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**Hint:**Draw a perpendicular from the aeroplane to the horizontal ground. Assume the distance between this point and stone 1 be ‘x’ and that between this point and stone 2 be ‘y’. Now, assume that the height of the plane in miles above the ground is ‘h’. Use a tangent of the given angles to find the value of ‘h’. Use the value of distance between the two stones as 1 mile because it is given that they are milestones.

**Complete step by step answer:**

The term angle of depression denotes the angle from the horizontal downward to an object an observer’s line of sight would be below the horizontal.

Now, let us come to the question. Let us assume that AM is the height of the aeroplane above the ground and stone 1 is at B and stone 2 is at C. Here, we have assumed $BM=x$ and $CM=y$. It is given that $\angle BAO=\alpha $ and $\angle CAN=\beta $.

From the above figure, we have,

$\angle ABM=\alpha $ and $\angle ACM=\beta $ because they are alternate angles of angle BAO and angle CAN respectively.

Now, in right angle triangle AMB, using $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$, we have,

$\begin{align}

& \tan \alpha =\dfrac{AM}{BM} \\

& \Rightarrow \tan \alpha =\dfrac{h}{x} \\

& \Rightarrow x=\dfrac{h}{\tan \alpha }........................(i) \\

\end{align}$

In right angle triangle AMC, using $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$, we have,

$\begin{align}

& \tan \beta =\dfrac{h}{y} \\

& \Rightarrow y=\dfrac{h}{\tan \beta }.......................(ii) \\

\end{align}$

Adding equations (i) and (ii) we get,

$x+y=\dfrac{h}{\tan \alpha }+\dfrac{h}{\tan \beta }...................(iii)$

Now, it is given that the two stones are milestones. That means the distance between the two stones is 1 mile. Therefore, x + y = 1. Substituting this in equation (iii) we get,

$\begin{align}

& 1=\dfrac{h}{\tan \alpha }+\dfrac{h}{\tan \beta } \\

& \Rightarrow h\left( \dfrac{1}{\tan \alpha }+\dfrac{1}{\tan \beta } \right)=1 \\

\end{align}$

Taking L.C.M we get,

$\begin{align}

& h\left( \dfrac{\tan \alpha +\tan \beta }{\tan \alpha \tan \beta } \right)=1 \\

& \Rightarrow h=\dfrac{\tan \alpha \tan \beta }{\tan \alpha +\tan \beta } \\

\end{align}$

**Note:**It is important to draw the diagram of the situation, so that it can be easy to understand what we are going to find out. Also, here tangent of the angle is used to solve the question because we have been provided with the base of the triangle and the value of perpendicular is required.

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