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