(D) Let the height of the tower $NM = H$ and the distance $LM = x$.In $\triangle NQM$,$\tan 60^{\circ} = \frac{NM}{QM} \implies \sqrt{3} = \frac{H}{QM

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(D) Let the height of the tower $NM = H$ and the distance $LM = x$.In $\triangle NQM$,$\tan 60^{\circ} = \frac{NM}{QM} \implies \sqrt{3} = \frac{H}{QM} \implies QM = \frac{H}{\sqrt{3}}$.In $\triangle NLM$,$\tan 30^{\circ} = \frac{NM}{LM} \implies \frac{1}{\sqrt{3}} = \frac{H}{x} \implies x = H\sqrt{3}$.Since the height $H$ of the tower is not provided,the exact numerical value of $x$ cannot be determined.Therefore,the data is inadequate.

Competitive Quantitative Aptitude Heights and Distances Heights and Distances

Easy

$A$ man standing at a point $P$ (or $L$ in the diagram) is watching the top of a tower $N$,which makes an angle of elevation of $30^{\circ}$ with the man's eye. The man walks some distance towards the tower to point $Q$ to watch its top and the angle of elevation becomes $60^{\circ}$. What is the distance (in units) between the base of the tower $M$ and the point $P$ (or $L$)?

A
$4 \sqrt{3}$
B
$8$
C
$12$
D
Data inadequate

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Heights and Distances

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Heights and Distances

An observer on the top of a cliff,$200 \, m$ above the sea level,observes the angle of depression of two ships at anchor to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the distance (in $m$) between the ships if the line joining them stretches to the base of the cliff.

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The angle of elevation of the top of an unfinished tower at a point distant $120 \ m$ from its base is $45^{\circ}$. If the elevation of the top at the same point is to be $60^{\circ}$,the tower must be raised to a height of:

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If $x = a \cos \theta + b \sin \theta$ and $y = b \cos \theta - a \sin \theta,$ then $x^{2} + y^{2}$ is equal to

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On the level ground,the angle of elevation of the top of the tower is $30^{\circ}$. On moving $20 \ m$ nearer,the angle of elevation is $60^{\circ}$. The height of the tower is (in $m$):

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$A$ man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of $45^{\circ}$ with the man's eye when at a distance of $60 \ m$ from the tower. After $5 \ s$,the angle of depression becomes $30^{\circ}$. What is the approximate speed (in $km/hr$) of the boat,assuming that it is running in still water?

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An observer on the top of a cliff,$200 \, m$ above the sea level,observes the angle of depression of two ships at anchor to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the distance (in $m$) between the ships if the line joining them stretches to the base of the cliff.

The angle of elevation of the top of an unfinished tower at a point distant $120 \ m$ from its base is $45^{\circ}$. If the elevation of the top at the same point is to be $60^{\circ}$,the tower must be raised to a height of:

If $x = a \cos \theta + b \sin \theta$ and $y = b \cos \theta - a \sin \theta,$ then $x^{2} + y^{2}$ is equal to

On the level ground,the angle of elevation of the top of the tower is $30^{\circ}$. On moving $20 \ m$ nearer,the angle of elevation is $60^{\circ}$. The height of the tower is (in $m$):

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