Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 26

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Question 1
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An electric pole $$10$$ m high is tied to three steel wires which are fixed to the ground at points which are at equal distances from the foot of the pole. These points, if joined form an equilateral triangle. If the steel wires are each inclined to the ground at $$60$$ $$^{\circ}$$, then each side of the equilateral triangle measures

A
$$10$$ m

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

C
$$10$$ $$\sqrt{3}$$ m

D
None of the above

Solution

Let C, D, F be the points at which are steel wires are fixed to the ground. B is the foot of the pole.
From fig. (ii), $$\cot 60^{\circ} = \cfrac{CB}{AB} \therefore \cfrac{1}{\sqrt{3}} = \cfrac{CB}{10 }$$ or $$ CB = \cfrac{10}{\sqrt{3}}$$
From fig. (i), $$\cfrac{CE}{CB} = \cos 30^{\circ}$$
$$\therefore CE = \cfrac{CB \times \sqrt{3}}{2} = \cfrac{10}{2} = 5, CD = 10$$
Question 2
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Directions For Questions

Trignometry is helpful in triangle based mathematical analysis. One of various trigonometric relation is tan (A+B) $$= \frac{tan A + tan B}{1- tan A tan B}$$.
Now consider a diagram representing a triangular roof frame ABC with a window frame EFC. BC and EF are horizontal and AD and FC are vertical.

...view full instructions

Calculate the height AD.

A
12.12

B
8.96

C
5.14

D
18.24

Solution

$$tan 75^{\circ} = \frac{AD}{2.4}$$
$$tan 75^{\circ} = tan (45^{\circ} + 30^{\circ})$$
$$\displaystyle =\frac{tan 45^{\circ} + tan 30^{\circ}}{1 - tan 45^{\circ} tan 30^{\circ}} = \frac{1+\frac{1}{\sqrt{3}}}{1- \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$$
$$=\displaystyle \frac{(\sqrt{3} + 1)^2}{3-1} = \frac{1+3 + 2 \sqrt{3}}{2} = 2 + \sqrt{3} = 3.732$$
$$AD = 2.4 \times tan 75^{\circ} = 8.96$$
Question 3
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When a eucalyptus tree is broken by strong wind, its top strikes the ground at an angle of $$30^{\circ}$$ and at a distance of $$15$$ m from the foot. What is the height of the tree?

A
$$15$$ $$\sqrt{3}$$

B
$$10$$ $$\sqrt{3}$$m

C
$$20$$ m

D
$$10$$ m

Solution

Let the height of the tree be $$A'C.$$
$$'h'$$ is the height of the tree from the broken point to the bottom.
Now, In $$\triangle BAC,$$
$$\Rightarrow \tan { 30° } =\dfrac { AC }{ BC } =\dfrac { h }{ 15 } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ 15 } $$
$$\Rightarrow h=\dfrac { 15 }{ \sqrt { 3 } } m$$
Again, $$\cos { 30° } =\dfrac { BC }{ BA } =\dfrac { 15 }{ BA } $$
$$\Rightarrow \dfrac { \sqrt { 3 } }{ 2 } =\dfrac { 15 }{ BA } $$

$$\Rightarrow BA=\dfrac { 30 }{ \sqrt { 3 } } m$$
Now, the height of tree $$=AB+AC$$
                                       $$=\dfrac { 30 }{ \sqrt { 3 } } +\dfrac { 15 }{ \sqrt { 3 } } $$

                                       $$=\dfrac { 45 }{ \sqrt { 3 } } $$

                                       $$=\dfrac { 45\sqrt { 3 } }{ 3 } $$

                                       $$=15\sqrt { 3 } m$$
Hence, the answer is $$15\sqrt { 3 } m.$$
Question 4
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STATEMENT - $$1$$: The angle of elevation of a point viewed is the angle formed by the line of sight with the horizontal when the point being view is above the horizontal level.
STATEMENT - $$2$$: Then the angle of depression of a point view is the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.

A
Statement - $$1$$ is True, Statement - $$2$$ is True, Statement - $$2$$ is a correct explanation for Statement - $$1$$

B
Statement - $$1$$ is True, Statement - $$2$$ is True : Statement $$2$$ is NOT a correct explanation for statement - $$1$$

C
Statement - $$1$$ is True, Statement - $$2$$ is False

D
Statement - $$1$$ is False, Statement - $$2$$ is True

Solution

Both the statements are true but the second statement is not a correct explanation of the first statement.
Hence, the answer is option $$B.$$
Question 5
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In a school hall, the stage is lit by a spotlight fixed to a wall. The spotlight is 4.35 metres up the wall and is set to shine on a spot on the stage 5.2 metres away from the wall, as shown in the diagram. Calculate the size of the angle marked x$$^{\circ}$$.

