question_answer

ABCD is a rectangular field. A vertical lamp post of height\[12\,\,m\]stands at the corner\[A\]. If the angle of elevation of its top from\[B\]is\[{{60}^{o}}\]and from\[C\]is \[{{45}^{o}}\], then the area of the field is

A) \[48\sqrt{2}\,\,{{m}^{2}}\]                          

B) \[12\sqrt{2}\,\,{{m}^{2}}\]

C) \[48\,\,{{m}^{2}}\]         

D)        \[12\sqrt{3}\,\,{{m}^{2}}\]

Correct Answer: A

Solution :

Let\[AE\]is a vertical lamp post In\[\Delta \,\,AEC\],                 \[\tan {{45}^{o}}=\frac{AE}{AC}\] \[\Rightarrow \]               \[AC=AE=12\,\,m\] and in\[\Delta \,\,ABE\],                 \[\tan {{60}^{o}}=\frac{AE}{AB}\] \[\Rightarrow \]               \[AB=\frac{AE}{\sqrt{3}}=4\sqrt{3}\] Now,     \[BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}\]                        \[=\sqrt{144-48}=\sqrt{96}\]                        \[=4\sqrt{6}\] \[\therefore \]Area of rectangle ABCD                        \[=AB\times BC\]                        \[=4\sqrt{3}\times 4\sqrt{6}\]                        \[=48\sqrt{2}\,\,{{m}^{2}}\]