Watching from the top of a building 40 , m hi | vedclass

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Question

(B) Let $\overline{AB}$ be the building of height $40 \, m$. Let $C$ and $D$ be the two positions of the moving car.From the figure,$AB = 40 \, m$ and

Vedclass · Accepted Answer

$29.2$

Vedclass · Answer

(B) Let $\overline{AB}$ be the building of height $40 \, m$. Let $C$ and $D$ be the two positions of the moving car.From the figure,$AB = 40 \, m$ and $\angle B = 90^{\circ}$.The angles of depression are $\angle XAC = 45^{\circ}$ and $\angle XAD = 30^{\circ}$.Since the line $AX$ is parallel to $BD$,the alternate interior angles are equal:$\angle ACB = \angle XAC = 45^{\circ}$ and $\angle ADB = \angle XAD = 30^{\circ}$.In right-angled $	riangle ABC$:$	an(45^{\circ}) = \frac{AB}{BC} \implies 1 = \frac{40}{BC} \implies BC = 40 \, m$.In right-angled $	riangle ABD$:$	an(30^{\circ}) = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{40}{BD} \implies BD = 40\sqrt{3} \, m$.Using $\sqrt{3} \approx 1.73$,we get $BD = 40 	imes 1.73 = 69.2 \, m$.The distance covered by the car is $CD = BD - BC$.$CD = 69.2 - 40 = 29.2 \, m$.Thus,the distance covered by the car is $29.2 \, m$.

Watching from the top of a building $40 \, m$ high,the angle of depression of a moving car increases from $30^{\circ}$ to $45^{\circ}$ in some interval of time. Find the distance covered by the car in that interval of time. (in $m$)

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Write 'True' or 'False' and justify your answer.
If the height of a tower and the distance of the point of observation from its foot,both,are increased by $10 \%$,then the angle of elevation of its top remains unchanged.

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