The length of the shadow is increased by 10 m on ground level of vertical building when the angle of elevation is changed from \[45{}^\circ \]to \[30{}^\circ .\] Find the height of the building. [SSC (10+2) 2014] |
A) \[5\,(\sqrt{3}+1)\,m\]
B) \[5\,(\sqrt{3}-1)\,m\]
C) \[5\sqrt{3}\,m\]
D) \[\frac{5}{\sqrt{3}}\,m\]
Correct Answer: A
Solution :
Let AB be the building and height of building be h m. |
Now, in \[\Delta ABC,\] |
\[\tan 45{}^\circ =\frac{AB}{BC}=\frac{h}{x}\]\[\Rightarrow \]\[1=\frac{h}{x}\] |
\[\Rightarrow \] \[h=x\] (i) |
Again in \[\Delta ABD,\]\[\tan 30{}^\circ =\frac{AB}{BD}=\frac{h}{x+10}\] |
\[\frac{1}{\sqrt{3}}=\frac{h}{x+10}\] \[\Rightarrow \] \[h\sqrt{3}=x+10\] |
\[\Rightarrow \] \[h\sqrt{3}=h+10\] [from Eq.(i)] |
\[\Rightarrow \] \[h\sqrt{3}-h=10\] |
\[\Rightarrow \] \[h\,(\sqrt{3}-1)=10\]\[\Rightarrow \]\[h=\frac{10}{(\sqrt{3}-1)}\] |
\[\Rightarrow \] \[h=\frac{10(\sqrt{3}+1)}{{{(\sqrt{3})}^{2}}-{{(1)}^{2}}}=\frac{10(\sqrt{3}+1)}{2}\] |
\[\Rightarrow \] \[h=5\,(\sqrt{3}+1)\] |
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