Let the required distance BH = x metres. HK is the cliff and ∠LKB is the angle of depression of B from K.
Then, ∠KBH = ∠LKB = 30° ( ∵ LM || BH, alt. ∠s)
∴ In ΔBKH,
tan 30° = \(\frac{KH}{BH} = \frac{300}{x}\)
x = \(\frac{300}{tan\,30°}=\frac{300}{\frac{1}{\sqrt3}}\) = (300 x √3) meters 300√3 m.