question_answer

The angle of elevation of the top of a tower from a point on the ground is \[45{}^\circ .\] Moving 21m directly towards the base of the tower, the angle of elevation changes to \[60{}^\circ .\]What is the height of the tower, to the nearest metre?

A) 48 m                

B) 49 m 

C) 50 m    

D) 51 m

Correct Answer: C

Solution :

In \[\Delta PBC,\]

\[\tan 60{}^\circ =\frac{h}{x}\]

\[\Rightarrow \]   \[\frac{h}{x}=\sqrt{3}\]

\[\Rightarrow \]   \[x=\frac{h}{\sqrt{3}}\]

In \[\Delta PAC,\]\[\tan 45{}^\circ =\frac{h}{21+x}=1\]

\[\Rightarrow \]\[h=21+x\]

\[\Rightarrow \]\[h=21+\frac{h}{\sqrt{3}}\]                                  [from Eq. (i)]

\[\Rightarrow \]\[h\left( 1-\frac{1}{\sqrt{3}} \right)=21\]\[\Rightarrow \]\[h=\frac{21\sqrt{3}}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}\]

\[\Rightarrow \]\[h=\frac{21\sqrt{3}\,(\sqrt{3}+1)}{2}=49.68\approx 50\,\,m\]