In triangle ABC, angle C is 90 degrees. CH height. AB = 34. tqА = 4. Find AH.

Tg A = BC / AC = 4, that is, the BC leg is four times larger than the AC leg. Hence, the Pythagorean theorem for a given triangle can be written as follows

AC² + (4AC) ² = AB².

Let’s substitute its value instead of AB and solve the resulting equation:

AC² + (4AC) ² = 34²,

AC² + 16AC² = 1156,

17AC² = 1156,

AC² = 1156/17,

AC² = 68,

AC = √68,

AC = 2√17.

Find BC

BC = 4 * 2√17 = 8√17.

Find the area of ​​triangle ABC:

S = ½ * AC * BC = ½ * 2√17 * 8√17 = 8 * 17 = 136.

The area of ​​the same triangle can be found with a different formula:

S = ½ * AB * CH.

From here

CH = 2S / AB.

Find the height CH:

CH = 2 * 136/34 = 8.

Since the angle A in triangle ACH remained the same as in triangle ABC, its tangent did not change either. This means that the AH leg is four times smaller than the CH leg. Let’s find its value:

AH = 8/4 = 2.

Answer: AH is 2.