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A parachute is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself.

Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.

Answer:

**Solution:**

A common notion in trigonometry, specifically, is the angle of elevation, which has to do with height and distance. It is described as an angle formed by the horizontal plane and an oblique line between the observer's eye and a target above it.

Let the parachute at highest point A and let C and D be points which are 100 m apart on ground where from then CD = 100 m

Angle of elevation from point D = 45° = α

Angle of elevation from point C = 60° = β

Let B be the point just vertically down the parachute

Let us draw figure according to above data then it forms the figure as shown in which ABC and ABD are two right angled triangles.

Maximum height of the parachute from the ground

AB = H m

Distance of point where parachute falls to just nearest observation point = x m

If in right angle triangle one of the included angles is θ then

\begin{aligned} &\tan \alpha=\frac{\mathrm{AB}}{\mathrm{BD}} \\ &\tan 45^{\circ}=\frac{\mathrm{H}}{\mathrm{x}+100} \\ &100+\mathrm{x}=\mathrm{H} \ldots \text { (a) } \\ &\tan \beta=\frac{\mathrm{AB}}{\mathrm{BC}} \\ &\tan 60^{\circ}=\frac{\mathrm{H}}{\mathrm{x}} \\ &\mathrm{H}=\sqrt{3} \mathrm{x} \ldots \text { (b) } \end{aligned}

From (a) and (b)

\begin{aligned} &\sqrt{3} x=100+x \\ &(\sqrt{3}-1) x=100 \\ &x=\frac{100}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ &x=\frac{100(\sqrt{3}+1)}{2} \\ &x=50(\sqrt{3}+1) \end{aligned}

x = 136.6 m in (b)

H = √3 × 136.6 = 236.6m

Therefore,

The maximum height of the parachute from the ground, H = 236.6m

**Distance between the two points where parachute falls on the ground and just the observation is x = 136.6 m**

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