A) \[20\,(\sqrt{3}-1)\,m\]
B) \[20\,(\sqrt{3}+1)\,m\]
C) \[10\,(\sqrt{3}-1)\,m\]
D) \[10\,(\sqrt{3}+1)\,m\]
Correct Answer: D
Solution :
[d] Let AB be a pillar of height h m. If BC = length of shadow = x Then, \[BD=(x+20)\,\,m\] From \[\Delta \Alpha \Beta C,\] \[\tan 45{}^\circ =\frac{h}{x}\]\[\Rightarrow \]\[h=x\] (i) From \[\Delta \Alpha \Beta D,\] \[\tan 30{}^\circ =\frac{AB}{BD}\]\[\Rightarrow \]\[\frac{1}{\sqrt{3}}=\frac{h}{x+20}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{h}{h+20}\] [from E. (i)] \[\Rightarrow \] \[\sqrt{3}h=h+20\] \[\Rightarrow \] \[(\sqrt{3}-1)\,h=20\] \[\Rightarrow \] \[h=\frac{20}{\sqrt{3}-1}=\frac{20}{\sqrt{3-1}}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\] \[=\frac{20\,(\sqrt{3}+1)}{2}\] \[=10\,(\sqrt{3}+1)\,m\] |
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