A
50$$^{\circ}$$.

B
45$$^{\circ}$$.

C
35$$^{\circ}$$.

D
30$$^{\circ}$$.

Solution

$$tan x = \cfrac{5.2}{4.35} = 1.195 \Rightarrow x = 50.08 = 50.1$$ or $$50$$
Some students may feel they do not know $$\tan \theta$$ value for angle $$50^{\circ}$$.
The options do not require that you remember the value, as the value of $$\tan x$$ is greater than 1.
Angle $$x$$ must be greater than $$45^{\circ}$$
Question 6
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A person, standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is $$60^{\circ}$$ when he retreats $$20$$ m from the bank, he finds the angle to be $$30^{\circ}$$. The height of the tree and the breadth of the river.

A
$$10$$ $$\sqrt{3}$$ m, $$10$$ m

B
$$10$$; $$10$$ $$\sqrt{3}$$ m

C
$$20$$ m, $$30$$ m

D
None of these

Solution

$$\textbf{Step-1: Draw the appropriate diagram & label it.}$$
$$\text{Let AB be the breadth of the river and BC be the height of the}$$ $$\text{tree which makes a}$$ $$\angle $$ $$\text{of 60}$$ $$^{\circ}$$
$$\text{at a point A on the opposite bank.}$$
$$\text{Let D be the position of the person after retreating 20 m from the bank.}$$
$$\text{Let AB}$$ $$= x$$ $$\text{metres and BC}$$ $$= h$$ $$\text{metres.}$$
$$\text{From right angled}$$ $$\triangle$$ $$\text{ABC and DBC,}$$

$$\text{we have,}$$ $$\tan 60^{\circ} = \cfrac{BC}{AB} $$ $$\text{and}$$ $$\tan 30^{\circ} = \cfrac{h}{20+x}$$
$$\Rightarrow \sqrt{3} = \cfrac{h}{x} $$ $$\text{and}$$ $$ \cfrac{1}{\sqrt{3}} = \cfrac{h}{x+20}$$
$$\textbf{Step-2: Use the above results & get the required unknown.}$$

$$\Rightarrow h = x \sqrt{3}$$ $$\text{and}$$ $$ h = \cfrac{x+20}{\sqrt{3}}$$

$$\Rightarrow x \sqrt{3} = \cfrac{x+20}{\sqrt{3}} \Rightarrow 3x = x+20 \Rightarrow x=10 m$$

$$\text{Putting}$$ $$x=10$$ $$\text{in}$$ $$h = \sqrt{3}x$$, $$\text{we get,}$$

$$h= 10 \sqrt{3} = 17.32 m$$

$$\textbf{Hence, the height of the tree = 17.32 m and the breadth of the river =10 m.}$$
Question 7
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Two poles of heights $$6$$ m and $$11$$ m stand vertically on a plane ground. If the distance between their feet is $$12$$ m, find the distance between their tips.

A
$$13$$ m

B
$$15$$ m

C
$$14$$ m

D
none of the above

Solution

From the given figure, the distance between the tips of the poles forms a right angled triangle with $$ 5 $$ m and $$ 12 m $$ as its hypotenuse.
Pythagoras theorem for right triangle,
$${Hyp}^{2} = ({Side1})^{2} + ({Side2})^{2}$$
So, required distance $$ = \sqrt {{5}^{2} + {12}^{2} } = \sqrt {25 + 144} = \sqrt {169} = 13 m $$
Question 8
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Assume the distance of the earth from the moon to be $$38,400 $$ km and the angle subtended by the moon at the eye of a person on the earth to be $${31}'$$, find the diameter of the moon.

