The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is \[60{}^\circ \]and the angle of elevation of the top of the second tower from the foot of the first tower is \[30{}^\circ .\]The distance between the two towers is n times the height of the shorter tower. What is n equal to? |
A) \[\sqrt{2}\]
B) \[\sqrt{3}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
Let h be the height of shorter tower, |
Then, the distance between the two towers is given by nh m. |
In \[\Delta CBD,\]\[\tan 30{}^\circ =\frac{h}{nh}\] |
\[\Rightarrow \]\[\frac{1}{\sqrt{3}}=\frac{1}{n}\]\[\Rightarrow \]\[n=\sqrt{3}\] |
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