question_answer

A kite flying at a height of 45 m from the level ground is attached to a string inclined at \[60{}^\circ \]to the horizontal. The length of the string is

A) \[\frac{30}{\sqrt{2}}m\]            

B) \[30\sqrt{5}\,m\]

C) \[30\sqrt{3}\,m\]

D) \[30\sqrt{2}\,m\]

Correct Answer: C

Solution :

Let length of string be x m.

In \[\Delta ABC,\]\[\sin 60{}^\circ =\frac{AB}{AC}=\frac{45}{x}\]

\[\Rightarrow \]   \[\frac{\sqrt{3}}{2}=\frac{45}{x}\]

\[\Rightarrow \]   \[x=\frac{2\times 45}{\sqrt{3}}\]

\[\Rightarrow \]   \[x=\frac{2\times 45\times \sqrt{3}}{3}\]

\[\therefore \]      \[x=30\sqrt{3}\,m\]

Hence, length of string is \[30\sqrt{3}\,m.\]