5. Zadatak

$$\begin{gathered}|AC|=|BC|=b=7.5\text{ cm}\\ |\angle BAC|=|\angle ABC|=75^\circ\\ \Longrightarrow\quad|\angle ACB|=30^\circ\end{gathered}$$

Odaberimo točku $D$ na $\overline{BC}$ tako da je $\overline{AD}\perp\overline{BC}$. $\overline{AD}$ je visina trokuta $ABC$ i kateta pravokutnog trokuta $\triangle ADC$ nasuprot kutu od $30^\circ$ te je:

$$|AD|=\frac12|AC|=\frac12\cdot7.5=3.75\text{ cm}.$$ $$P_{\triangle ABC}=\frac12|BC|\cdot|AD|=\frac12\cdot7.5\cdot3.75=14.0625\text{ cm}^2.$$