A tree is $x$ meters high. The angle of elevation of its top from a point $P$ on the ground is 23 degrees. From another point $Q$, 10 meters from $P$ and in line with $P$ and the foot of the tree, the angle of elevation is 32 degrees. Find $x$.
[OP points out in the comments: if $QR=y$, then $\tan32=x/y$, so $y=x/\tan32$, and $\tan23=x/(10+y)$.]
Let the length $PR$ be $y$. Then we have the following two equations:
In $\triangle PRS$, we have $\tan(23^\circ)=\frac{x}{y}$ $(1)$ and $\tan(32^\circ)=\frac{x}{y-10}$ $(2)$.
Solving for $y$ in equation $2$, we get $y=\frac{x}{\tan(32^\circ)}+10$. Plugging this into equation $1$ and solving gives $x\approx13.236$.
It is easy to get \begin{equation} 10 = \frac{x}{\tan 23^{o}}-\frac{x}{\tan 32^{o}} \end{equation} and then you can obtain that \begin{equation} x = \frac{10}{\frac{1}{\tan 23^{o}} - \frac{1}{\tan 32^{o}}} \end{equation}
