Question

A tower subtends an angle $\alpha $ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b meters just above A is $\beta $ . Prove that the height of the tower is $b\tan \alpha \cot \beta $ .

Answer 1

Hint: First we will draw the required diagram and with the help of that we will mark the angles and find the value of height of triangle using trigonometry formulas like $\tan \alpha = \dfrac{height}{base}$ and $\cot \beta =\dfrac{base}{height}$.

Complete step-by-step answer:
Let’s first draw the required diagram,

In the above diagram,
BC = h (height of tower)
AB = base
AD = b
Now we can see that we need to prove h = $b\tan \alpha \cot \beta $,
So, first let’s find the value of tan$\alpha $,
We will use the formula,
$\tan \alpha =\dfrac{height}{base}$
Now in triangle ABC, substituting the values of height and base we get,
$\begin{align}
  & \tan \alpha =\dfrac{h}{AB} \\
 & \Rightarrow AB=\dfrac{h}{\tan \alpha }............(1) \\
\end{align}$
Now we have found the value of base AB, with help of that we will find the value of cot$\beta $ in triangle ABD,
The formula for finding the value of cot$\beta $ is,
$\cot \beta =\dfrac{base}{height}$
Now in triangle ABD, substituting the values of height and base we get,
$\cot \beta =\dfrac{AB}{b}$
Now substituting the value of AB from equation (1) we get,
$\begin{align}
  & \Rightarrow \cot \beta =\dfrac{\dfrac{h}{\tan \alpha }}{b} \\
 & \Rightarrow \cot \beta =\dfrac{h}{b\tan \alpha } \\
 & \Rightarrow h=b\tan \alpha \cot \beta \\
\end{align}$
Hence, we have proved that the height of the tower is $b\tan \alpha \cot \beta $.

Note: In this question we have used the formula of tan and cot, which is important and students must be careful with the terms like angle of elevation and angle of depression to avoid any mistakes while solving the question.