The angle of elevation of the top of a tower from the bottom of a building is twice that from its top. What is the height of the building, if the height of the tower is 75 m and the angle of elevation of the top of the tower from the bottom of the building is\[60{}^\circ \]? |
A) 25 m
B) 37.5 m
C) 50 m
D) 60 m
Correct Answer: C
Solution :
We have to find DC. |
Given, \[2x=60{}^\circ \] |
\[\therefore \] \[x=30{}^\circ \] |
In \[\Delta ABC,\]\[\tan 60{}^\circ =\frac{AB}{BC}\] |
\[\Rightarrow \] \[\frac{\sqrt{3}}{1}=\frac{75}{BC}\] |
\[\therefore \] \[BC=\frac{75}{\sqrt{3}}=\frac{75\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\] |
\[=25\sqrt{3}\,\,cm\] |
In \[\Delta AED,\]\[\tan 30{}^\circ =\frac{AE}{ED}=\frac{AE}{25\sqrt{3}}\] \[[\because BC=ED]\] |
\[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{AE}{25\sqrt{3}}\] |
\[\therefore \] \[AE=25\,\,m\] |
\[\therefore \]\[DC=EB=AB-AE=75-25=50\,\,cm\] |
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