In triangle ABC, angle C is 90 degrees, CH-height, AC = 6 roots out of 10, tg A = 1/3. Find BH.

In triangle ABC it is known:

Angle C is 90 °;
Height CH;
AC = 6√10;
tg a = 1/3.
Find BH.

1) If tg a and AC are known, then we can find BC.

tg a = BC / AC;

Hence BC = AC * tg a;

BC = 6√10 * 1/3 = 6/3 * √10 = 2 * √10 = 2√10;

2) Find the hypotenuse AB.

AB = √ (AC ^ 2 + BC ^ 2);

AB = √ ((6√10) ^ 2 + (2√10) ^ 2) = √ (36 * 10 + 4 * 10) = √ (360 + 40) = √400 = 20;

3) Find the cos b of the triangle ABC.

cos b = BC / AB;

cos b = 2√10 / 20 = √10 / 10;

4) Consider a triangle CHB, where the angle H = 90 °.

If the sun hypotenuse and cos b are known, then:

cos b = BH / BC;

Hence, BH = BC * cos b;

BH = 2√10 * √10 / 10 = 2/10 * √100 = 2/10 * 10 = 2;

Answer: BH = 2.