In triangle ABC, angle C is 90 °, CH = height, AB = 34, tg = 3/5. Find BH.

Let’s apply the formula Sin2BAC + Cos2BAC = 1 and divide both parts by Cos2BAC.

(Sin2BAC / Cos2BAC) + (Cos2BAC / Cos2BAC) = 1 / Cos2BAC.

tg2BAC + 1 = 1 / Cos2BAC.

1 / Cos2BAC = (3/5) 2 + 1 = 9/25 + 25/25 = 34/25.

Cos2BAC = 25/34.

CosBAC = 5 / √34.

Then CosBAC = √34 / 5 = AC / AB.

AC = 34 * 5 / √34 = 5 * √34 cm.

By the Pythagorean theorem BC ^ 2 = AB ^ 2 – AC ^ 2 = 1156 – 850 = 306.

BC = 3 * √34 cm.

Triangles ACB and СВН are similar in acute angle, then:

AB / BC = BC /ВН.

BH = BC ^ 2 / AB = 306/34 = 9 cm.

Answer: The length of the ВН segment is 9 cm.