 ^ 5 – 2 ^ 5 = 600
AC = 10 * √6
3. Consider a triangle ACH: sinA = CH / AC
Let’s find CH.
CH = AC * sinA = 0.2 * 10 * √6 = 2 * √ 6.
4. Consider the CHB triangle:
by the Pythagorean theorem: CB ^ 2 = BH ^ 2 + CH ^ 2
From here we will find BH.
BH2 = CB2 – CH2 = 25 – (4 * 6) = 25 – 24 = 1
BH = 1.
Answer: BH = 1.