Given,

Let the angles be $A, B \,and\, C$. $\sin A=\frac{5}{13}$ and es $\sin B=\frac{99}{101}$

we know;

$\cos \theta=\sqrt{1-\sin ^2 \theta}$

Hence,

$\cos A =\sqrt{1-\sin ^2 A}=\sqrt{1-\left(\frac{5}{13}\right)^2}$

$=\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$

simitarly, $\cos B=\sqrt{1-\sin ^2 B}=\frac{20}{101}$.

In a $\Delta$,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \quad \angle C=180^{\circ}-(\angle A+\angle B)$

$\Rightarrow \cos C=\cos (180-(A+B))$

$\Rightarrow \cos C=-\cos (A+B)$

${[\cos (\pi-\theta)=-\cos \theta] }$

$\Rightarrow \cos C=-[\cos A \cos B-\sin A \cdot \sin B]$

$\Rightarrow \cos C=-\left[\frac{12}{13} \times \frac{20}{101}-\frac{5}{13} \times \frac{99}{101}\right]$

$\Rightarrow \cos C=\frac{255}{1313}$