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NIMCET 2025 Paper

Section: Mathematics

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Question : 8 of 120

Marks: +1, -0

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

metres above the first, the depression of the foot of the tower is

60^{\circ}

. What is the horizontal distance of the tower from the point?

$2 h \tan 60^{\circ}$
$h \tan 60^{\circ}$
$2 h \cot 60^{\circ}$
$h \cot 60^{\circ}$

Solution:

Use

\tan =

opposite

/

adjacent .

The angle subtended by the tower gives a relation between height and distance.

The depression angle from a higher point gives another relation.

Solve the two equations to find the horizontal distance.

Solution

Let the horizontal distance be

d

.

From ground point:

\tan 30^{\circ}= \; \frac{H}{d} \Rightarrow H= \; \frac{d}{\sqrt{3}}

From point at height

h

:

\tan 60^{\circ}= \; \frac{h}{d} \Rightarrow d= \; \frac{h}{\sqrt{3}}

Thus,

d=h \cot 60^{\circ}

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