(A) Let $PQ = x$ be the height of the tower and $A$ be the point on the same level as the foot of the tower $Q$. Given $\angle PAQ = 30^{\circ}$.Let $

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(A) Let $PQ = x$ be the height of the tower and $A$ be the point on the same level as the foot of the tower $Q$. Given $\angle PAQ = 30^{\circ}$.Let $B$ be a point $h \ m$ vertically above $A$,so $AB = h$. The angle of depression of the foot of the tower $Q$ from $B$ is $60^{\circ}$,which implies $\angle BQA = 60^{\circ}$.In the right-angled triangle $\triangle ABQ$,we have:$\tan 60^{\circ} = \frac{AB}{AQ}$$\sqrt{3} = \frac{h}{AQ}$$AQ = \frac{h}{\sqrt{3}} = h \cot 60^{\circ}$.Thus,the horizontal distance of the tower from the point $A$ is $h \cot 60^{\circ}$.

Competitive Quantitative Aptitude Heights and Distances Heights and Distances

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$A$ tower subtends an angle of $30^{\circ}$ at a point on the same level as the foot of the tower. At a second point,$h \ m$ above the first,the angle of depression of the foot of the tower is $60^{\circ}$. The horizontal distance of the tower from the point is

A
$h \cot 60^{\circ}$
B
$h \cot 30^{\circ}$
C
$\frac{h}{2} \cot 60^{\circ}$
D
$\frac{h}{2} \cot 30^{\circ}$

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Two towers are situated on opposite banks of a river. One tower is $60 \ m$ high. From the top of this tower,the angles of depression of the top and the bottom of the other tower are $30^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river and the height of the other tower (in $m$).

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$A$ tower standing on a horizontal plane subtends a certain angle at a point $160 \; m$ apart from the foot of the tower. On advancing $100 \; m$ towards it,the tower is found to subtend an angle twice as before. The height of the tower is (in $m$):

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The angle of elevation of an aeroplane as observed from a point $30 \ m$ above the transparent water surface of a lake is $30^{\circ}$ and the angle of depression of the image of the aeroplane in the water of the lake is $60^{\circ}$. The height of the aeroplane from the water surface of the lake is (in $m$):

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The angle of elevation of an aeroplane from a point on the ground is $60^{\circ}$. After flying for $30$ seconds, the angle of elevation changes to $30^{\circ}$. If the aeroplane is flying at a constant height of $4500 \text{ m}$, what is the speed (in $\text{m/s}$) of the aeroplane (in $\sqrt{3}$)?

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From a point $20 \; m$ away from the foot of a tower,the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is:

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$A$ tower subtends an angle of $30^{\circ}$ at a point on the same level as the foot of the tower. At a second point,$h \ m$ above the first,the angle of depression of the foot of the tower is $60^{\circ}$. The horizontal distance of the tower from the point is

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Two towers are situated on opposite banks of a river. One tower is $60 \ m$ high. From the top of this tower,the angles of depression of the top and the bottom of the other tower are $30^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river and the height of the other tower (in $m$).

$A$ tower standing on a horizontal plane subtends a certain angle at a point $160 \; m$ apart from the foot of the tower. On advancing $100 \; m$ towards it,the tower is found to subtend an angle twice as before. The height of the tower is (in $m$):

From a point $20 \; m$ away from the foot of a tower,the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is:

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