A) 10 m
B) 20 m
C) 30 m
D) 40 m
Correct Answer: D
Solution :
Let x be the height of the tower. |
AB = 60m, BD = y m |
In \[\Delta ABD,\] |
\[\frac{AB}{BD}=\tan 60{}^\circ \]or\[\frac{60}{y}=\sqrt{3}\]\[\Rightarrow \]\[y=\frac{60}{\sqrt{3}}\] ?(i) |
In\[\Delta AEC,\]\[\frac{AE}{EC}=\tan 30{}^\circ \] |
\[\Rightarrow \] \[\frac{AE}{BD}=\frac{1}{\sqrt{3}}\]\[\Rightarrow \]\[\frac{AE}{Y}=\frac{1}{\sqrt{3}}\]\[\Rightarrow \]\[\frac{AE}{\frac{60}{\sqrt{3}}}=\frac{1}{\sqrt{3}}\] |
\[AE=\frac{1}{\sqrt{3}}\times \frac{60}{\sqrt{3}}=20\,\,m\] |
\[\therefore \] \[x=AB-AE=60-20=40\,\,m\] |
So, height of tower = 40 m |
You need to login to perform this action.
You will be redirected in
3 sec