Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 43

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Question 1
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The top of a hill when observed from the top and bottom of a building of height h is at angles of elevation p and q respectively. What is the height of the hill?

A
$$\dfrac{h \cot q}{\cot q - \cot p}$$

B
$$\dfrac{h \cot p}{\cot p - \cot q}$$

C
$$\dfrac{2 h \tan p}{\tan p - \tan q}$$

D
$$\dfrac{2 h \tan q}{\tan q - \tan p}$$

Solution

Let the height of hill $$=H$$
The distance between hill and building $$=x$$
From $$\triangle ABC$$
$$\tan p=\dfrac{H-h}{x}$$
From $$\triangle CED$$
$$\tan q=\dfrac{H}{x}$$
$$\therefore \dfrac{\tan q}{\tan p}=\dfrac{\dfrac{H}{x}}{\dfrac{H-h}{x}}$$

$$\therefore \dfrac{\tan q}{\tan p}=\dfrac{H}{H-h}$$

$$\therefore \dfrac{\tan q}{\tan p}-1=\dfrac{H}{H-h}-1$$ (subtracting $$1$$ from both the sides)

$$\therefore \dfrac{h}{H-h}=\dfrac{\tan q-\tan p}{\tan p}$$

$$\therefore \dfrac{H-h}{h}=\dfrac{\tan p}{\tan q-\tan p}$$

$$\therefore \dfrac{H-h}{h}+1=\dfrac{\tan p}{\tan q-\tan p}+1$$ (adding $$1$$ both the sides)

$$ \therefore \dfrac{H}{h}=\dfrac{\tan p+(\tan q-\tan p)}{\tan q-\tan p}$$

$$\therefore \dfrac{H}{h}=\dfrac{\tan q}{\tan q-\tan p}$$

$$\therefore \dfrac{H}{h}=\dfrac{\cot p}{\cot p-\cot q}$$ (on putting $$\tan \theta=\dfrac{1}{\cot \theta}$$)

$$\therefore H=\dfrac{h\cot p}{\cot p-\cot q}$$

Answer is option (B)
Question 2
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The angle of elevation of the tip of a flag staff from a point 10 m due South of its base is $$60^o$$ . What is the height of the flag staff correct to the nearest meter ?

A
15 m

B
16 m

C
17 m

D
18 m

Solution

Here, let $$h$$ be the height of the flag, then
$$\tan(60^{o})=\dfrac { h }{ 10 } \Rightarrow h=10\sqrt { 3 } =17.3 $$
$$h$$ is nearest to $$17$$
Hence, height of the flag is $$17$$ m.
Question 3
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The angles of elevation of the top of a tower from two places situated on the opposite sides of the base at distances $$21 \text{m}$$ and $$x \ \text{m} $$ from the base of the tower are $${45}^{o}$$ and $${60}^{o}$$ respectively. What is the value of $$x$$?

A
$$7\sqrt { 3 } $$

B
$$7 - \sqrt { 3 } $$

C
$$7 + \sqrt { 3 } $$

D
$$14$$

Solution

Let B and C be those two points situated at the given distance and A be the top of the tower.
In $$\triangle ABD,$$
$$\tan 45^{o}= \dfrac{AD}{BD}$$

$$1 = \dfrac { h }{ 21 }$$

$$\Rightarrow h=21m$$

In $$\triangle ADC,$$
$$\tan 60^{o}= \dfrac{AD}{DC}= \dfrac { h }{ x } $$

$$\Rightarrow x=\dfrac { h }{\tan 60^{o}}$$

$$x = \dfrac { 21 }{ \sqrt { 3 } } =7\sqrt { 3 } $$
Hence, option A is correct.
Question 4
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What is the angle subtended by 1 m pole at a distance 1 km on the ground in sexagesimal measure ?

A
$$\frac{9} {50 \pi}$$ degree

B
$$\frac{9} {5 \pi}$$ degree

C
3.4 minute

D
3.5 minute
Question 5
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From the top of a lighthouse $$70m$$ high with its base at sea level, the angle of depression of a boat is $${ 15 }^{ o }$$. The distance of the boat from the foot of the lighthouse is:

A
$$70(2-\sqrt { 3 } )m$$

B
$$70(2+\sqrt { 3 } )m$$

C
$$70(3-\sqrt { 3 } )m$$

D
$$70(3+\sqrt { 3 } )m$$

Solution

Let AB be the light house and the boat is at point C
Angle BCA is also equal to angle of depression $$=15^{o}$$
$$\Rightarrow \angle BCA=15^{o}$$
Now, $$\tan (15^{o})=\dfrac { AB }{ BC }$$
$$\Rightarrow BC=AB(\tan (15^{o}))\\ \tan (15^{o})=\tan (60^{o}-45^{o})\\ =\dfrac { \tan 60^{o}-\tan 45^{o} }{ 1+\tan 60^{o}\tan 45^{o} } =\dfrac { \sqrt { 3 } -1 }{ 1+\sqrt { 3 } } =\dfrac { { (\sqrt { 3 } -1) }^{ 2 } }{ 3-1 } =2-\sqrt { 3 }$$
So, $$BC=70(2-\sqrt { 3 } )m$$
Option A is correct
Question 6
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The angle of elevation of the top of a tower from a point $$20\ m$$ away from its base is $$45^{\circ}$$. What is the height of the tower?

