Question : The distance between the two pillars is 120 metres. The height of one pillar is three times the other. The angles of elevation of their tops from the midpoint of the line connecting their feet are complementary to each other. The height (in metres) of the taller pillar is:
Option 1: 34.64
Option 2: 51.96
Option 3: 69.28
Option 4: 103.92
Correct Answer: 103.92
Solution :
Given: The distance between the two pillars is 120 metres.
The height of one pillar is three times the other.
Let the height of the taller pillar be $H_{2}$ and the height of the smaller pillar be $H_{1}$ and the angle of elevation to the top of the taller pillar is $(90°– \theta)$ and for the smaller pillar is $\theta$ from the midpoint of 120 metres.
$H_{2}=3x$
$H_{1}=x$
$\tan\theta = \frac{\text{Height}}{\text{Base}}$
$⇒\tan(90°-\theta) = \frac{H_{2}}{60}$
$⇒\cot\theta = \frac{3x}{60}$ (equation 1)
$⇒\tan\theta = \frac{H_{1}}{60}$
$⇒\tan\theta = \frac{x}{60}$ (equation 2)
Now divide equation 2 by equation 1, we get,
$\frac{\tan\ \theta}{\cot\ \theta}= \frac{60x}{60×3x}$
$⇒\frac{\tan\ \theta}{\cot\ \theta}= \frac{1}{3}$
$⇒ \tan^{2} \theta= \frac{1}{3}$
$⇒\tan\ \theta= \frac{1}{\sqrt{3}}=\tan 30°$
$\therefore\theta = 30°$
Now, the height of the taller pillar is $\tan(90°-\theta) = \frac{H_{2}}{60}$
$⇒\tan(90°- 30°) = \frac{H_{2}}{60}$
$⇒\tan60° = \frac{H_{2}}{60}$
$\therefore H_{2} = 60\sqrt{3}= 103.92$ m
Hence, the correct answer is 103.92.
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