In triangle ABC, angle C is 90 °, CH is height, AB = 12, tgA = 3. Find AH.

Let’s designate the segment we need AH – x, then BH will be (12 – x).
Consider a right-angled triangle ACN, and write in it the definition of the tangent of angle A:
Tg A = CH / AH → CH = tg A * AH = 3x.
In the same triangle, according to the Pythagorean theorem, we find AC:
AC² = CH² + AH² = 9x² + x² = 10x².
Consider the BCH triangle and find the BC leg:
BC² = CH² + BH² = 9x² + (12 – x) ² = 9x² + 144 – 24x + x² = 10x² – 24x + 144.
In triangle ABC we write down the Pythagorean theorem:
AC² + BC² = AB²
10x² + 10x² – 24x + 144 = 12²
20x² – 24x = 0
5x² – 6x = 0
x (5x – 6) = 0
5x – 6 = 0
x = 6/5 = 1.2
Answer: the length of AH is 1.2.