In triangle ABC: Angle C = 90 *, height CH is 3, angle ctg B = 0.75 Find AH.

In triangle ABC it is known:

Angle C = 90 °;

Height CH = 3;

ctg B = 0.75.

Find AH.

1) Consider a triangle CHB with a right angle H, since the height of CH is perpendicular to the hypotenuse AB of triangle ABC.

ctg B = BH / CH;

BH = CH * ctg B;

Substitute the known values and find the HV value.

BH = 3 * 0.75 = 3 * 3/4 = 9/4;

Hence, BH = 9/4.

2) CH ^ 2 = AH * BH;

From here we express BH.

BH = CH ^ 2 / AH;

Substitute the known values.

BH = 3 ^ 2 / (9/4) = 9 / (9/4) = 9 * 4/9 = 1 * 4/1 = 4;

As a result, we got that the side BH of the triangle ABC is equal to BH = 4.