7. Zadatak

Vrijedi

$$24^2+32^2=576+1024=1600=40^2$$

pa je trokut $\triangle ABC$ pravokutan s pravim kutom u vrhu $A$.

Označimo: $\alpha=|\angle BAD|$, $\beta=|\angle DAE|$, $\gamma=|\angle EAC|$.

$$\alpha+\beta+\gamma=90^\circ.$$

Sada imamo

$$|DE|=40-16-8=16\ \text{cm}.$$ $$\begin{aligned} &\Rightarrow |BE|=8+16=24\ \text{cm} \Rightarrow \triangle ABE\ \text{je jednakokračni}\\ &\Rightarrow |\angle AEB|=\alpha+\beta, \end{aligned}$$ $$\begin{aligned} &\Rightarrow |CD|=16+16=32\ \text{cm} \Rightarrow \triangle ACD\ \text{je jednakokračni}\\ &\Rightarrow |\angle ADC|=\beta+\gamma. \end{aligned}$$

Za kutove $\triangle ADE$ sad vrijedi:

$$\begin{aligned} \beta+(\alpha+\beta)+(\beta+\gamma)&=180^\circ,\\ 90^\circ+2\beta&=180^\circ. \end{aligned}$$

Traženi kut iznosi: $\beta=45^\circ$.