A triangle's sides, perimeter, and area with the usual notations: $a, b, c, 2s, T$. What is the angle opposite to the side $c$ of the triangle if

$$
T+\frac{a b}{2}=s(s-c) ?
$$
Express the area of a triangle using two sides and the included angle, and substitute $s=\frac{a+b+c}{2}$ into the equation. Then (1) becomes:

$$
\frac{a b \sin \gamma}{2}+\frac{a b}{2}=\frac{a+b+c}{2} \cdot \frac{a+b-c}{2}
$$

Simplifying by 2 and rearranging, we get

$$
2 a b(\sin \gamma+1)=a^{2}+b^{2}-c^{2}+2 a b
$$

Since $a, b>0$, we can divide through by $2 a b$. Thus,

$$
\sin \gamma+1=\frac{a^{2}+b^{2}-c^{2}}{2 a b}+1
$$

Notice that the fraction on the right is nothing but $\cos \gamma$ (from the cosine rule), so the equation simplifies to

$$
\sin \gamma=\cos \gamma, \quad \text { where } \quad 0^{\circ}<\gamma<180^{\circ}
$$

The equality holds only when $\gamma=45^{\circ}$, so the angle of the triangle we are looking for is $45^{\circ}$.

Since we performed equivalent transformations throughout, this is indeed a solution to the original equation (which we can also verify by substitution).