question_answer

A 1-2 m girl see a balloon moving in horizontal line, at height of 88-2 m, at an angle of inclination of \[60{}^\circ \] from her eyes. After some time, the angle of elevation is \[30{}^\circ \] Find the distance traveled by the balloon in this interval?

A) \[91\sqrt{3}\]meters                   

B) \[91\sqrt{3}\] meters

C) \[58\sqrt{3}\] meters      

D) \[15\sqrt{3}\] meters

Correct Answer: C

Solution :

In\[\Delta ADE\]     \[\tan \,\,60{}^\circ =\frac{88.2-1.2}{AE}\] \[\sqrt{3}=\frac{87}{AE}\] \[\Rightarrow \]   \[AE=\frac{87}{\sqrt{3}}\] And in\[\Delta AFH\] \[\tan \,\,30{}^\circ =\frac{88.2-1.2}{AH}\] \[\frac{1}{\sqrt{3}}=\frac{87}{AH}\] \[AH=87\sqrt{3}\] Now, the distance travelled by the balloon in this interval \[=AH-AE\] \[=BG-BC\] \[=87\sqrt{3}-\frac{87}{\sqrt{3}}\] \[=\frac{87[3-1]}{\sqrt{3}}\]             \[=\frac{87\times 2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\] \[=\frac{174\times \sqrt{3}}{3}\] \[=58\sqrt{3\,\,meters}\]