Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 24

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Question 1
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The angles of elevation of the top of a tower from two points at distance m and n metres are complementary. If the two points and the base of the tower are on the same straight line, then the height of the tower is

A
$$\sqrt {mn}$$

B
mn

C
$$\frac {m}{n}$$

D
None of these

Solution

From $$\Delta ADB,$$
$$\dfrac {h}{m}=tan\theta$$

From $$\Delta ABC,$$
$$\dfrac {h}{n}=tan(90-\theta)$$
$$=cot\theta$$

$$\therefore \dfrac {h}{m}\times \dfrac {h}{n}=tan\theta \times cot\theta$$
$$=1$$

$$\Rightarrow h=\sqrt {mn}$$
Question 2
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What is the height of a tower if the angles of elevation of its top from two points $$x$$ and $$y$$ at distance of $$a$$ and $$b$$ respectively from the base and on the same straight line with the tower, are complementary?

A
$$\dfrac{\sqrt b}{a}$$

B
$$\dfrac{\sqrt a}{b}$$

C
$$\sqrt {ab}$$

D
None of these

Solution

Let $$h$$ be the height of the tower $$AB$$
$$\therefore \dfrac {h}{a}=tan p$$.....(1)
$$\dfrac {h}{b}=tan(90-p)=cot p$$......(2)
Multiply (1) and (2), we get
$$\therefore \dfrac {h}{a}\times \dfrac {h}{b}=tan p\times cot p=1$$
$$\Rightarrow h^2=ab\Rightarrow h=\sqrt {ab}$$
Question 3
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The angle of elevation of the sun when the length of the shadow of a pole is $$\sqrt 3$$ times the height of the pole is

A
$$30^0$$

B
$$45^0$$

C
$$60^0$$

D
$$75^0$$

Solution

$$tan\angle MOP=\frac {MP}{MO}=\frac {1}{\sqrt 3}=tan 30^0$$
$$\therefore\ \angle MOP=30^0$$
Question 4
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A vertical pole breaks due to storm and the broken part bends so that the top of the pole touches the ground, making an angle of $$30^{\circ}$$ with the ground. The distance from the foot of the pole to the point where the top touches the ground is $$10 \text{ m}$$. The height of the pole is

A
$$10(\sqrt 3+1)m$$

B
$$10\sqrt 3m$$

C
$$10(\sqrt 3-1)m$$

D
$$\dfrac {10}{\sqrt 3}m$$

Solution

$$ The\quad pole\quad breaks\quad at\quad A.$$

$$\therefore \ AP=AR.\quad \quad \quad QR=10m$$

$$\therefore \ AP+AQ=PQ\quad \quad or\quad AR+AQ=PQ$$

$$ \dfrac { AQ }{ QR } =\tan{ 30 }^{ \circ \quad }$$

$$\therefore \ AQ=QR \tan{ 30 }^{ \circ }=\dfrac { 10 }{ \sqrt { 3 } } m.$$

$$Again, $$

$$\dfrac { QR }{ AR } =\cos{ 30 }^{ \circ }=\dfrac { \sqrt { 3 } }{ 2 } $$

$$\therefore AR=QR\times \dfrac { 2 }{ \sqrt { 3 } } =\dfrac { 20 }{ \sqrt { 3 } } m.$$
$$\therefore PQ=AQ+AR = \dfrac { 10 }{ \sqrt { 3 } } + \dfrac { 20 }{ \sqrt { 3 } } m = 10 \sqrt{3} m $$

$$ Hence,\quad the\quad length\quad of \quad the\quad pole\quad is\quad 10\sqrt{3} m.$$
Question 5
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From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be $$\alpha $$ and $$\beta $$ where $$\alpha >\beta. $$ The height of the tower is $$130\ m,$$ $$\alpha =60^\circ$$ and $$\beta =30^\circ.$$ The distance of the extreme object from the top of the tower is ______.

A
$$65\ m$$

B
$$130\ m$$

C
$$260\ m$$

D
None of the above

Solution

Let $$AB$$ be the tower

Given: Height of the tower $$h=130$$ m

and $$\alpha=60^\circ,\ \beta=30^\circ$$

The extreme object from the top of the tower has low angle of depression.

$$\therefore $$ The angle of elevation of the point $$=30^{\circ}$$

Let the distance of the point from the top of the tower $$=p$$

Thus, $$\sin 30^\circ = \dfrac{h}{p}$$

$$\Rightarrow \dfrac{1}{2} = \dfrac{130}{p}$$

$$\Rightarrow p = 260$$ m
Question 6
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The angles of elevation of an artificial satellite measured from two earth stations are $$30^0$$ and $$60^0$$ respectively. If the distance between the earth stations is 4000 km, then the height of the satellite is

A
2000 km

B
6000 km

C
3464 km

D
2828 km
Question 7
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The Qutub Minar casts a shadow $$150~ \text{m}$$ long, at the same time the Vikas Minar casts a shadow $$120 ~\text{m}$$ long on the ground. If the height of the Vikas Minar is $$80~ \text{m}$$, find the height of the Qutub Minar.

