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1

### JEE Main 2014 (Offline)

A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $${45^ \circ }$$. It files off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $${30^ \circ }$$. Then the speed (in m/s) of the bird is
A
$$20\sqrt 2$$
B
$$20\left( {\sqrt 3 - 1} \right)$$
C
$$40\left( {\sqrt 2 - 1} \right)$$
D
$$40\left( {\sqrt 3 - \sqrt 2 } \right)$$

## Explanation

Let the speed be $$y$$ $$m/sec$$.

Let $$AC$$ be the vertical pole of height $$20$$ $$m.$$

Let $$O$$ be the point on the ground such that $$\angle AOC = {45^ \circ }$$

Let $$OC = x$$

Time $$t=1$$ $$s$$

From $$\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right)$$

and from $$\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right)$$

From $$(i)$$ and $$(ii),$$ we have $$x=20$$

and $${1 \over {\sqrt 3 }} = {{20} \over {x + y}}$$

$$\Rightarrow {1 \over {\sqrt 3 }} = {{20} \over {20 + y}}$$

$$\Rightarrow 20 + y = 20\sqrt 3$$

So, $$y = 20\left( {\sqrt 3 - 1} \right)\,\,i.e.,$$

speed $$= 20\left( {\sqrt 3 - 1} \right)m/s$$
2

### AIEEE 2010

For a regular polygon, let $$r$$ and $$R$$ be the radii of the inscribed and the circumscribed circles. A $$false$$ statement among the following is
A
There is a regular polygon with $${r \over R} = {1 \over {\sqrt 2 }}$$
B
There is a regular polygon with $${r \over R} = {2 \over 3}$$
C
There is a regular polygon with $${r \over R} = {{\sqrt 3 } \over 2}$$
D
There is a regular polygon with $${r \over R} = {1 \over 2}$$

## Explanation

If $$O$$ is center of polygon and

$$AB$$ is one of the side, then by figure

$$\cos {\pi \over n} = {r \over R}$$

$$\Rightarrow {r \over R} = {1 \over 2},{1 \over {\sqrt 2 }},{{\sqrt 3 } \over 2}\,\,for$$

$$n = 3,4,6$$ respectively.
3

### AIEEE 2008

$$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $${45^ \circ }$$. Then the height of the pole is
A
$${{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 - 1} }}m$$
B
$${{7\sqrt 3 } \over 2}\left( {\sqrt {3 + 1} } \right)m$$
C
$${{7\sqrt 3 } \over 2}\left( {\sqrt {3 - 1} } \right)m$$
D
$${{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 + 1} }}m$$

## Explanation

In $$\Delta ABC$$

$${h \over x} = \tan {60^ \circ } = \sqrt 3$$

$$\Rightarrow x = {h \over {\sqrt 3 }}$$

In $$\Delta ABD{h \over {x + 7}}$$

$$= \tan {45^ \circ } = 1$$

$$\Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7$$

$$\Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$$

$$\Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3 + 1\,m} \right)$$
4

### AIEEE 2007

A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is
A
$$a/\sqrt 3$$
B
$$a\sqrt 3$$
C
$$2a/\sqrt 3$$
D
$$2a\sqrt 3$$

## Explanation

In the $$\Delta AOB,\,\,\angle AOB = {60^ \circ },$$ and

$$\angle OBA = \angle OAB$$

(since $$OA=OB=AB$$ radius of same circle).

$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.

Let the height of tower is $$h$$

$$m.$$ Given distance between two points $$A$$ & $$B$$ lie on boundary of

circular park, subtends an angle of $${60^ \circ }$$ at the foot of the tower

is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular

park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \circ }$$ .

In $$\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$$

$$\Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$$

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