10.387 In triangle $A B C$, each altitude $h_{c}$ and $h_{b}$ is not less than the side to which it is dropped. Find the angles of the triangle.
Solution. Let $\alpha=\angle B A C, c=A B, b=A C$. Then

$S_{A B C}=\frac{b \cdot c \cdot \sin \alpha}{2}=\frac{b \cdot h_{b}}{2}=\frac{c \cdot h_{c}}{2} \Rightarrow b \cdot \sin \alpha=h_{c}, c \cdot \sin \alpha=h_{b} \Rightarrow$

$\Rightarrow \sin \alpha \cdot(b+c)=h_{b}+h_{c}$. By the condition, $h_{b} \geq b, h_{c} \geq c \Rightarrow$

$\Rightarrow \sin \alpha \cdot(b+c) \geq b+c \Rightarrow \sin \alpha=1, \alpha=90^{\circ}, h_{c}=b, h_{b}=c \Rightarrow$

$\Rightarrow c=h_{b} \geq b=h_{c} \geq c \Rightarrow b=c$.

Answer: $90^{\circ}, 45^{\circ}, 45^{\circ}$.