Question

From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive kilometer stones on (1) opposite (2) same sides of the aeroplane are observed to be ${30}^{o}$ and ${60}^{o}$. Find the height of the aeroplane above the road in two cases. ${30}^{o}$.

Answer 1

Hint:
(1) Let P be the point of the plane. Let points A and C denote the two consecutive kilometer stones on the road AC. Also, we must Assume the perpendicular height of the plane as PB. Then, we will construct a line DE parallel to AC and mark the angles of depression as given to us in question. Then, as DE is parallel to AC find the Angles A and C using the given information. Further, use Trigonometric Ratios in the triangles PAB and PBC to find the distances AB and BC and finally the height PB.

(2) Let P be the point of the plane. Let points B and C denote the two consecutive kilometer stones on the road AC. Also, we must Assume the perpendicular height of the plane as PA. Then, we will construct a line parallel to AC and mark the angles of depression as given to us in question. Then, mark the angles A and C using the given information. Further, use Trigonometric Ratios in the triangles PAB and PAC to find the distances AB and finally the height PB.

Complete step by step solution:
(i)

Now, let the height PB of the plane from the road AC be $h$ kms. Also, let’s assume the distance AB be $x$ kms. As given to us in question the points A and C denote two consecutive milestones, which means they are 1 km apart. So, we can assume the distance BC be \[1 - x\] kms.
Also, as DE is parallel to AC
Angle \[A = {60^0}\] and Angle \[C = {30^0}\]
Now in Triangle PBA
$
  \tan A = \dfrac{{PB}}{{AB}} \\
   \Rightarrow \tan {60^0} = \dfrac{h}{x} \\
   \Rightarrow h = \sqrt 3 x \\
 $
And now see the Triangle PBC
$
  \tan B = \dfrac{{PB}}{{BC}} \\
   \Rightarrow \tan {30^0} = \dfrac{h}{{1 - x}} \\
   \Rightarrow h = \dfrac{{1 - x}}{{\sqrt 3 }} \\
$
 Now, we equate the value of $h$ calculated from both triangles
\[
  \dfrac{{1 - x}}{{\sqrt 3 }} = \sqrt 3 x \\
   \Rightarrow 1 - x = 3x \\
   \Rightarrow 4x = 1 \\
   \Rightarrow x = \dfrac{1}{4} \\
 \]
Therefore, the distance \[AB = x = \dfrac{1}{4}\] kms. So, the distance \[BC = 1 - x = \dfrac{3}{4}\] kms
Also, the height $h = \sqrt 3 x \Rightarrow h = \dfrac{{\sqrt 3 }}{4}$

So, the height of the plane from the road is \[\dfrac{{\sqrt 3 }}{4}\] kms.

(2)

Now, let the height PA of the plane from the road AC be $h$ kms. Also, let’s assume the distance AB be $x$ kms. As given to us in question the points B and C denote two consecutive milestones, which means they are 1 km apart. So, we can assume the distance BC be \[1 + x\] kms.
Also, \[A = {60^0}\] and Angle \[C = {30^0}\]
Now in Triangle PAB
$
  \tan B = \dfrac{{PA}}{{AB}} \\
   \Rightarrow \tan {60^0} = \dfrac{h}{x} \\
   \Rightarrow h = \sqrt 3 x \\
$
And now see the Triangle PAC
$
  \tan C = \dfrac{{PA}}{{AC}} \\
   \Rightarrow \tan {30^0} = \dfrac{h}{{1 + x}} \\
   \Rightarrow h = \dfrac{{1 + x}}{{\sqrt 3 }} \\
 $
 Now, we equate the value of $h$ calculated from both triangles
\[
  \dfrac{{1 + x}}{{\sqrt 3 }} = \sqrt 3 x \\
   \Rightarrow 1 + x = 3x \\
   \Rightarrow 2x = 1 \\
   \Rightarrow x = \dfrac{1}{2} \\
\]
Therefore, the distance \[AB = x = \dfrac{1}{2}\] kms.
Also, the height $h = \sqrt 3 x \Rightarrow h = \dfrac{{\sqrt 3 }}{2}$

So, the height of the plane from the road is \[\dfrac{{\sqrt 3 }}{2}\] kms.

Note:
The question cannot be proceeded further after a certain point if you don’t crack that the distance between the two kilometer stones is 1 km. Also, before starting to solve the problem visualize it and sketch a rough diagram of the problem for better understanding.