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 $48^{\circ}$ and $B$ of $57^{\circ} .$ What is the distance between the climbers?

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Solution

In $Δ$ ACD-AD=40 tan48°

In $Δ B C D$
$h$ +40tan48°=40tan57° $h = 40 (t a n 57 ° - t a n 48 °) h = 17.17 m$

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Two climbers are at points A and B on a vertical cliff face. To an observer C,40m 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?

In Δ ACD-AD=40 tan48° In ΔBCDh+40tan48°=40tan57°h=40(tan57°−tan48°)h=17.17m

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 $48^{\circ}$ and $B$ of $57^{\circ} .$ What is the distance between the climbers?

In $Δ$ ACD-AD=40 tan48°

In $Δ B C D$
$h$ +40tan48°=40tan57° $h = 40 (t a n 57 ° - t a n 48 °) h = 17.17 m$