On the same side of tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is.
A. 63.5 m B. 76.9 m C. 86.7 m D. 90 m Answer: Option AShow Answer
Solution(By Apex Team)
Given that height of the tower is 150 m
Let the two objects be placed at point C and D respectively
In △ABC
$\begin{array}{l}
\tan 60^{\circ}=\frac{150}{\mathrm{BC}} \\
\Rightarrow \mathrm{BC}=\frac{150}{\sqrt{3}}\quad(\text { Take } \sqrt{3}=1.73)
\end{array}$
and in △ABD
$\tan 45^{\circ}=\frac{150}{\mathrm{BD}} \Rightarrow \mathrm{BD}=150 \mathrm{~m}$
Now, Distance between objects is;
$\mathrm{CD}=\mathrm{BD}-\mathrm{BC}$
$\begin{array}{l}
=\left(150-\frac{150}{\sqrt{3}}\right) \mathrm{m} \\
=\left[\frac{150(\sqrt{3}-1)}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\right] \mathrm{m} \\
=50(3-\sqrt{3}) \mathrm{m} \\
=(50 \times 1.27) \mathrm{m} \\
=63.5 \mathrm{~m}
\end{array}$
