A parachutist is descending vertically and makes angles of elevation of \[{{45}^{\circ }}\] and \[{{60}^{\circ }}\] 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 the point where he falls on the ground from the just observation points.

Hint: consider the maximum height as AB and draw the angle of elevations at two different points and apply \[\tan \theta \]to the two right angled triangles and we will get two equations and then we have to compute the maximum height from which he falls.

Complete step-by-step answer:



Given, the angle of elevations are \[{{45}^{\circ }}\]and\[{{60}^{\circ }}\]at two observing points 100m apart from each other on the left side of himself.

Let the maximum height be AB and BD be x.

In the right angled triangle ABC

\[\tan {{45}^{\circ }}=\dfrac{h}{x+100}\]. . . . . . . . . . . . . . . . . . . (1)

\[h=x+100\]. . . . . . . . . . . . . . . . . . . . . . . . (2)

In the right angled triangle ABD

\[\tan {{60}^{\circ }}=\dfrac{h}{x}\]. . . . . . . . . . . . . . . . . . . . . . (3)

\[x=\dfrac{h}{\sqrt{3}}\]. . . . . . . . . . . . . . . . . . . . . . . . . (4)

Substitute \[x=\dfrac{h}{\sqrt{3}}\]in equation (2)

\[h=\dfrac{h}{\sqrt{3}}+100\]

\[h\left( 1-\dfrac{1}{\sqrt{3}} \right)=100\]

\[0.423h=100\]

\[h=236.4m\]

So, the maximum height from which he falls is AB=\[h=236.4m\]

Note: The angle of elevation is the angle between the horizontal line from the observer and the line of sight to an object that is above the horizontal line. As the person moves from one point to another angle of elevation varies. If we move closer to the object the angle of elevation increases and vice versa.