Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 46

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Question 1
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The angle of elevation of a cloud from a point 100 meter above the surface of a lake is $$30^0$$ and the angle of depression of its image in the lake is $$60^0$$ then height of the cloud above the lake is

A
100 m

B
50 m

C
200 m

D
150 m

Solution

$$tan 30^0=\dfrac{h-100}{d}=\dfrac{1}{\sqrt{3}}$$......(1)
$$tan 60^0=\dfrac{h+100}{d}=\sqrt{3}$$......................(2)
Dividing equation (1) by (2) , we get
$$\dfrac{h-100}{h+100}= \dfrac{1}{3}$$
$$\Rightarrow 3h-300=h+100$$
$$\Rightarrow 2h=400$$
$$h=200 $$ metre
$$\therefore$$ Height above the lake=200 m
Question 2
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The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of $$30^o$$ with horizontal, then the length of the wire is

A
$$12$$ m

B
$$10$$ m

C
$$8$$ m

D
$$6$$ m

Solution

Given that : Heights of two poles are $$20$$ and $$14$$m respectively.
Angle wire makes with horizontal $$= 30^{\circ}$$

Let distance between two poles be $$x$$ and length of wire be $$y$$
$$tan 30^{\circ} = \dfrac{20-14}{x} = \dfrac{6}{x}$$

$$\dfrac{1}{\sqrt3} = \dfrac{6}{x}$$

$$\Rightarrow x = 6\sqrt{3}$$

As length of the wire will form the hypotenuse of right angled triangle thus formed,
$$y^{2} = 6^{2}+(6\sqrt3)^{2} = 144$$
$$\therefore y = 12$$ m
Question 3
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The angle of elevation of stationary cloud from a point 25 ml above the lake is $$ 15^0$$ and the angle of depression of reflection in the lake is $$45^0$$ .Then the height of the cloud above the level

A
$$25\sqrt{3} mt$$

B
$$ \dfrac{25\sqrt{3}}{3} mt$$

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

D
$$\dfrac{50\sqrt{3}}{\sqrt{3}+1} mt$$

Solution

$$tan15^{\circ}=\dfrac{x}{25}$$
$$\therefore x=25tan15^{\circ}$$
$$=25\left ( \dfrac{\sqrt{3}-1}{\sqrt{3}+1} \right )$$
$$\therefore $$ Height $$=25+25\left ( \dfrac{\sqrt{3}-1}{\sqrt{3}+1} \right )$$
$$=\dfrac{50\sqrt{3}}{\sqrt{3}+1}$$
Question 4
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The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle 0 with level ground such that $$tan\, \theta = \frac {15}{8}$$, how high is the kite ?

A
$$75$$ m

B
$$78.05$$ m

C
$$226$$ m

D
$$70$$ m

Solution

Given that : $$AC = 90$$m and $$tan \theta = \dfrac{15}{8} = \dfrac{BC}{AB} = \dfrac{h}{AB}$$

$$h = AB tan \theta....(1)$$

Now, In $$\triangle$$ ABC,
$$(AC)^{2} = (AB)^{2}+(BC)^{2}$$

$$\Rightarrow 90^{2} = AB^{2} + h^{2}$$
$$\Rightarrow AB = \sqrt{90^{2}-h^{2}}$$

Substituting in $$(1)$$, we get
$$h = \sqrt{90^{2}-h^{2}} tan \theta$$

$$\Rightarrow h^{2}(1+tan^{2}\theta) = 90^{2}tan^{2}\theta$$

Now substituting the value of $$tan \theta$$, we get
$$\Rightarrow h^{2} = \dfrac{8100\times225}{289}$$

$$\Rightarrow h = \dfrac{90\times15}{17} = 79.4 \approx 75$$m
Question 5
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A man observes two objects in a line in the west. On walking a distance $$x $$ towards the north, the objects subtends an angle $$\alpha$$ in front of him and on walking a further distance $$x$$ to north they subtend an angle, $$\beta$$, then the distance between the objects is

