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 \[{{60}^{o}}\] and from C is\[{{45}^{o}}\], then the area of the field is [Kerala (Engg.) 2005]

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

B) \[48\sqrt{3}sq.m\]

C) \[48sq.m\]

D) \[12\sqrt{2}sq.m\]

E) \[12\sqrt{3}sq.m\]

Correct Answer: A

Solution :

Let AE is a vertical lamp-post.  Given, \[AE=12\]m \[\tan {{45}^{o}}=\frac{AE}{AC}\] \[AC=AE=12m\] \[\tan {{60}^{o}}=\frac{AE}{AB}\] \[AB=\frac{AE}{\sqrt{3}}=4\sqrt{3}\] \[BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}\]\[=\sqrt{144-48}\] \[=\sqrt{96}\] \[=4\sqrt{6}\] Area =\[AB\times BC=4\sqrt{3}\times 4\sqrt{6}\] =\[48\sqrt{2}\]sq. cm.