Из Δ АBC
[m]tg \alpha =\frac{BC}{AC}[/m]
[m]BC=AC\cdot tg α =x\cdot tg α [/m]
По теореме Пифагора
[m]AC^2+BC^2=AB^2[/m]
[m]x^2+(x\cdot tg \alpha)^2=k^2[/m]
[m]x^2=\frac{k^2}{1+tg^2\alpha}[/m]
[m]x=\sqrt{\frac{k^2}{1+tg^2\alpha}}[/m]
[m]x=\frac{k}{\sqrt{1+tg^2\alpha}}[/m]
[m]AC=\frac{k}{\sqrt{1+tg^2\alpha}}[/m]
[m]BC=\frac{k\cdot tg\alpha }{\sqrt{1+tg^2\alpha}}[/m]
Из Δ BCD
[m]сos\alpha =\frac{AD}{AC}[/m] ⇒[m] AD=AC\cdot cos α =x\cdot cos α[/m]
[m] AD=\frac{k\cdot cos\alpha}{\sqrt{1+tg^2\alpha}}[/m]