Question

A 6 feet tall man finds that the angle of elevation of the top of a 24 feet high pillar and the angle of depression of its base are complementary angles. The distance of the man from the pillar is
\[\begin{align}
  & A.2\sqrt{3}ft \\
 & B.8\sqrt{3}ft \\
 & C.6\sqrt{3}ft \\
 & D.\text{None of these} \\
\end{align}\]

Answer 1

Hint: In this question, we need to find distance between man and pillar if information is given related to the angle of elevation and angle of depression. We will first draw a diagram for understanding the question better. Then we will use trigonometric properties in the right angle triangle which are $\tan \theta =\dfrac{\text{Side perpendicular to }\theta }{\text{Side adjacent to }\theta }$. Complementary angles are the angles whose sum is ${{90}^{\circ }}$. Also, $\tan \left( {{90}^{\circ }}-\theta \right)$ is equal to $\cot \theta \text{ and }\tan \theta =\dfrac{1}{\cot \theta }$.

Complete step by step answer:
Let us draw a diagram according to the question.
Here AB represent the length of the pillar and PQ represent the height of the man. Angle of elevation from the top of man to the top of the tower is considered as $\alpha $. Since angle of depression to the base of the tower is complementary with angle of elevation to the top, so angle of depression will be \[{{90}^{\circ }}-\alpha \].
Diagram looks like this:

AB = 24ft, PQ = 6ft, AC = 6ft, $\angle BQC=\alpha \text{ and }\angle CQA={{90}^{\circ }}-\alpha $.
Let us suppose the distance between the man and the pillar is x. Therefore, CQ = AP = x ft.
Now as we can see from the figure, we have formed two right angled triangles $\Delta BCQ\text{ and }\Delta ACQ$.
We know that, in a right angled triangle, $\tan \theta =\dfrac{\text{Side perpendicular to }\theta }{\text{Side adjacent to }\theta }$.
So in right angled triangle $\Delta BCQ$,
\[\begin{align}
  & \tan \alpha =\dfrac{\text{Side perpendicular to }\alpha }{\text{Side adjacent to }\alpha } \\
 & \Rightarrow \tan \alpha =\dfrac{BC}{CQ} \\
\end{align}\]
We can write BC as AB-AC so we get:
\[\Rightarrow \tan \alpha =\dfrac{AB-AC}{CQ}\]
Since AB = 24 ft, AC = 6ft and CQ = x ft, so we get:
\[\begin{align}
  & \Rightarrow \tan \alpha =\dfrac{24-6}{x} \\
 & \Rightarrow \tan \alpha =\dfrac{18}{x}\cdots \cdots \cdots \left( 1 \right) \\
\end{align}\]
Now in right angled triangle $\Delta ACQ$,
\[\begin{align}
  & \tan \left( {{90}^{\circ }}-\alpha \right)=\dfrac{\text{Side perpendicular to }\left( {{90}^{\circ }}-\alpha \right)}{\text{Side adjacent to }\left( {{90}^{\circ }}-\alpha \right)} \\
 & \Rightarrow \tan \left( {{90}^{\circ }}-\alpha \right)=\dfrac{AC}{CQ} \\
\end{align}\]
We know that, $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ so we get: $\cot \alpha =\dfrac{AC}{CQ}$.
Since AC = 6ft and CQ = x ft. So we get:
\[\Rightarrow \cot \alpha =\dfrac{6}{x}\cdots \cdots \cdots \left( 2 \right)\]
Multiplying (1) by (2) we get:
\[\Rightarrow \tan \alpha \cdot \cot \alpha =\dfrac{18}{x}\cdot \dfrac{6}{x}\]
We know that $\tan \theta =\dfrac{1}{\cot \theta }$ so we get:
\[\Rightarrow \dfrac{\cot \alpha }{\cot \alpha }=\dfrac{108}{{{x}^{2}}}\]
Cancelling $\cot \alpha $ we get:
\[\Rightarrow 1=\dfrac{108}{{{x}^{2}}}\]
Cross multiplying we get:
\[\Rightarrow {{x}^{2}}=108\]
Taking square root both sides we get:
\[\Rightarrow x=\sqrt{108}ft\]
Now 108 can be written as $2\times 2\times 3\times 3\times 3$ so $\sqrt{108}=\sqrt{2\times 2\times 3\times 3\times 3}=2\times 3\times \sqrt{3}=6\sqrt{3}$.
Therefore $x=6\sqrt{3}ft$.
Hence, the distance between man and pillar is $6\sqrt{3}ft$.

So, the correct answer is “Option C”.

Note: Students should note that diagrams are compulsory for these questions. They should know all values of trigonometric function in a right angled triangle. Make sure that units used are the same as given units. Students can get confused between complementary and supplementary angles.