question_answer

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

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

B)  \[48\sqrt{3}\,sq\,m\]

C)  \[48\,sq\,m\]   

D)         \[12\sqrt{2}\,sq\,m\]

E)  \[12\sqrt{3}\,sq\,m\]

Correct Answer: A

Solution :

In \[\Delta ABE,\] \[\tan 60{}^\circ =\frac{AE}{AB}\Rightarrow AB=\frac{12}{\sqrt{3}}=4\sqrt{3}m\] and in \[\Delta ACE\]                 \[\tan 45{}^\circ =\frac{12}{AC}\Rightarrow AC=12\] \[\therefore \] \[BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}=\sqrt{144-48}=\sqrt{96}=4\sqrt{6}\] \[\therefore \]Area of rectangular field\[=AB\times BC\]                 \[=4\sqrt{3}\times 4\sqrt{6}\]                 \[=48\sqrt{2}\,sq\,m\]