Let the required distance BH be equal to x metres. HK is the cliff and ∠LKB is the angle of depression of B from K. Then,
∠KBH = ∠LKB = 40º [alt ∠s, LM || BH]
∴ ∠BKH = 90° – 40° = 50°
From ∆BKH, we get, \(\frac{x}{300}\) = tan 50°
\(x\) = 300 × tan 50° = 300 × 1.1918 = 357.54
= 358 m, correct to the nearest metre.