From the top of a 60 m high tower, the angle of depression of the top and the bottom of a building are observed to be \[30{}^\circ \] and \[60{}^\circ \] respectively. The height of the building is [WBSSC (CGL) Pre 2014] |
A) \[60\sqrt{3}\]
B) \[40\sqrt{3}\]
C) 40 m
D) 20 m
Correct Answer: C
Solution :
Let DC is the tower and AB is the building. |
In \[\Delta ADE,\]\[\tan 30{}^\circ =\frac{60-h}{x}\] |
\[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{(60-h)}{x}\] |
\[\Rightarrow \] \[x=(60-h)\sqrt{3}\] ... (i) |
Now, in \[\Delta BDC,\] |
\[\tan 60{}^\circ =\frac{60}{x}\] \[\Rightarrow \] \[\sqrt{3}=\frac{60}{x}\] |
\[\Rightarrow \] \[x\sqrt{3}=60\] |
\[\Rightarrow \]\[[(60-h)\sqrt{3}]\,\sqrt{3}=60\] [from Eq.(i)] |
\[\Rightarrow \] \[(60-h)\times 3=60\] |
\[\Rightarrow \] \[60-h=20\] |
\[\Rightarrow \] \[h=60-20=40\] |
\[\therefore \] \[h=40\,m\] |
So, height of the building is 40 m. |
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