10. Let $0<\theta<\pi$, then the maximum value of $\sin \frac{\theta}{2}(1+\cos \theta)$ is $\qquad$ .
$$
\begin{array}{l}
\sin \frac{\theta}{2}(1+\cos \theta)=2 \sin \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}=\sqrt{2} \cdot \sqrt{2 \sin ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2}} \\
\leqslant \sqrt{2} \cdot \sqrt{\left(\frac{2}{3}\right)^{3}}=\frac{4 \sqrt{3}}{9} \text {, equality holds when } 2 \sin ^{2} \frac{\theta}{2}=\cos ^{2} \frac{\theta}{2} \\
\Rightarrow \sin \frac{\theta}{2}=\frac{\sqrt{3}}{3}, \cos \theta=\frac{1}{3} \text {. Therefore, the maximum value of } \sin \frac{\theta}{2}(1+\cos \theta) \text { is } \frac{4 \sqrt{3}}{9} .
\end{array}
$$