Three people $A$, $B$ and $C$ can see each other using telescope across a valley. The horizontal distance between $A$ and $B$ is $8 \; km$ and the horizontal distance between $B$ and $C$ is $12 \; km$. The angle of depression of $B$ from $A$ is $20^\circ$ and the angle of elevation of $C$ from $B$ is $30^\circ$. Calculate the vertical height between $A$, $B$ and $B$, $C$. [Given: $\tan 20^\circ = 0.3640$, $\sqrt{3} = 1.732$]
$A'B = 8 \; km =$ Horizontal distance between $A$ and $B$
$BC' = 12 \; km =$ Horizontal distance between $B$ and $C$
$AA' =$ Vertical height between $A$ and $B$
$CC' =$ Vertical height between $B$ and $C$
In $\triangle AA'B$, $\;$ $\dfrac{AA'}{AB} = \tan 20^\circ$
$\therefore \;$ $AA' = AB \; \tan 20^\circ = 8 \times 0.3640 = 2.912$
In $\triangle CC'B$, $\;$ $\dfrac{CC'}{BC'} = \tan 30^\circ$
$\implies$ $CC' = BC' \; \tan 30^\circ = 12 \times \dfrac{1}{\sqrt{3}} = 4 \sqrt{3} 4 \times 1.732 = 6.925$
$\therefore \;$ Vertical height between $A$ and $B$ $= 2.912 \; km$
Vertical height between $B$ and $C$ $= 6.925 \; km$