G2.4 In $\triangle A B C, \cos A=\frac{4}{5}$ and $\cos B=\frac{7}{25}$. If $\cos C=d$, find the value of $d$.
$\begin{aligned} \sin A & =\frac{3}{5}, \sin B=\frac{24}{25} \\ \cos C & =\cos \left(180^{\circ}-A-B\right)=-\cos (A+B)=-\cos A \cos B+\sin A \sin B \\ & =-\frac{4}{5} \frac{7}{25}+\frac{3}{5} \frac{24}{25}=\frac{44}{125}\end{aligned}$