The angle of elevation of the top of a tree
at a point $B$due south of it is 60° and at a point $C$due north
of it is 30°. $D$is a point due north of $C$where the angle
of elevation is 15°, then given $\sqrt{3}=1\frac{8}{11}\text{ and }BC\times CD=$

$2^3\times 3^2\times 19\times 11,$ the height of the tree is

Let $A$be the top of the tree $OA = h$ (Fig. 27.50)
$B, C, D$ be the three point of observation such then $\angle ABO =$
$60°, \angle ACO = 30° and \angle ADO = 15°$

Then $BC=BO+OC=h\left(\cot {60}^{\circ }+\cot 30\right)$

$\begin{array}{l}=h(\sqrt{3}+1/\sqrt{3})=(4/\sqrt{3})h\\ CD=h\cot {15}^{\circ }-h\cot {30}^{\circ }\\ =h(2+\sqrt{3}-\sqrt{3})=2h\end{array}$

So that   $2^3\times 3^2\times 19\times 11$

            $\begin{array}{l}=\frac{4}{\sqrt{3}}\times 2h^2=\frac{4\times 2\times 11}{19}{h}^2\left(\because \sqrt{3}=\frac{19}{11}\right)\\ \Rightarrow h^2=3^2\times {19}^2\Rightarrow h=57\end{array}$