question_answer

The shadow of a tower is found to be 60 m shorter when the suns altitude changes from \[30{}^\circ \]to\[60{}^\circ \]. The height of the tower from the ground is approximately equal to:

A)  62 m                                    

B)  301 m

C)  101 m                  

D)         75 m

E)  52 m

Correct Answer: E

Solution :

In\[\Delta ABC,\] \[\tan 60{}^\circ =\frac{h}{x}\]                                   \[h=\sqrt{3}x\]                         ...(i) and in\[\Delta ABD,\]                 \[\tan 30{}^\circ =\frac{h}{x+60}\] \[\Rightarrow \]               \[x+60=\sqrt{3}h\]                          ?. (ii) From Eqs. (i) and (ii) \[x+60=3x\] \[\Rightarrow \]               \[x=30\text{ }m\] \[\therefore \]  From Eq. (i) \[h=\sqrt{3}\times 30=1.732\times 30=51.96\text{ }m\] \[=52\text{ }m\]    (approx.)