Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 18

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Question 1
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The angles of depression of the foot and the top of a pole at the top of a tower of height $$100 $$ metres are $$45^{ }$$ and $$30^{0}$$ respectively. The height of the pole is

A
$$\displaystyle \frac{100(3+\sqrt{3})}{2}$$

B
$$\displaystyle \frac{100(3-\sqrt{3})}{3}$$

C
$$100(\sqrt{3}+1)$$

D
$$100(\sqrt{3}-1)$$

Solution

$$\tan 45 = \dfrac{100}{d}$$
$$\tan 30 ^{0} = \dfrac{100 -h}{d}$$
$$\sqrt{3} = \dfrac{100}{100 -h}$$
$$\sqrt{3}100-100 = h \sqrt{3}$$
$$ h = \dfrac{100(3-\sqrt{3})}{{3}} m.$$
Question 2
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A person walking along a straight road towards a hill observes at two points distance $$\sqrt{3}$$ km, the angle of elevation of the hill to be $$30^{0}$$ and $$60^{0}$$. The height of the hill is

A
$$\dfrac{3}{2}$$ km

B
$$\sqrt{\dfrac{2{1}}{3{2}}}$$

C
$$\displaystyle \frac{(\sqrt{3}+1)}{2}$$ km

D
$$\sqrt{3}$$ km

Solution

$$\tan 60= \dfrac{h}{d} $$
$$\tan 30= \dfrac{h}{ \sqrt{3}+d } $$
$$3 = \dfrac{\sqrt{3}+d}{d}$$
$$2 = \dfrac{ \sqrt{3} }{d}$$
$$d = \dfrac{ \sqrt{3} }{2}$$
$$h = \sqrt{3} \times \dfrac{ \sqrt{3} }{2} $$
$$ = \dfrac{3}{2} km$$
Question 3
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$$A$$ man standing on a level plane observes the elevation of the top of a pole to be $$\alpha$$. He then walks a distance equal to double the height of the pole and then finds that the elevation is now $$2\alpha$$. Then $$\alpha=$$

A
$$30^{\circ}$$

B
$$15^{\circ}$$

C
$$60^{\circ}$$

D
$$45^{\circ}$$

Solution

Let, $$OP$$ be the height of tower $$h$$ and $$\angle OAP=\alpha,\angle OBP=2\alpha$$
$$AB=2h$$ (Given) and
$$AB=OA-OB\dots 1$$

In $$\triangle OBP$$:
$$\dfrac{OB}{OP}=\cot 2\alpha $$
$$OB=OP \cot 2\alpha $$
$$OB=h\cot2\alpha$$

In $$\triangle OAP$$:
$$\dfrac{OA}{OP}=\cot \alpha $$
$$OA=OP \cot \alpha $$
$$OA=h\cot \alpha$$

Substituting the values of $$AB$$, $$OA$$ and $$OB$$ in equation $$1$$:
$$2h=h\cot { \alpha } -h\cot { 2\alpha } $$

$$2=\dfrac{\cos \alpha}{\sin \alpha}-\dfrac{\cos 2\alpha}{\sin 2\alpha}$$

$$ \cos { \alpha } \sin { 2\alpha } -\sin { \alpha } \cos { 2\alpha } =2\sin { \alpha } \sin { 2\alpha } $$

$$\sin { \alpha } =2\sin { \alpha } \sin { 2\alpha } $$

$$\displaystyle \sin { 2\alpha } =\frac { 1 }{ 2 } \quad \quad \quad \left( \because \sin { \alpha } \neq 0 \right) $$

$$\displaystyle 2\alpha =\frac { \pi }{ 6 } \Rightarrow \alpha =\frac { \pi }{ 12 } ={ 15 }^{ \circ}$$
Question 4
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Two pillars of equal height stand at a distance of $$100$$ metres. $$A$$t a point between them the elevation of their tops are found to be $$30^{o }$$ and $$60^{0}$$ Then height of the each pillar in metres is

A
$$25\sqrt{3}$$

B
$$20\sqrt{3}$$

C
$$50\sqrt{3}$$

D
$$35\sqrt{3}$$

Solution

Let's $$'h'$$ be the height of each pillar and $$d$$ is distance from one pillar to the point.

$$\tan 30 ^{0} = \dfrac{h}{d}$$

$$d = h\sqrt{3}$$

$$\tan 60^o = \dfrac{h}{100-d}$$

$$100- h\sqrt{3} = \dfrac{h}{\sqrt{3}}$$

Simplify further,

$$100\sqrt{3}-3h=h$$

Thus,

$$h = 25\sqrt{3} m.$$
Question 5
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$$A$$ flag staff stands upon the top of a building. $$A$$t a distance of 40 $$m$$. the angles of elevation of the tops of the flag staff and building are $$60^{ 0}$$ and $$30^{0}$$ then the height of the flag staff in metres is

A
$$40\sqrt{3}$$

B
$$\displaystyle \frac{40}{\sqrt{3}}$$

C
$$\displaystyle \frac{160}{\sqrt{3}}$$

D
$$\displaystyle \frac{80}{\sqrt{3}}$$
Question 6
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A tower subtends an angle $$\alpha$$ at a point$$A$$ on the same level as the foot of the tower $$B$$ is a point vertically above $$A$$ and $$AB=h$$ metres. The angle of depression of the foot of the tower from $$B$$ is $$\beta$$. The height of the tower is

