Question

An aero plane at an altitude of $1200m$ finds that two ships are sailing towards it in the same direction. The angle of depression of the ships as observed from the planner are ${60^ \circ }$ and ${30^ \circ }$ respectively. Find the distance between the two ships?

Answer 1

Hint: This question is again a common type of question form the height and distance. First draw the suitable diagram showing all the terms and their values. Then consider two right angled triangles one by one. In each triangle apply a suitable trigonometry ratio. Find two different parameters from both the triangles. With their help compute the distance between two ships.

Complete step-by-step answer:
The diagram is given below. In the diagram, at point A aeroplane is there. At points D and C two ships are there. Let AB is the altitude of the aero plane, which is equal to 1200 m.

Let us suppose that the distance between two ships means the length of the CD is x meter.
Now, in right angle triangle ABC, we have
$
  \tan \angle ACB = \dfrac{{Perpendicular}}{{Base}} \\
   \Rightarrow \tan \angle ACB = \dfrac{{AB}}{{BC}} \\
 $
Substituting the values in above equation, we get
$ \Rightarrow \tan {60^ \circ } = \dfrac{{1200}}{{BC}}$
As, $\tan {60^0} = \sqrt 3 $ , so we have
 $
   \Rightarrow \sqrt 3 = \dfrac{{1200}}{{BC}} \\
   \Rightarrow BC = \dfrac{{1200}}{{\sqrt 3 }} \\
 $ ….(1)
Now, in the right angled triangle ABD, we have
$
  \tan \angle ADB = \dfrac{{Perpendicular}}{{Base}} \\
   \Rightarrow \tan \angle ADB = \dfrac{{AB}}{{BD}} \\
 $
Substituting the values in above equation, we get
$ \Rightarrow \tan {30^ \circ } = \dfrac{{1200}}{{BD}}$
As, $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$ , so we have
$ \Rightarrow \tan {30^ \circ } = \dfrac{{1200}}{{BD}}$ …(2)
From the diagram we see that, BD = DC + BC
So, BD = x + $\dfrac{{1200}}{{\sqrt 3 }}$ (from equation 1 put value of BC and also DC is assumed as x)
So, we have now in equation (2),
$
   \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{1200}}{{BD}} \\
   \Rightarrow x + \dfrac{{1200}}{{\sqrt 3 }} = 1200 \times \sqrt 3 \\
   \Rightarrow x = 1200\sqrt 3 - \dfrac{{1200}}{{\sqrt 3 }} \\
   \Rightarrow x = \dfrac{{2400}}{{1.732}} \\
   \Rightarrow BD \simeq 1386\;m \\
 $

$\therefore $Distance between the two ships will be 1386 m in nearest integer value.

Note: For geometrical shapes, especially in triangles, trigonometry ratios are very much applicable, for making relationships among the sides and angles of the triangle. There are six such trigonometry ratios have been defined. These are $\sin \theta ,\;\cos \theta ,\;\tan \theta ,\;\sec \theta ,\;\cos ec\theta ,\;and\;\cot \theta .$ These ratios can be used suitably to get the value of unknown terms.