Answer

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**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.

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