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

$\tan 45 = \dfrac{100}{d}$
$\tan 30 ^{0} = \dfrac{100 -h}{d}$
$\sqrt{3} = \dfrac{100}{100 -h}$
$\sqrt{3}100-100 = h \sqrt{3}$
$h = \dfrac{100(3-\sqrt{3})}{{3}} m.$

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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

$\tan 60= \dfrac{h}{d}$
$\tan 30= \dfrac{h}{ \sqrt{3}+d }$
$3 = \dfrac{\sqrt{3}+d}{d}$
$2 = \dfrac{ \sqrt{3} }{d}$
$d = \dfrac{ \sqrt{3} }{2}$
$h = \sqrt{3} \times \dfrac{ \sqrt{3} }{2}$
$= \dfrac{3}{2} km$

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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 $1$ :
$2h=h\cot { \alpha } -h\cot { 2\alpha }$

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

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

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

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

$\displaystyle 2\alpha =\frac { \pi }{ 6 } \Rightarrow \alpha =\frac { \pi }{ 12 } ={ 15 }^{ \circ}$

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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,

$h = 25\sqrt{3} m.$

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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

$\tan\alpha =\dfrac{h_{1}}{d}$
$\tan\beta =\dfrac{h}{d}$
$\dfrac{\tan\alpha }{\tan\beta }=\dfrac{h_{1}}{h}$
$h_1=h \tan \alpha \cot \beta$

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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}

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

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

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

$\Rightarrow h_{1}=20\sqrt{3} \,m$

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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

100(3+3)2\displaystyle \frac{100(3+\sqrt{3})}{2}2100(3+3​)​

100(3−3)3\displaystyle \frac{100(3-\sqrt{3})}{3}3100(3−3​)​

100(3+1)100(\sqrt{3}+1)100(3​+1)

100(3−1)100(\sqrt{3}-1)100(3​−1)

Solution

Question 2

32\dfrac{3}{2}23​ km

2132\sqrt{\dfrac{2{1}}{3{2}}}3221​​

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

3\sqrt{3}3​ km

Solution

Question 3

30∘30^{\circ}30∘

15∘15^{\circ}15∘

60∘60^{\circ}60∘

45∘45^{\circ}45∘

Solution

Question 4

25325\sqrt{3}253​

20320\sqrt{3}203​

50350\sqrt{3}503​

35335\sqrt{3}353​

Solution

Question 5

40340\sqrt{3}403​

403\displaystyle \frac{40}{\sqrt{3}}3​40​

1603\displaystyle \frac{160}{\sqrt{3}}3​160​

803\displaystyle \frac{80}{\sqrt{3}}3​80​

Question 6

htan⁡αcot⁡βh\tan\alpha\cot\betahtanαcotβ

htan⁡αtan⁡βh\tan\alpha\tan\betahtanαtanβ

hcot⁡αcot⁡βh\cot\alpha\cot\betahcotαcotβ

hcot⁡αtan⁡βh\cot\alpha\tan\betahcotαtanβ

Solution

Question 7

20(3+1)20(\sqrt{3}+1)20(3​+1) mts

20(3−1)20(\sqrt{3}-1)20(3​−1) mts

20320\sqrt{3}203​ mts

202020 mts

Solution

Question 8

45045^{0}450

60060^{0}600

75075^{0}750

30030^{0}300

Question 9

32hcot⁡α\displaystyle \frac{\sqrt{3}}{2}h\cot\alpha23​​hcotα

23hcot⁡α\displaystyle \frac{2}{\sqrt{3}}h\cot\alpha3​2​hcotα

32htan⁡α\displaystyle \frac{\sqrt{3}}{2}h\tan\alpha23​​htanα

23htan⁡α\displaystyle \frac{2}{\sqrt{3}}h\tan\alpha3​2​htanα

Question 10

20320 \sqrt{3}203​

40340 \sqrt{3}403​

60360 \sqrt{3}603​

403\displaystyle \frac{40}{\sqrt{3}}3​40​

Solution

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

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

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

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

$\dfrac{3}{2}$ km

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

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

$\sqrt{3}$ km

$30^{\circ}$

$15^{\circ}$

$60^{\circ}$

$45^{\circ}$

$25\sqrt{3}$

$20\sqrt{3}$

$50\sqrt{3}$

$35\sqrt{3}$

$40\sqrt{3}$

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

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

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

$h\tan\alpha\cot\beta$

$h\tan\alpha\tan\beta$

$h\cot\alpha\cot\beta$

$h\cot\alpha\tan\beta$

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

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

$20\sqrt{3}$ mts

$20$ mts

$45^{0}$

$60^{0}$

$75^{0}$

$30^{0}$

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

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

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

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

$20 \sqrt{3}$

$40 \sqrt{3}$

$60 \sqrt{3}$

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