A
$$\cfrac { 2x }{ 2\cot { \beta } -\cot { \alpha } } $$

B
$$\cfrac { 3x }{ 2\cot { \beta } -\cot { \alpha } } $$

C
$$\cfrac { 3x }{ 2\cot { \beta } +\cot { \alpha } } $$

D
none of these

Solution

$$ \tan p = \dfrac{2x}c$$ $$ \tan q = \dfrac{x}c$$
$$\tan a = \dfrac{2x}{c+d}$$ $$ \tan b = \dfrac{x}{c+d}$$

$$\beta + a = p$$

$$\alpha + b = q$$

hence $$ \tan\beta$$ = $$\tan ({p-a})$$

$$ \tan\alpha$$ = $$\tan ({q-b})$$

so

$$\tan\beta = \dfrac{\tan p-\tan a}{1-\tan p\tan a}$$

$$ = \dfrac{\dfrac{2x}{c}-\dfrac{2x}{c+d}}{1+\dfrac{2x}c\dfrac{2x}{c+d}}$$

$$=\dfrac{2xd}{c^2+dc+4x^2}$$

$$c^2+dc+4x^2= 2xd\cot\beta$$ ...........(1)

$$\tan\alpha = \dfrac{\tan q-\tan b}{1-\tan q\tan b}$$

$$ = \dfrac{\dfrac{x}{c}-\dfrac{x}{c+d}}{1+\dfrac{x}c\dfrac{x}{c+d}}$$

$$=\dfrac{xd}{c^2+dc+x^2}$$

$$c^2+dc+x^2= xd\cot\alpha$$ ...........(2)

From 1 and 2

$$3x^2= xd(2\cot\beta-\cot\alpha)$$

$$\Rightarrow d = \dfrac{3x}{2\cot\beta - \cot\alpha}$$
Question 6
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A sphere with centre $$O$$ sits atop of pole as shown in the figure. An observer on the ground is at a distance $$50m$$ from the foot of the pole. She notes that the angles of elevation from the observer to points $$P$$ and $$Q$$ on the sphere are $$30^{\circ}$$ and $$60^{\circ}$$, respectively. Then, the radius of the sphere in meters is

A
$$100\left (1 - \dfrac {1}{\sqrt {3}}\right )$$

B
$$\dfrac {50\sqrt {6}}{3}$$

C
$$50\left (1 - \dfrac {1}{\sqrt {3}}\right )$$

D
$$\dfrac {100\sqrt {6}}{3}$$

Solution

in $$\triangle PST$$
$$\tan30^o=\dfrac{PS}{50}\Rightarrow PS=\dfrac{50}{\sqrt3}$$

Now, in $$\triangle QUT$$
$$QU=PS+r$$ and $$UT=50-r$$
$$\tan60^o=\dfrac{QU}{US}\Rightarrow \sqrt3=\dfrac{PS+r}{50-r}\Rightarrow r=\dfrac{50\sqrt3-PS}{\sqrt3+1}\Rightarrow r=\dfrac{50\sqrt3-\dfrac{50}{\sqrt3}}{\sqrt3+1}=50\Bigg(1-\dfrac{1}{\sqrt3}\Bigg)$$

Therefore, Answer is $$50\Bigg(1-\dfrac{1}{\sqrt3}\Bigg)$$
Question 7
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One side of a rectangular piece of paper is $$6\text{ cm}$$, the adjacent sides being longer than $$6\text{ cm}$$. One corner of the paper is folded so that it sets on the opposite longer side. If the length of the crease is $$l$$ $$\text{cm}$$ and it makes an angle $$\theta$$ with the long side as shown, then $$l$$ is

A
$$\cfrac { 3 }{ \sin ^{ 2 }{ \theta } \cos { \theta } } $$

B
$$\cfrac { 6 }{ \sin ^{ 2 }{ \theta } \cos { \theta } } $$

C
$$\cfrac { 3 }{ \sin { \theta } \cos { \theta } } $$

D
$$\cfrac { 3 }{ \sin ^{ 3 }{ \theta } } $$

Solution

Let EC $$=x$$
In $$\Delta BCE$$,
$$\displaystyle\frac{x}{l}=\sin \theta \Rightarrow x=l\sin \theta$$
$$EF=x=l\sin \theta$$
In $$\Delta ABF,$$
$$\angle ABF=90-2\theta$$
$$\angle AFB=2\theta$$
$$\sin 2\theta =\displaystyle\frac{6}{l\sin \theta}$$
$$\Rightarrow l=\displaystyle\frac{6}{\sin\theta .\sin 2\theta}$$
$$=\displaystyle\frac{6}{\sin \theta(2\sin \theta\cdot \cos\theta)}$$
$$l=\displaystyle\frac{3}{\sin^2\theta \cdot \cos\theta}$$.
Question 8
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An observer standing at a point $$P$$ on the top of a hill near the sea-shore notices that the angle of depression of a ship moving towards the hill in a straight line at constant speed is $${30}^{o}$$. After 45 minutes, this angle becomes $${45}^{o}$$. If T (in minutes) is the total time taken by the ship to move to a point in the sea where the angle of depression from $$P$$ of the ship is $${60}^{o}$$, then T is equal to:

