ABCD is a rectangular field. A vertical lamp | vedclass

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Question

(A) Let $AE$ be the vertical lamp post of height $12\,m$. In $	riangle EAB$,$\angle EBA = 60^\circ$ and $\angle EAB = 90^\circ$. Thus,$	an 60^\circ

Vedclass · Accepted Answer

$48\sqrt{2}\,sq.\,m$

Vedclass · Answer

(A) Let $AE$ be the vertical lamp post of height $12\,m$. In $	riangle EAB$,$\angle EBA = 60^\circ$ and $\angle EAB = 90^\circ$. Thus,$	an 60^\circ = \frac{AE}{AB} \implies \sqrt{3} = \frac{12}{AB} \implies AB = \frac{12}{\sqrt{3}} = 4\sqrt{3}\,m$.In $	riangle EAC$,$\angle ECA = 45^\circ$ and $\angle EAC = 90^\circ$. Thus,$	an 45^\circ = \frac{AE}{AC} \implies 1 = \frac{12}{AC} \implies AC = 12\,m$.In the rectangular field $ABCD$,$	riangle ABC$ is a right-angled triangle at $B$. By Pythagoras theorem,$BC^2 = AC^2 - AB^2 = 12^2 - (4\sqrt{3})^2 = 144 - 48 = 96$.So,$BC = \sqrt{96} = 4\sqrt{6}\,m$.The area of the rectangular field $ABCD = AB 	imes BC = (4\sqrt{3}) 	imes (4\sqrt{6}) = 16\sqrt{18} = 16 	imes 3\sqrt{2} = 48\sqrt{2}\,sq.\,m$.

$ABCD$ 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 $60^\circ$ and from $C$ is $45^\circ$,then the area of the field is

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