Trigonometric problem: Elevation angle The elevation of the top of a tower $KT$ from a point $A$ is $27^\circ$. At another point $B$, $50$ meters nearer to the foot of the tower where $ABK$ is a straight line, the angle of elevation is $40^\circ$. Find the height of the tower $KT$.

Let $AK=d$, $BK=d-50$, and $KT=h$. Then
$$
\tan 27^\circ=\frac{h}{d}\qquad\rightarrow\qquad d=\frac{h}{\tan 27^\circ}\tag1
$$
and
$$
\tan 40^\circ=\frac{h}{d-50}\qquad\rightarrow\qquad d-50=\frac{h}{\tan 40^\circ}\tag2
$$
Substitute $(1)$ to $(2)$, yield
\begin{align}
\frac{h}{\tan 27^\circ}-50&=\frac{h}{\tan 40^\circ}\\
\frac{h}{\tan 27^\circ}-\frac{h}{\tan 40^\circ}&=50\\
h\left(\frac{1}{\tan 27^\circ}-\frac{1}{\tan 40^\circ}\right)&=50\\
h\left(\frac{\tan 40^\circ-\tan 27^\circ}{\tan 27^\circ\tan 40^\circ}\right)&=50\\
h&=50\left(\frac{\tan 27^\circ\tan 40^\circ}{\tan 40^\circ-\tan 27^\circ}\right)
\end{align}