A
$$ \dfrac{45}{\sqrt{3}}-45$$

B
$$45\left( 1+\cfrac { 2 }{ \sqrt { 3 } } \right) $$

C
$$45\left( 2+\cfrac { 1 }{ \sqrt { 3 } } \right) $$

D
$$45\left( 1+\cfrac { 1 }{ \sqrt { 3 } } \right) $$

Solution

We know that, time taken to travel between any points is proportional to distance covered.
Hence, $$\dfrac {T_{AC}}{T_{AB}}=\dfrac {AC}{AB}$$ [uniform speed]

In $$\triangle PQA$$, $$\dfrac {PQ}{AQ}=\tan 30$$
$$\Rightarrow AQ=\sqrt 3h$$ ....(1)

In $$\triangle PBQ$$, $$\tan 45=\dfrac {PQ}{BQ}$$
$$\Rightarrow BQ=h$$ ....(2)

Similarly in $$\triangle PQC$$, $$\tan 60=\dfrac {PQ}{CQ}$$
$$\Rightarrow CQ=\dfrac {h}{\sqrt 3}$$

Now $$\dfrac {AC}{AB}=\dfrac {AQ-CQ}{AQ-BQ}$$

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

$$=\dfrac {\sqrt 3-\frac {1}{\sqrt 3}}{\sqrt 3-1}$$

$$=1+\dfrac {1}{\sqrt 3}$$

Thus $$T_{AC}=T_{AB}\times \left(1+\dfrac {1}{\sqrt 3}\right)$$

$$\Rightarrow T=45\left(1+\dfrac {1}{\sqrt 3}\right)$$

Hence, the time taken is $$45\left(1+\dfrac {1}{\sqrt 3}\right)$$.
Question 9
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The height of a house subtends a right angle at opposite window from the base of the house is $$60^0$$. If the width of the road be 6 metres, then the height of the house is

A
$$6 \sqrt 3 m$$

B
$$8 \sqrt 3 m$$

C
$$10 \sqrt 3 m$$

D
$$3 \sqrt 6$$

Solution

Given $$\angle DBC=60^0$$ and $$\angle ADB=90^0$$ from figure we know that $$\angle EDB=\angle DBC=60^0$$
Since $$\angle ADB=\angle ADE+\angle EDB=90^0$$
$$\angle ADE =90^0-60^0=30^0$$
From AED ,
$$\tan30=\dfrac{AE}{6}$$
$$AE-2\sqrt3 m$$
From EDC,
$$\tan60=\dfrac{EB}{6}$$
$$EB=6\sqrt 3m$$
Height of the house =$$AE+EB=2\sqrt3+6\sqrt3=8\sqrt3 m$$
Question 10
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From the top of aspire the angle of depression of the top and bottom of a tower of height h are $$\theta$$ and $$\phi$$ respectively. Then height of the spire and its horizontal distance from the tower are respectively.

A
$$\dfrac{h \cos\theta \sin\phi}{\sin(\phi- \theta)}$$ and $$\dfrac{h \cos\theta \cos\phi}{\sin(\phi- \theta)}$$

B
$$\dfrac{h \cos\theta \cos\phi}{\sin(\theta + \phi)}$$, $$\dfrac{h \tan\theta cos\phi}{\sin(\theta + \phi)}$$

C
$$\dfrac{h \sin\theta sin\phi}{\sin(\theta + \phi)}$$, $$\dfrac{h \cos\theta \cos\phi}{\sin(\theta + \phi)}$$

D
None of these

Solution

Given that $$\angle ECD=\theta,\angle ECA=\phi,AD=h $$
From Geometry $$\angle ECD=\theta =\angle FDC ,\angle ECA=\phi =\angle BAC $$
$$\tan \theta =\dfrac{BC-h}{AB}$$ .............. $$(1)$$
$$\tan \phi=\dfrac{BC}{AB}\Rightarrow BC=AB\tan\phi$$ .......... $$(2)$$
Substitute equation $$(2)$$ in $$(1)$$
$$AB\tan\theta=AB\tan\phi -h$$
Distance betweeen tower and sphire $$AB=\dfrac{h}{\tan\phi-\tan \theta}=\dfrac{h}{\dfrac{\sin\phi}{\cos\phi}-\dfrac{\sin\theta}{\cos\theta}}=\dfrac{h\cos\theta\cos\phi}{\sin(\phi-\theta)}$$
Height of the spire $$BC=AB\tan\phi=\dfrac{h\cos\theta \cos\phi}{\sin(\phi-\theta)}\times \tan\phi=\dfrac{h\cos\theta\sin\phi}{\sin(\phi-\theta)}$$