b. In the right angled ∆ABD
\(tan60^\circ=\frac{AD}{AB}\)
\(AB=\frac{AD}{tan60^\circ}\)
\(x=\frac{h}{\sqrt{3}}\)
In the right angled ∆ACD
\(tan30^\circ=\frac{AD}{AC}\)
\(AC=\frac{AD}{tan30^\circ}\)
\(x+300=\frac{h}{tan30^\circ}\)
\(x+300=h\sqrt{3}\)
\(\therefore \frac{h}{\sqrt{3}}+300=h\sqrt{3}\)
\(h+300\sqrt{3}=3h\)
\(2h=300\sqrt{3}\)
\(h=150\sqrt{3}=259.8m\)