Question

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

Answer 1

Hint: For this kind of problem we will convert the given data into the diagrammatical format. From that diagram list all the parameters that we have and list the parameters that we have to calculate to find the result. For the given data we need to find the value of the distance between the foot of the tower to the given point $A$ by using the trigonometric ratios. By using the value of the distance between the foot of the tower to the given point $A$ is will prove that the height of the tower is $\text{b tan}\alpha \text{ cot}\beta \text{.}$ by using trigonometric ratios.

Complete step by step answer:
Given that, A tower subtends an angle $\alpha $ at a point $\text{A}$ in the plane of its base and the angle of depression of the foot of the tower at a point $\text{b}$ meters just above $\text{A}$ is $\beta $. So, the diagrammatical representation of the given data is given below

Here
     $DC$ is the height of the tower
     $A$ is the base point
     $B$ is the point which is above the point $A$ at a distance of $b$ meters, then $AB=b$
     $DAC$ is the angle subtended by the tower at point $A$, then $\angle DAC=\alpha $
     $BCA$ is the angle of depression of foot of the tower to the point $B$, then $\angle BCA=\beta $
Now we have to prove that the height of the tower is $\text{b tan}\alpha \text{ cot}\beta \text{.}$ i.e. We have to prove $DC=\text{b tan}\alpha \text{ cot}\beta \text{.}$For that find the value of $AC$ from the triangle $ABC$ using trigonometric ratios. And then we will make a relation between $DC,AC$ by using trigonometric ratios. Then substitute the value of $AC$ to get the value of $DC$.

The value of $AC$ from the triangle $ABC$ is
Consider the value of $\tan \beta $. We know
$\begin{align}
  & \tan \beta =\dfrac{\text{Opposite side to }\beta }{\text{Adjacent side to }\beta } \\
 & \tan \beta =\dfrac{AB}{AC} \\
 & AC=\dfrac{AB}{\tan \beta }
\end{align}$
Substituting the value of $AB=b$ and $\dfrac{1}{\tan \beta }=\cot \beta $ in the above equation then
$AC=b\cot \beta .....\left( \text{i} \right)$

From the triangle $DCA$ the value of $\tan \alpha $ is
$\begin{align}
  & \tan \alpha =\dfrac{\text{Opposite side to }\alpha }{\text{Adjacent side to }\alpha } \\
 & \tan \alpha =\dfrac{CD}{AC} \\
 & CD=AC\tan \alpha
\end{align}$
From the equation $\left( \text{i} \right)$ substituting the value of $AC$ in the above equation then we have
$\begin{align}
  & CD=b\cot \beta .\tan \alpha \\
 & =b\tan \alpha .\cot \beta
\end{align}$
Here $CD$ is the height of the tower, hence that the height of the tower is $b\tan \alpha .\cot \beta $.

Note:
For this kind of problems the diagrammatical representation is important. You need check for the common key from which we will get the result. For the value of common key use the given data and apply the trigonometric ratios/formulas. After finding the value of common key establish a trigonometric relation between the common key and the required value and substitute the all the values, we have then we will get the result. Some of the trigonometric ratios/formulas are given below
$\begin{align}
  & \sin \theta =\dfrac{\text{Opposite side to }\theta }{\text{Hypotenuse}} \\
 & \cos \theta =\dfrac{\text{Adjacent side to }\theta }{\text{Hypotenuse}} \\
 & \tan \theta =\dfrac{\text{Opposite side to }\theta }{\text{Adjacent side to }\theta }=\dfrac{\sin \theta }{\cos \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{\cos \theta }{\sin \theta } \\
 & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \csc \theta =\dfrac{1}{\sin \theta }
\end{align}$