question_answer

A person observed the angle of elevation of the top of a tower as \[{{30}^{o}}\]. He walked 10 m towards the foot of the tower along ground level and found the angle of elevation of the top of the tower as \[{{60}^{o}}\]. Find the height of the tower.

A)  \[9.66\text{ }m\]                    

B)  \[7.89\text{ }m\]        

C)  \[8.66\text{ }m\]       

D)  \[7.64\text{ }m\]

Correct Answer: C

Solution :

Suppose height of the tower AB = x m and distance BC = ym. ln rt.   \[\Delta \,ABC,\,\,\frac{AB}{BC}=\tan {{60}^{o}}\] \[\Rightarrow \]  \[\frac{x}{y}=\sqrt{3}\Rightarrow y=\frac{x}{\sqrt{3}}\] In rt. \[\Delta \,ABC,\frac{AB}{BC}=\tan {{60}^{o}}\] \[\Rightarrow \]\[\frac{x}{y}=\sqrt{3}\Rightarrow y=\frac{x}{\sqrt{3}}\] In rt, \[\Delta \,ABD,\,\,\frac{AB}{DB}=\tan {{30}^{o}}\] \[\Rightarrow \] \[\frac{x}{10+y}=\frac{1}{\sqrt{3}}\Rightarrow \sqrt{3}x=10+y\]  ?.(2) Putting the value of y in (2), we get \[\sqrt{3}x=10+\frac{x}{\sqrt{3}}\]