In triangle ABC, angle A = α, angle C = β, side BC = 7 cm, BH – height. Find AH.

In △ ABC, side AB lies opposite ∠C, side BC is opposite ∠A.
1. From △ ABC By the sine theorem:
BC / sin∠A = AB / sin∠C;
7 / sinα = AB / sinβ;
AB = 7sinβ / sinα (proportional).
2. From △ BHC:
BC / sin∠BHC = BH / sin∠C;
7 / sin90 ° = BH / sinβ;
7/1 = BH / sinβ;
BH = 7sinβ.
3. From △ AHC by the Pythagorean theorem:
AH = √ (AB² – BH²);
AH = √ ((7sinβ / sinα) ² – (7sinβ) ²) = √ (49sin²β / sin²α – 49sin²β) = (bring the numbers to a common denominator sin²α, multiplying 49sin²β by sin²α) = √ ((49sin²β – 49sin²β * sin²α) / sin²α) = (put 49sin²β in the numerator of the fraction outside the brackets) = √ (49sin²β (1 –
sin²α) / sin²α) = (1 – sin²α = cos²α – basic trigonometric identity) = √ (49sin²β * cos²α / sin²α) = (cos²α / sin²α = ctg²α) = √ (49sin²β * ctg²α) = (extract the root) = 7sinβ * c …
Answer: AH = 7sinβ * ctgα.