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?

Let P be the foot of cliff.

From figure,

In △BCP,

$\text{tan 57°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 1.539 = \dfrac{BP}{PC} \\[1em] \Rightarrow BP = PC \times 1.539 \\[1em] \Rightarrow BP = 40 \times 1.539 \\[1em] \Rightarrow BP = 61.57 \text{ meters}.$

In △ACP,

$\text{tan 48°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 1.110 = \dfrac{AP}{PC} \\[1em] \Rightarrow AP = PC \times 1.110 \\[1em] \Rightarrow AP = 40 \times 1.110 \\[1em] \Rightarrow AP = 44.40 \text{ meters}.$

Distance between climbers (BA) = BP - AP

= 61.57 - 44.40

= 17.17 meters.

Hence, distance between two climbers = 17.17 meters.