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in Trigonometry by (23.5k points)
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The angle of depression of a boat B from the top K of cliff HK, 300 metres high, is 40°. Find the distance of the boat from the foot H of the cliff.

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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.

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