A
$$180~\text{m}$$

B
$$100~\text{m}$$

C
$$150~\text{m}$$

D
$$120~\text{m}$$

Solution

$$\text{Since both the Qutub Minar and the Vikas Minar cast the shadow at the same time,the elevation } \alpha \text{ is same for both so}\\ \text{far as the shadow cast by the sun is concerned}.\\ \text{Given V=80 m, height of Vikas Minar}\\ \quad \quad \quad\text{ R=120 m, shadow of Vikas Minar}\\ \quad \quad \quad \text{S=150 m, shadow of Qutub Minar}\\ \quad \quad \quad \text{Q=?, height of Qutub Minar}\\ \text{now},\dfrac { \text{V} }{ \text{R} } =\tan { \alpha } \quad \quad \\ \therefore \tan { \alpha } =\dfrac { 80 }{ 120 } =\dfrac { 2 }{ 3 } (\text{for Vikas Minar})\\ \therefore \text{for Qutab Minar}, \frac { \text{Q} }{ \text{S} } =\tan { \alpha } =\dfrac { 2 }{ 3\\ } \\ \implies \text{Q}= \dfrac { 2 }{ 3 } \times \text{S}\\ \implies \text{Q}=\dfrac{2}{3}\times 150=100~\text{m}\\ $$
Question 8
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The distance between the tops of two trees $$20\ m$$ and $$28\ m$$ high is $$17\ m$$. The horizontal distance between the two tree is

A
$$9\ m$$

B
$$11\ m$$

C
$$15\ m$$

D
$$31\ m$$

Solution

In right $$\triangle ABE$$
$$AB^2=BE^2+AE^2$$
$$AE^2=AB^2-BE^2$$
$$={17}^2-8^2$$
$$=289-64$$
$$=225$$
$$\therefore AE=\sqrt {225}$$
$$\Rightarrow AE=15\Rightarrow CD=15\ m$$
Question 9
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The angles of elevation of the top of a vertical tower from two points. 30 metros apart, and on the same straight line passing through the base of tower, are $$30^0$$ and $$60^0$$ respectively. The height
of the tower is

A
$$10$$ m

B
$$15$$ m

C
$$15\sqrt{3}$$

D
$$30$$ m

Solution

Let AB is tower, the angles of elevation of the top of a vertical tower AB from two points D and C are $$30^0$$ and $$60^0$$ respectively.

Let the height of tower be h metres

and $$BC=x m$$

In triangle $$ABC,\frac{AB}{BC}=tan 60^0$$

$$\therefore \frac{h}{x}=\sqrt{3}$$...........(i)

In triangle ABD,$$\frac{AB}{BD}=tan 30^0$$

$$\therefore \frac{h}{(x+30)}=\frac{1}{\sqrt{3}}$$

or $$\sqrt{3}h=x+30$$

Put value of x from equation Ci) in equation (ii), we get

$$\sqrt{3}h=\frac{h}{\sqrt{3}}+30$$

or $$\sqrt{3}h-\frac{h}{\sqrt{3}}=30$$

or $$\frac{2h}{\sqrt{3}}=30$$

$$\therefore h=\frac{30\sqrt{3}}{2}=15\sqrt{3}$$

So, height of the tower is $$15\sqrt{3}$$metre.
Question 10
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A man observes the elevation of a tower to be $$30^0.$$ After advancing 11 cm towards it, he finds that the elevation is $$45^0.$$ The height of the tower to the nearest metro is

A
10

B
15

C
20

D
22

Solution

Let AB the height of the tower
$$\angle ACB=45^0$$ and $$\angle D=30^0$$ ...(given)
Consider triangle ABC
$$tan45^0=1=\frac{AB}{BC}$$
Therefore $$AB=BC$$ ...(i)
Consider Triangle ABD
$$tan30^0=\frac{1}{\sqrt{3}}=\frac{AB}{BC+CD}$$
$$BC+CD=\sqrt{3}AB$$
$$AB+CD=\sqrt{3}AB$$ ...from i
$$11=AB(\sqrt{3}-1)$$
$$AB=\frac{11(\sqrt{3}+1)}{2}$$ (after rationalization)
$$AB=15.026$$
Hence answer is B