a a parachutist is descending vertically and makes anggles of depression of 45degree and 60 degree at two observation points100m - Maths - Some Applications of Trigonometry

Question

a a parachutist is descending vertically and makes anggles of
depression of 45degree and 60 degree at two observation points100m
apart from each other on the left side of himself find ,in
metres,the approximate height from which he falls and also find ,in
metres the approximate distance of the point where he falls on the
ground from the first observation point?????????????

Ankush Jain · Accepted Answer

Let the parachutist be at a height h m above the ground and distance where he falls on the ground from the first observation point be x m.

Suppose A and B are first and second observation points respectively.

∴ CD = h m and AC = x m

Given, AB = 100 m

In Δ ACD,

$\tan 60 ° = \frac{CD}{AC} (∵ \tan θ = \frac{Perpendicular}{Base}) ∴ \sqrt{3} = \frac{h}{x} \Rightarrow h = \sqrt{3} x ...(1)$

In Δ BCD,

$\tan 45 ° = \frac{CD}{BC} ∴ 1 = \frac{h}{100 + x}$

⇒ h = 100 + x ...(2)

From (1) and (2), we have

$\sqrt{3} x = 100 + x ∴ (\sqrt{3} - 1) x = 100 \Rightarrow x = \frac{100}{\sqrt{3} - 1} \Rightarrow x = \frac{100 (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} - 1)} = \frac{100 (\sqrt{3} + 1)}{3 - 1} = \frac{100 (\sqrt{3} + 1)}{2} = 50 (1.73 + 1) = 50 \times 2.73 m = 136.5 m (approx)$

From (1), we have

$h = \sqrt{3} \times (\frac{100}{\sqrt{3} - 1}) \Rightarrow h = \frac{100 \sqrt{3}}{\sqrt{3} - 1} = \frac{100 \sqrt{3} (\sqrt{3} + 1)}{3 - 1} = \frac{100 \sqrt{3} (\sqrt{3} + 1)}{2} = 50 \sqrt{3} (\sqrt{3} + 1) = 150 + 50 \sqrt{3} = 236.5 m (approx)$

Thus, the parachutist falls from a height of 236.5 m (approx) and he falls at a distance of 136.5 m (approx) from the first observation point.