In triangle ABC, angle C is 90 degrees, tgA = 0.2, AB = 13. Find AH.

In triangle ABC it is known:

Angle C = 90 °;
tg A = 0.2;
AB = 13.
Find AH.

Solution:

1) Find tg a.

1 + tg ^ 2 a = 1 / cos ^ 2 a;

1 + (0.2) ^ 2 = 1 / cos ^ 2 a;

1 / cos ^ 2 a = 1 + 0.04;

1 / cos ^ 2 a = 1.04;

cos ^ 2 a = 1 / 1.04;

cos ^ 2 a = 1 / (1 4/100);

cos ^ 2 a = 1 / (1 1/25);

cos ^ 2 a = 1 / (26/25);

cos ^ 2a = 25/26;

cos a = 5 / √26;

2) cos a = AC / AB;

AC = AB * cos a = 13 * 5 / √26 = 65 / √26;

3) Consider a triangle ANS, where the angle H = 90 °;

cos A = AH / AC;

AH = cos A * AC = 65 / √26 * 5 / √26 = 65 * 5/26 = 13 * 5 * 5/26 = 5 * 5/2 = 25/2 = 12.5.

Answer: AH = 12.5.