$\cos \alpha =\frac{4}{5}$

${\sin }^2\alpha +{\cos }^2\alpha =1$

${\sin }^2\alpha +{\left(\frac{4}{5}\right)}^2=1$

${\sin }^2\alpha +\frac{16}{25}=1|-\frac{16}{25}$

${\sin }^2\alpha =\frac{9}{25}\stackrel{\alpha \in \left(0^{\circ },180^{\circ }\right)}{\to }\sin \alpha =\frac{3}{5}$

$P=\frac{1}{2}\cdot \left|AC\right|\cdot \left|AB\right|\cdot \sin \alpha =\frac{1}{2}\cdot 4\cdot 3\cdot \frac{3}{5}=\frac{36}{10}=3,6$

Odp.: Pole trójkąta $ABC$ wynosi $3,6$.