question_answer

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)]

\[\Rightarrow \]   \[h\sqrt{3}-h=10\]

\[\Rightarrow \]   \[h\,(\sqrt{3}-1)=10\]\[\Rightarrow \]\[h=\frac{10}{(\sqrt{3}-1)}\]

\[\Rightarrow \]   \[h=\frac{10(\sqrt{3}+1)}{{{(\sqrt{3})}^{2}}-{{(1)}^{2}}}=\frac{10(\sqrt{3}+1)}{2}\]

\[\Rightarrow \]   \[h=5\,(\sqrt{3}+1)\]