In $\triangle A B C, A B=10$ and $\sin 3 A+3 \sin C=4$. What is the length of the altitude from $C$ to $A B$ ?
For every real number $x,-1 \leq \sin x \leq 1$, which means that $\sin 3 A$ and $\sin C$ are both at most 1 . Therefore, $3 \sin C$ has a maximum possible value of 3 , so $\sin 3 A+3 \sin C \leq 4$.

The only way that $\sin 3 A+3 \sin C=4$ is if $\sin 3 A=1$ and $\sin C=1$.

Since $C$ corresponds to an angle in a triangle, $0^{\circ}<C<180^{\circ}$.

This and $\sin C=1$ together imply that $C=90^{\circ}$. We conclude that $\triangle A B C$ has a right angle at $C$.

Since $A$ is an angle in a triangle, we have that $0^{\circ}<A<90^{\circ}$. However, we have already shown that this triangle has a right angle not at $A$, which means $0^{\circ}<A<90^{\circ}$.

Multiplying this inequality through by 3 gives $0^{\circ}<3 A<270^{\circ}$.

This and $\sin 3 A=1$ together imply that $3 A=90^{\circ}$, which means $A=30^{\circ}$.

It follows that $\triangle A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with hypotenuse $A B=10$.

The side $B C$ is opposite the $30^{\circ}$ angle, so its length is half that of the hypotenuse, or $B C=5$.

The side $A C$ is opposite the $60^{\circ}$ angle, so its length is $A C=\frac{\sqrt{3}}{2} \times A B=5 \sqrt{3}$.

The area of $\triangle A B C$ is $\frac{1}{2} \times A C \times B C=\frac{1}{2} \times 5 \times 5 \sqrt{3}=\frac{25 \sqrt{3}}{2}$.

Suppose the length of the altitude from $C$ to $A B$ is $h$. Then the area of $\triangle A B C$ is also equal to $\frac{1}{2} \times A B \times h=5 h$.

Therefore, $5 h=\frac{25 \sqrt{3}}{2}$, and so $h=\frac{5 \sqrt{3}}{2}$.