Question

The angle of elevation of the top of a tower standing on a horizontal plane from point A is $\alpha$. After walking a distance d towards the foot of the tower the angle of elevation is found to be $\beta$. The height of the tower is:
(a) $\dfrac{d}{\cot \alpha +\cot \beta }$
(b) $\dfrac{d}{\cot \alpha -\cot \beta }$
(c) $\dfrac{d}{\tan \beta -\tan \alpha }$
(d) $\dfrac{d}{\tan \beta +\tan \alpha }$

Answer 1

Hint: First of all, draw a tower AB of height h and its angle of elevation at point of C on the ground. Now move distance ‘d’ towards the tower and draw an angle of elevation from the top of the tower at this point as $\beta$. Now take $\tan \alpha$ and $\tan \beta$ and use $\tan \theta =\dfrac{perpendicular}{base}$ on two triangles to the desired required value.

Complete step-by-step answer:
Here, we are given that the angle of elevation of the top of a tower standing on a horizontal plane from a point A is $\alpha$. After walking a distance d towards the foot of the tower the angle of elevation is found to be $\beta$. We have to find the value of the height of the tower.
First of all, let us draw a tower AB and angle of its elevation as $\alpha$.

After walking a distance ‘d’ towards the tower, the angle of elevation is $\beta$.

Let us assume that the height of the tower as h. Let us consider $\Delta ABC$. We know that,
$\tan \theta =\dfrac{perpendicular}{base}$
So, in $\Delta ABC$, we get,
$\tan \alpha =\dfrac{AB}{BC}....\left( i \right)$
From the diagram, we can see that, AB =h, BC = BE + EC = BE + d
By substituting these values in equation (i), we get,
$\tan \alpha =\dfrac{h}{BE+d}....\left( ii \right)$
Now, let us consider $\Delta ABE$, again we know that,
$\tan \theta =\dfrac{perpendicular}{base}$
So, in $\Delta ABE$, we get,
$\tan \beta =\dfrac{AB}{BE}$
We know that AB = h, so we get,
$\tan \beta =\dfrac{h}{BE}$
By multiplying BE on both sides of the above equation, we get,
$BE\tan \beta =h$
$\Rightarrow BE=\dfrac{h}{\tan \beta }$
By substituting the value of BE in equation (ii), we get,
$\tan \alpha =\dfrac{h}{\dfrac{h}{\tan \beta }+d}$
By cross multiplying the above equation, we get,
$\tan \alpha \left( \dfrac{h}{\tan \beta }+d \right)=h$
$\dfrac{\left( \tan \alpha \right)h}{\left( \tan \beta \right)}+\left( \tan \alpha \right)d=h$
$\left( \tan \alpha \right)d=h-h\left( \dfrac{\tan \alpha }{\tan \beta } \right)$
By taking out h common, we get,
$\left( \tan \alpha \right)d=h\left( 1-\dfrac{\tan \alpha }{\tan \beta } \right)$
$h=\dfrac{d\left( \tan \alpha \right)}{1-\dfrac{\tan \alpha }{\tan \beta }}$
$h=\dfrac{d\tan \alpha \tan \beta }{\left( \tan \beta -\tan \alpha \right)}$
By dividing the numerator and denominator by $\tan \alpha \tan \beta$, we get,
$h=\dfrac{d}{\dfrac{\tan \beta -\tan \alpha }{\tan \alpha \tan \beta }}$
$h=\dfrac{d}{\dfrac{1}{\tan \alpha }-\dfrac{1}{\tan \beta }}$
We know that $\dfrac{1}{\tan \theta }=\cot \theta$. By using this, we get,
$h=\dfrac{d}{\cot \alpha -\cot \beta }$
So, we get the height of the tower as $\dfrac{d}{\cot \alpha -\cot \beta }$
Hence, option (b) is the right answer.

Note: In this question, some students make this mistake of interchanging the position of $\alpha \text{ and }\beta$. So, this must be taken care of because a person is walking towards the tower and not away from it. Also, students must be careful while substituting $\dfrac{h}{\tan \beta }$ in the expression of $\tan \alpha$ because sometimes calculation mistakes arise.