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Question
A parachutist is descending vertically and makes angles of elevation of 45° and 60° from two observing points 100 m apart to his right. Find the height from which he falls and the distance of the point where he falls on the ground from the nearest observation pcint.
Solution
Let A be the position of the parachutist and C and D be the two observation points.
In ΔABC,
`tan 60^circ = "AB"/"BC"`
⇒ `sqrt(3) = "h"/"x"`
⇒ `"h" = sqrt(3"x")`
In ΔABD,
`tan 45^circ = "AB"/"BD"`
⇒ `1 = ("h"/("x" + 100))`
⇒ x + 100 = h
⇒ x + 100 = `sqrt(3)`x
⇒ `"x"(sqrt(3) - 1) = 100`
⇒ x = `100 xx 1/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1)`
⇒ x = `100 xx ((sqrt(3) + 1))/(3-1) = 50(sqrt(3) + 1) = 50 xx 2.732 = 136.6`
Thus , the distance of the point where he falls on the ground from the nearest observation point (C) is 136.6 m.
Height from which the parachutist fall
= h = `sqrt(3)"x" = 1.732 xx 136.6 = 236.59 ≈ 236.6` m.
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