Question

A tree standing on a horizontal plane is leaning towards the east. At two points situated at a distance a and b exactly due west on it, the angles of elevation of the top are respectively $\alpha $ and $\beta $. Prove that the height of the top from the ground is $\dfrac{(b-a)\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$.

Answer 1

Hint: Assume that the height of the top of the tree from the ground is ‘h’. First draw a rough diagram of the given conditions and then use the formula $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$ in the different right angle triangles and substitute the given values to get the height.

Complete step by step answer:

According to the above figure:
Let AC be the tree with its top as A. Assume that the two points are D and E due west at a distance of a and b respectively from the base of the tree, that is point C.
We have assumed the height of the top of the tree from the ground as ‘h’. Therefore, AB = h. Also, assume that the distance BC is ‘x’.
Now, in right angle triangle ADB,
$\angle ADB=\alpha $
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
$\begin{align}
  & \tan \alpha =\dfrac{AB}{BD} \\
 & \Rightarrow \tan \alpha =\dfrac{h}{x+a} \\
 & \Rightarrow x+a=\dfrac{h}{\tan \alpha } \\
 & \Rightarrow x=\dfrac{h}{\tan \alpha }-a.........................(i) \\
\end{align}$
Now, in right angle triangle AEB,
\[\angle AEB=\beta \]
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
$\begin{align}
  & \tan \beta =\dfrac{AB}{BE} \\
 & \Rightarrow \tan \beta =\dfrac{h}{x+b} \\
 & \Rightarrow x+b=\dfrac{h}{\tan \beta } \\
 & \Rightarrow x=\dfrac{h}{\tan \beta }-b.........................(ii) \\
\end{align}$
From equations (i) and (ii), we get,
$\begin{align}
  & \dfrac{h}{\tan \alpha }-a=\dfrac{h}{\tan \beta }-b \\
 & \Rightarrow b-a=\dfrac{h}{\tan \beta }-\dfrac{h}{\tan \alpha } \\
 & \Rightarrow b-a=h\left( \dfrac{1}{\tan \beta }-\dfrac{1}{\tan \alpha } \right) \\
\end{align}$
Taking L.C.M we get,
$b-a=h\left( \dfrac{\tan \alpha -\tan \beta }{\tan \beta \tan \alpha } \right)$
By cross-multiplication, we get,
\[\begin{align}
  & \dfrac{(b-a)\tan \beta \tan \alpha }{(\tan \alpha -\tan \beta )}=h \\
 & \Rightarrow h=\dfrac{(b-a)\tan \beta \tan \alpha }{(\tan \alpha -\tan \beta )} \\
\end{align}\]

Note: Do not use any other trigonometric function like sine or cosine of the given angle because the information which is provided to us is the base of the triangle and we have to find the height. So, the length of hypotenuse is of no use. Therefore, formula of the tangent of the angle is used. We can use sine or cosine of the given angles but then the process of finding the height will be lengthy.