question_answer

A flagpole stands on a building of height 450 ft and an observer on a level ground is 300 ft from the base of the building. The angle of elevation of the bottom of the flagpole is\[30{}^\circ \]and the height of the flagpole is 50 ft. If\[\theta \]is the angle of elevation of the top of the flagpole, then\[\tan \theta \]is equal to

A)  \[\frac{4}{3\sqrt{3}}\]  

B)         \[\frac{\sqrt{3}}{2}\]

C)  \[\frac{9}{2}\]                  

D)         \[\frac{\sqrt{3}}{5}\]

E)  \[\frac{4\sqrt{3}+1}{6}\]

Correct Answer: A

Solution :

In\[\Delta DCE,\] \[\tan 30{}^\circ =\frac{150}{CD}\Rightarrow CD=\frac{150}{1/\sqrt{3}}\] \[\Rightarrow \]               \[CD=\sqrt{3}\times 150\] Now, in\[\Delta DCF,\] \[\tan \theta =\frac{DF}{CD}=\frac{200}{\sqrt{3}.150}=\frac{4}{3\sqrt{3}}\]