ABCD is a rectangular field. A vertical lamp | vedclass

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(A) Let the height of the lamp post at $A$ be $AE = 12 \ m$. Since the post is vertical,$AE \perp AB$ and $AE \perp AD$.In $\Delta ABE$,$	an 60^{\cir

Vedclass · Accepted Answer

$48 \sqrt{2} \ m^2$

Vedclass · Answer

(A) Let the height of the lamp post at $A$ be $AE = 12 \ m$. Since the post is vertical,$AE \perp AB$ and $AE \perp AD$.In $\Delta ABE$,$	an 60^{\circ} = \frac{AE}{AB}$ $\Rightarrow \sqrt{3} = \frac{12}{AB}$ $\Rightarrow AB = \frac{12}{\sqrt{3}} = 4\sqrt{3} \ m$.In $\Delta ACE$,$	an 45^{\circ} = \frac{AE}{AC}$ $\Rightarrow 1 = \frac{12}{AC}$ $\Rightarrow AC = 12 \ m$.In right-angled $\Delta ABC$,by Pythagoras theorem,$BC^2 = AC^2 - AB^2 = 12^2 - (4\sqrt{3})^2 = 144 - 48 = 96$.$BC = \sqrt{96} = 4\sqrt{6} \ m$.Area of the rectangular field $ABCD = AB 	imes BC = (4\sqrt{3}) 	imes (4\sqrt{6}) = 16\sqrt{18} = 16 	imes 3\sqrt{2} = 48\sqrt{2} \ m^2$.

$ABCD$ 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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