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 $KT$.

Draw a right angled triangle $\;\Delta AKT\;,\;\;\angle K=90^\circ\;,\;\;\angle A=27^\circ$ 
We also have the right angled triangle $\;\Delta BKT\;\;\;\angle KBT=40^\circ\;,\;AB=50$
If we put $\;AK=x\;$, then $\;BJ=x-50\;$ , so by trigonometry:
$$\begin{align*}\text{On}\;\;\Delta AKT:&\;\;\tan 27^\circ=\frac{KT}x\\{}\\\text{On}\;\;\Delta BKT:&\;\;\tan40^\circ=\frac{KT}{x-50}\end{align*}$$
use now both those non-linear equations in $\;x\,,\;KT\;$ to find $\;KT\;$ (and you can find $\;x\;$ as well if you want to)