A
$$h\tan\alpha\cot\beta$$

B
$$h\tan\alpha\tan\beta$$

C
$$h\cot\alpha\cot\beta$$

D
$$h\cot\alpha\tan\beta$$

Solution

$$\tan\alpha =\dfrac{h_{1}}{d}$$
$$\tan\beta =\dfrac{h}{d}$$
$$\dfrac{\tan\alpha }{\tan\beta }=\dfrac{h_{1}}{h}$$
$$h_1=h \tan \alpha \cot \beta$$
Question 7
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The horizontal distance between two towers is $$30$$ meters. From the foot of the first tower the angle of elevation of the top of the second tower is $$60^{o}$$. From the top of the second tower the angle of depression of the top of the first is $$30^{o}$$. The height of the small tower is:

A
$$20(\sqrt{3}+1)$$ mts

B
$$20(\sqrt{3}-1)$$ mts

C
$$20\sqrt{3}$$ mts

D
$$20$$ mts

Solution

$$\tan 60^o= \dfrac{h_{2}}{30}$$ ...{1}

$$\tan 30^o=\dfrac{h_{2}-h_{1}}{30}$$ ...{2}

$$\tan 30^{o}=\dfrac{h_{2}}{30}-\dfrac{h_{1}}{30}$$

From {1} and {2}

$$\Rightarrow \dfrac{h_{1}}{30}=\tan 60^o - \tan 30^o$$

$$\Rightarrow \dfrac{h_{1}}{30}=\dfrac{3-1}{\sqrt{3}}$$

$$\Rightarrow h_1=\dfrac{30\times2}{\sqrt{3}}$$

$$\Rightarrow h_{1}=20\sqrt{3} \,m$$
Question 8
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C is the mid point of the line joining two pionts A,B on the ground. A tower at C slightly leans towards B. If the angles of elevation of the top of the tower from A and B are $$30^{0},60^{0}$$ respectvely, the angle made by the tower with the horizontal is

A
$$45^{0}$$

B
$$60^{0}$$

C
$$75^{0}$$

D
$$30^{0}$$
Question 9
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A pole of height $$h$$ stands at one corner of a park in the shape of an equilateral triangle. If $$\alpha$$ is the angle which the pole subtends at the midpoint of the opposite side, the length of each side of the park is:

A
$$\displaystyle \frac{\sqrt{3}}{2}h\cot\alpha$$

B
$$\displaystyle \frac{2}{\sqrt{3}}h\cot\alpha$$

C
$$\displaystyle \frac{\sqrt{3}}{2}h\tan\alpha$$

D
$$\displaystyle \frac{2}{\sqrt{3}}h\tan\alpha$$
Question 10
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Straight pole(AB) subtends a right angle at a point $$D$$ of another pole at a distance of $$30$$ meters from $$A$$, the top of $$A$$ being $$60^{0}$$ above the horizontal line joining the point $$B$$ to the pole $$A$$. The length of the pole $$A$$ is, in meters

A
$$20 \sqrt{3}$$

B
$$40 \sqrt{3}$$

C
$$60 \sqrt{3}$$

D
$$\displaystyle \frac{40}{\sqrt{3}}$$

Solution

$$\tan60=\dfrac{h_{2}}{30}$$
$$h_{2}= 30\sqrt{3}$$
$$\tan30^{0}=\dfrac{h_{1}}{30}$$
$$h_{1}=10\sqrt{3}$$
$$h_{1}+h_{2} $$= length of the pole =$$40\sqrt{3}$$

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Some Applications of Trigonometry test - 18

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Some Applications of Trigonometry test - 18

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Question 1

$$\displaystyle \frac{100(3+\sqrt{3})}{2}$$

$$\displaystyle \frac{100(3-\sqrt{3})}{3}$$

$$100(\sqrt{3}+1)$$

$$100(\sqrt{3}-1)$$

Solution

Question 2

$$\dfrac{3}{2}$$ km

$$\sqrt{\dfrac{2{1}}{3{2}}}$$

$$\displaystyle \frac{(\sqrt{3}+1)}{2}$$ km

$$\sqrt{3}$$ km

Solution

Question 3

$$30^{\circ}$$

$$15^{\circ}$$

$$60^{\circ}$$

$$45^{\circ}$$

Solution

Question 4

$$25\sqrt{3}$$

$$20\sqrt{3}$$

$$50\sqrt{3}$$

$$35\sqrt{3}$$

Solution

Question 5

$$40\sqrt{3}$$

$$\displaystyle \frac{40}{\sqrt{3}}$$

$$\displaystyle \frac{160}{\sqrt{3}}$$

$$\displaystyle \frac{80}{\sqrt{3}}$$

Question 6

$$h\tan\alpha\cot\beta$$

$$h\tan\alpha\tan\beta$$

$$h\cot\alpha\cot\beta$$

$$h\cot\alpha\tan\beta$$

Solution

Question 7

$$20(\sqrt{3}+1)$$ mts

$$20(\sqrt{3}-1)$$ mts

$$20\sqrt{3}$$ mts

$$20$$ mts

Solution

Question 8

$$45^{0}$$

$$60^{0}$$

$$75^{0}$$

$$30^{0}$$

Question 9

$$\displaystyle \frac{\sqrt{3}}{2}h\cot\alpha$$

$$\displaystyle \frac{2}{\sqrt{3}}h\cot\alpha$$

$$\displaystyle \frac{\sqrt{3}}{2}h\tan\alpha$$

$$\displaystyle \frac{2}{\sqrt{3}}h\tan\alpha$$

Question 10

$$20 \sqrt{3}$$

$$40 \sqrt{3}$$

$$60 \sqrt{3}$$

$$\displaystyle \frac{40}{\sqrt{3}}$$

Solution

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