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Question
Two climbers are at points A and B on a vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers?
Solution
Let P be the foot of the cliff on level ground.
Then, ∠ACP = 48° and ∠BCP = 57°
∴ `(BP)/(PC) = tan 57^circ`
`=>` BP = 40 × 1.539 = 61.57 m
Also, `(AP)/(PC) = tan 48^circ`
`=>` AP = 40 × 1.110 = 44.4 m
Hence, distance between the climbers = AB = BP – AP = 17.17 m
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