A
$$10\ m$$

B
$$20\ m$$

C
$$30\ m$$

D
$$40\ m$$

Solution

$$BC$$ is the tower
In triangle $$ABC$$
$$\tan A=\dfrac { BC }{ AB } \\ \Longrightarrow \tan 45^{o}=\dfrac { x }{ 20 } \\ \Longrightarrow 1=\dfrac { x }{ 20 } \\ \Longrightarrow x=20m$$
Question 7
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A person standing on the bank of a river observes that the angle subtended by a tree on the opposite of bank is $${60}^{o}$$. When he retires $$40 m$$ from the bank, he finds the angle to be $${30}^{o}$$. What is the breadth of the river?

A
$$60 m$$

B
$$40 m$$

C
$$30 m$$

D
$$20 m$$

Solution

Let $$R$$ br the length of River
and $$T$$ be the length of tree

In $$\triangle ABC$$
$$\tan60^o=\dfrac{T}{R}\Rightarrow T=\sqrt3R.......(1)$$

in $$\triangle ABD$$
$$\tan30^o=\dfrac{T}{R+40}\Rightarrow T=\dfrac{R+40}{\sqrt3}.....(2)$$

From (1) and (2)
$$\sqrt3R=\dfrac{R+40}{\sqrt3}\Rightarrow 3R=R+40\Rightarrow R=20$$

Therefore, Answer is $$20m$$
Question 8
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The top of a hill observed from the top and bottom of a building of height $$h$$ is at angles of elevation $$\alpha$$ and $$\beta$$ respectively. The height of the hill is

A
$$\cfrac { h\cot { \beta } }{ \cot { \beta } -\cot { \alpha } } $$

B
$$\cfrac { h\cot { \alpha } }{ \cot { \alpha } -\cot { \beta } } $$

C
$$\cfrac { h\tan { \alpha } }{ \tan { \alpha } -\tan { \beta } } $$

D
None of the above

Solution

Refer the image
AB is the hill and CD is the building
$$\tan (\alpha )=\dfrac { AB }{ BE } \\ \implies BE=AB\cot (\alpha )\\ \tan (\beta )=\dfrac { AB }{ BD } \\ \implies BD=AB\cot (\beta )$$
Also, $$\tan (\alpha )=\dfrac { CD }{ DE }$$
$$\implies DE=CD\cot (\alpha )=h\cot (\alpha )$$
Now, $$BE=BD+DE$$
$$\Rightarrow AB\cot (\alpha )=AB\cot (\beta )+h\cot (\alpha )\\ \Rightarrow AB=\dfrac { h\cot (\alpha ) }{ \cot (\alpha )-\cot (\beta ) } $$
Option B is correct
Question 9
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The shadow of a pole is $$\sqrt {3}$$ times longer. The angle of elevation is equal to :

A
$$40^{\circ}$$

B
$$\dfrac {45^{\circ}}{2}$$

C
$$60^{\circ}$$

D
$$30^{\circ}$$

Solution

Height of pole $$= h$$
$$\therefore $$ shadow of pole $$=\sqrt { 3 } h$$
Angle of elevation, $$\theta $$
$$\tan\theta =\dfrac { h }{ \sqrt { 3 } h } =\dfrac { 1 }{ \sqrt { 3 } } $$

$$\theta =\tan^{ -1 }\dfrac { 1 }{ \sqrt { 3 } } $$
$$ { \theta =3{ 0 }^{ o } } $$
Question 10
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The upper $$\dfrac{3}{4}th$$ portion of a vertical pole subtends an angle $$\tan^{-1}\dfrac{3}{5}$$ at a point in the horizontal plane through its foot and at a distance $$40\ m$$ from the foot. The possible height of the vertical pole is

A
$$40\ m$$ or $$100\ m$$

B
$$160\ m$$ or $$40\ m$$

C
$$100\ m$$ or $$120\ m$$

D
$$200\ m$$ or $$80\ m$$

Solution

Given, $$\theta_2=\tan^{-1}\dfrac{3}{5}\Rightarrow \tan \theta_2=\dfrac{3}{5}$$ ...(i)

In $$\triangle AOC, \tan \theta_1=\dfrac{AC}{AO} = \dfrac{h}{160}$$ ...(ii)

and in $$\triangle AOB, \tan (\theta_1+\theta_2) = \dfrac{h}{40}$$

$$\Rightarrow \dfrac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1\tan\theta_2}=\dfrac{h}{40}$$

$$\Rightarrow \dfrac{\dfrac{h}{160}+\dfrac{3}{5}}{1-\dfrac{h}{160}\times \dfrac{3}{5}} = \dfrac{h}{40}$$ [from Eqs. (i) and (ii)]

$$\Rightarrow \dfrac{5(h+96)}{800-3h} = \dfrac{h}{40}$$

$$\Rightarrow h^2-200h+6400=0$$

$$\Rightarrow (h-160)(h-40)=0$$

$$\Rightarrow h = 160 \,or \,h = 40$$

Hence, the height of the vertical pole is $$40\ m$$