Problem 19. In triangle $ABC$, the difference between the internal angles $A-B=60^{\circ}$. It is known that the height $CH$ from $C$ to side $AB$ is equal to $CH=BC-AC$. Find the angles of the triangle.
Solution. Point $H$ can coincide with point $A$, lie inside segment $A B$, or lie on ray $B A$ (outside segment $B A$).

1) If point $H$ coincides with point $A$, then $A=90^{\circ}$, $B=30^{\circ}, C=60^{\circ}$.
2) If $A>90^{\circ}$, then as angle $A$ increases, angle $B$ also increases. In this case, $B C$ decreases, and $C A$ increases. Therefore, for a given height $C H$, the equation $C H=B C-A C$ has no more than one solution. It is easy to show that this equation (if $A>90^{\circ}$) has no solutions.
3) If point $H$ lies inside segment $A B$, then $A<90^{\circ}$ and $B<30^{\circ}$. In this case, the equation $\mathrm{CH}=$ $=B C-A C$ takes the form:

$$
\frac{1}{\sin B}-\frac{1}{\sin \left(B+60^{\circ}\right)}=1
$$

We transform it as follows:

$$
\begin{gathered}
\frac{\sin \left(B+60^{\circ}\right)-\sin B}{\sin B \sin \left(B+60^{\circ}\right)}=1 \\
\frac{\cos \left(B+30^{\circ}\right)}{\sin B \sin \left(B+60^{\circ}\right)}=1
\end{gathered}
$$

If $0<B<30^{\circ}$, then the function $y=\cos \left(B+60^{\circ}\right)$ is decreasing, and the function $y=\sin B \sin \left(B+60^{\circ}\right)$ is increasing. Therefore, equation (1) has no more than one solution. We prove that equation (1) has no solutions (if $0<B<30^{\circ}$).

Answer. $A=90^{\circ}, B=30^{\circ}, C=60^{\circ}$.

## Assignment 6. RATIONAL EQUATIONS AND INEQUALITIES