Question

A parachutist is descending vertically. At a certain height, his angle of elevation from a point on the ground is 60° and when he has descended 300 m further, the angle of elevation becomes 45° from the same point of observation. Find the distance of the point of observation from the place where the parachutist lands.

Answer 1

We will first draw the figure to get a clear image of what the question is about. Then, we will operate the trigonometric tangent function on the angles of elevation individually in the obtained triangles to calculate the value of the distance of point of observation from the place where the parachutist lands using the formula: $\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$ .

Complete step-by-step answer:
We are given a parachutist who is descending vertically.
At a certain point, say A, the angle of elevation of the parachutist on a point on the ground, say B, is 60°.
Afterwards, when he has descended 300 m down. Let us say he is now at a point P, the angle of elevation of the parachutist from the same point B is 45° now.
We are required to calculate the distance of the point of observation B from the place where the parachutist lands, say C. Let us assume the distance is x metres.
Let us draw the figure of this arrangement:

From the figure, in the right angled triangle DBC, we can say that
$\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$
Using the formula, we get
$\Rightarrow \tan {45^ \circ } = \dfrac{{DC}}{{BC}}$
On substituting the values of DC and BC, we get
$\Rightarrow \tan {45^ \circ } = \dfrac{x}{y} \\\ \Rightarrow 1 = \dfrac{x}{y} \\\$
$\Rightarrow x = y$ equation (1)
Now, in the right angled triangle ABC, using tan 60°, we get
$\Rightarrow \tan 60^\circ = \dfrac{{AC}}{{BC}}$
Substituting the values of AC and BC, we get
$\Rightarrow \tan 60^\circ = \dfrac{{300 + x}}{y} \\\ \Rightarrow \sqrt 3 = \dfrac{{300 + x}}{y} \\\ \Rightarrow \sqrt 3 y = 300 + x \\\$
Using equation (1), we get
$\Rightarrow \sqrt 3 y = 300 + y$
Now, solving it for the value of y, we get
$\Rightarrow \sqrt 3 y - y = 300 \\\ \Rightarrow \left( {\sqrt 3 - 1} \right)y = 300 \\\ \Rightarrow \left( {1.732 - 1} \right)y = 300 \\\$ (using $\sqrt 3$= 1.732)
$\Rightarrow \left( {0.732} \right)y = 300 \\\ \Rightarrow y = 409.83 \simeq 410{\text{m}} \\\$
Therefore, the distance between the point of the landing of the parachutist and the point of the observation is found to be 410 metres.

Note: You may get confused while drawing the figure using the word problem given because you are required to draw the descending parachutist from the same point of observation at two distant points. Be careful while solving the tangent of the angle ABC as you are required to use the relation when x = y to calculate the value of the distance of the point of observation and landing point of the parachutist.