Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 12

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Question 1
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By looking at the height of the chimney it is found that on walking towards it 50m in the horizontal line through its base, the angle of elevational of its top changes from $$30^o$$ to $$60^o$$. The height of the chimney is:

A
$$25m$$

B
$$25\, \sqrt{2}$$

C
$$25\, \sqrt{3}$$

D
None of these

Solution

Let feet of the chimney be point $$D$$

So by using trigonometry,

$$\tan 60^{\circ}=\dfrac{h}{BD}\implies BD=\dfrac{h}{\tan 60^{\circ}}$$

Also,

$$\tan 30^{\circ}=\dfrac{h}{AD}\implies AD=\dfrac{h}{\tan 30^{\circ}}$$

Now, $$(AD-BD)=50=\dfrac{h}{\tan 30^{\circ}}-\dfrac{h}{\tan 60^{\circ}}$$

$$50={\sqrt3h}{}-\dfrac{h}{\sqrt{3}}=\dfrac{2h}{\sqrt{3}}$$

$$h=25\sqrt{3} m$$
Question 2
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A vertical pole of height $$10$$ meters stands at one corner of a rectangular field. The angle of elevation of its top from the farthest corner is $$30^{\circ}$$, while that from another corner is $$60^{\circ}$$. The area (in $$m^{2})$$ of rectangular field is

A
$$\dfrac {200\sqrt {2}}{3}$$

B
$$\dfrac {400}{\sqrt {3}}$$

C
$$\dfrac {200\sqrt {2}}{\sqrt {3}}$$

D
$$\dfrac {400\sqrt {2}}{\sqrt {3}}$$

Solution

$$\tan 30^{\circ} = \dfrac {10}{AC} = \dfrac {1}{\sqrt {3}}$$
$$\therefore\sqrt{l^2+b^2}= AC = 10\sqrt {3} \text{ cm}$$ -----------(i)
$$\tan 60^{\circ} = \dfrac {10}{DC} = \sqrt {3}$$
$$\therefore DC = \dfrac {10}{\sqrt3}=l$$
In traingle DCE
$$l^2=\dfrac{100}{3}$$ -----------(ii)
$$b^2 = {300 - \dfrac {100}{3}}$$
$$b^2 = {\dfrac {800}{3}}$$
$$l^2.b^2=\dfrac{100}{3}\times \dfrac{800}{3}=\dfrac{80000}{9}$$
$$Area = lb=\dfrac{\sqrt{80000}}{{\sqrt9}} = \dfrac {200\sqrt {2}}{3} \text{m}^{2}$$.
Question 3
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A ladder $$17m$$ long reaches the window of a building $$15m$$ above the ground. Find the distance of the foot of the ladder from the building.

A
$$8m$$

B
$$5m$$

C
$$10m$$

D
$$12m$$

Solution

Let AC be the ladder, AB be the wall.

Then in $$\triangle ABC$$,

$$AC^2 = AB^2 + BC^2$$ (Pythagoras theorem)

$$17^2 = 15^2 + BC^2$$

$$BC^2 = 8^2$$

$$BC = 8$$ m

Hence, the foot of the ladder is 8 m from the wall.
Question 4
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The length of the string of a kite flying at $$100\ m$$ above the ground with the elevation of $$60^{\circ}$$ is :

A
$$100\ m$$

B
$$100\sqrt{2}m$$

C
$$\dfrac{200}{\sqrt{3}}m$$

D
$$200\ m$$

Solution

Let $$x$$ be the length of the string of kite.
In $$\triangle ABC,$$
$$\Rightarrow \sin { 60^{\circ} } =\dfrac { AC }{ AB } =\dfrac { 100 }{ x } $$
$$\Rightarrow \dfrac { \sqrt { 3 } }{ 2 } =\dfrac { 100 }{ x } $$
$$\Rightarrow x=\dfrac { 200 }{ \sqrt { 3 } } $$
Hence, the answer is $$\dfrac { 200}{\sqrt { 3 }} m.$$
Question 5
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The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal when it is below the horizontal level
that is -

A
the case when we lower our head to look at the object.

B
the case when we upper our head to look at the object.

C
the case when we look straight to the object.

D
irrelevant defination

Solution

option A is correct
the case when we lower our head to look at the object.
Question 6
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The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal when it is

A
Above the horizontal level

B
Below the horizontal level

C
At the horizontal level

D
None of the above

Solution

In the figure, $$AB$$ is a vertical object with $$A$$ as foot and $$B$$ as top.
$$P$$ is the point of observation.
So, $$PA$$ is the line of horizontal sight and we have to raise our head if we look at $$B$$.
The angle, described in this case $$=\angle PAB$$ when $$PB$$ is the line of sight.
But by definition, $$\angle APB$$ is the angle of elevation.
Hence, option A is correct.
Question 7
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Two chimneys 18 m and 13 m high stand upright in the ground. If their feet are 12 m apart, then the distance between their tops is

A
5 m

B
31 m

C
13 m

D
18 m

Solution

We have to find AC.
$$AC^2\,=\, AB^2\,=\,BC^2\, (\because\, \Delta ABC\, is\, right\, \Delta\,le)$$
$$AC^2\,=\, 5^2\, +\, 12^2$$
25 + 144 = 169
AC = $$\sqrt{169}\,=\, 13 m$$
Question 8
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The angle of depression of a boat from the top of a cliff 300 m high is $$\displaystyle 60^{\circ} $$ The distance of the boat from the foot of the cliff is

A
$$\displaystyle 100 \sqrt{3} $$

B
100

C
$$\displaystyle 300 \sqrt{3} $$

D
300
Question 9
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In the given figure, $$\beta$$ is _____.

A
Angle of elevation

B
Angle of depression

C
Line of sight

D
Horizontal level

Solution

The object A is below the level of eye O.
Thus, B is angle of depression.
So, option B is correct.
Question 10
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The given figure represents

A
Angle of elevation

B
Angle of depression

C
Right angle

D
Line of sight

Solution

Here, O represents the eye of an observer and A is an object above the level of eye.
B is any horizontal line passing from D.
Thus $$\angle$$AOB i.e. $$\alpha$$ is called the angle of elevation of object A as seen from O.
So, option A is correct.