An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from his/her eyes has measure 45 degrees. What is the height of the tower?

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28.5 m

30 m

27 m

1.5 m

answer is B.

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Detailed Solution

Given,
Height of observer, BD = CE = 1.5 m
Angle of elevation

∠ BCA = 45^{o}

In ΔABC,
∠CBA

= 90^{\circ}

(right angle)
Thus, by trigonometry properties,

\begin{matrix} \tan 45^{\circ} = \frac{AB}{BC} \\ 1 = \frac{AB}{28.5} (since, \tan 45^{\circ} = 1) \\ AB = 28.5 \end{matrix}

Now, height of the tower,
h=AB+BD
h=28.5+1.5
h=30 m
Correct option is 2.

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