question_answer

From the top of a \[60\text{ }m\] high building, the angles of depression of the top and bottom of a tower are measured to be \[30{}^\circ \] and\[60{}^\circ \]. Find the height of the tower:

A) \[20\text{ }m\]                        

B) \[30\text{ }m\]

C) \[40\text{ }m\] 

D) \[50\text{ }m\]

Correct Answer: C

Solution :

Let the height of tower \[AD=a\text{ }m\] From ACED, \[\tan \,30{}^\circ =\frac{60-a}{x}\] \[x=\sqrt{3}\,\,(60-a)\]                  ...(i) and from \[\Delta ABC\,\,\tan 60{}^\circ =\frac{60}{x}\] \[x=\frac{60}{\sqrt{3}}\]                    ..?(ii)             Equating equation (i) and (ii), we get \[\frac{60\sqrt{3}}{\sqrt{3}}=\sqrt{3\,\,(60-a)}\] \[60=\sqrt{3}\times \sqrt{3}\,(60-a)=180-3a\] \[3a=180-60=120=\]