In triangle ABC, angle C = 90 CH – height AB = 34 tg A = 4. Find AH.

In the ABC triangle it is known:

Angle C = 90 °;
CH – height;
AB = 34;
tg A = 4.
Find AH.

Solution:

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

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

cos ^ 2 a = 1/17;

cos a = 1 / √17;

2) sin a = √ (1 – cos ^ 2 a) = √ (1 – 1/17) = √ (17/17 – 1/17) = √16 / √17 = 4 / √17;

3) cos a = AC / AB;

AC = AB / cos a = 34 / (4 / √17) = 34 * √17 / 4 = 8.5 * √17;

4) Consider a triangle ACH with a right angle H.

cos A = AH / AC;

AH = AC * cos A;

Substitute the known values and calculate its value.

AH = 8.5 * √17 * 1 / √17 = 8.5 * √17 / √17 = 8.5.

Answer: AH = 8.5.