A
The diameter of the moon is $$\displaystyle 3464\frac{8}{63}$$ km

B
The diameter of the moon is $$\displaystyle 3564\frac{8}{63}$$ km

C
The diameter of the moon is $$\displaystyle 346\frac{8}{63}$$ km

D
The diameter of the moon is $$\displaystyle 3664\frac{8}{63}$$ km

Solution

Let $$PQ$$ be the diameter of the moon and $$A$$ be the observer.
Given, $$\displaystyle \angle PAQ={31}'=\frac{31}{60}\times \frac{\pi }{180}$$ rad
Since the angle subtended by the moon is very small, its diameter will be approximately equal to the small arc of a circle whose center is the eye of the observer and the radius is the distance of the earth from the moon. Also the moon subtends an angle of $${31}'$$ at the center of this circle.
We know, $$\displaystyle \theta =\frac{\ell}{r}$$
$$\implies \displaystyle \frac{31}{60}\times \frac{\pi }{180}=\frac{PQ}{38,4000}$$
$$\displaystyle PQ=\frac{31}{60}\times \frac{22}{7\times 180}\times 38,4000=3464\frac{8}{63}$$ km
Question 9
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A man standing between two vertical posts finds that the angle subtended at his eyes by the tops of the posts is a right angle. If the height of the two posts are two times and four times the height of the man and the distance between the poles is equal to the height of the largest pole, then the ratio of the distance of the man from the shorter and the longer post is

A
$$3 :1$$

B
$$2 : 3$$

C
$$3 : 2$$

D
$$1 : 3$$

Solution

Let $$AB$$ be the men of height $$h\ m$$,

$$CD$$ and $$EF$$ be the two towers of height $$2h\ m$$ and $$4h\ m$$ respectively.

$$\therefore \angle CAE={ 90 }^{ o }$$

Draw $$PQ\parallel DF$$ passing through $$A$$.

$$\Rightarrow CP=PD=h\ m,QF=h\ m$$

$$\therefore EQ=3h\ m$$

Let $$DB$$ be $$d\ m$$

$$\therefore DB=PA=d\ m$$ and $$BF=AQ=\left( 4h-d \right) m$$

PAQ is a straight line.
$$\therefore \angle PAC+\angle CAE+\angle EAQ={ 180 }^{ o }$$

$$\angle PAC+{ 90 }^{ o }+\angle EAQ={ 180 }^{ o }$$

$$\therefore \angle PAC+\angle EAQ={ 90 }^{ o }$$

Let $$\angle PAC$$ be $$x$$.

$$\therefore \angle EAQ={ 90 }^{ o }-x$$ ...(1)

In $$\triangle CPA,\angle PAC=x$$

$$\Rightarrow \angle CPA={ 90 }^{ o }-x$$ ...(2)

In $$\triangle CPA$$ and $$\triangle AQE$$

$$\angle PCE=\angle EAQ$$ $$\left[ each{ 90 }^{ o }-x \right] $$

$$\angle CPA=\angle EAQ$$ $$\left[ each{ 90 }^{ o } \right] $$

$$\therefore \triangle CPA\sim \triangle AQE$$

$$\displaystyle\Rightarrow \frac { CP }{ AQ } =\frac { PA }{ QE } =\frac { CA }{ AE } $$ (ESST)

$$\displaystyle\therefore \frac { h }{ 4h-d } =\frac { d }{ 3h } $$

$$\Rightarrow 3{ h }^{ 2 }=d\left( 4h-d \right) $$

$$\Rightarrow 3{ h }^{ 2 }=4dh-{ d }^{ 2 }$$

$$\left( d-3h \right) \left( d-h \right) =0$$

$$d-3h=0$$ or $$d=h$$

when $$d=3h$$

$$\displaystyle\frac { DB }{ BF } =\frac { 3h }{ h } =\frac { 3 }{ 1 } =3:1$$

when $$d=h$$

$$\displaystyle\frac { DB }{ BF } =\frac { h }{ 3h } =\frac { 1 }{ 3 } =1:3$$

Henc,e option $$D$$ is correct.
Question 10
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A vertical lamp-post of height 9 meters stands at the corner of a rectangular field.The angle of elevation of its top from the farthest corner is $$\displaystyle 30^{\circ}$$,while from another corner it is $$\displaystyle 45^{\circ}.$$ The area of the field is

A
$$\displaystyle 9\sqrt{2}\: meter\:^{2}$$

B
$$\displaystyle 81\sqrt{3} \:meter\:^{2}$$

C
$$\displaystyle 81\sqrt{2}\: meter\:^{2}$$

D
$$\displaystyle 9\sqrt{3}\: meter\:^{2}$$

Solution

Consider rectangle $$ABCD$$ be the rectangular field. Let $$PD$$ be the height of vertical lamp post.

it is given that Height of the Pole $$=PD=9$$m

Now in $$\triangle PDA$$
$$tan45^o=\dfrac{PD}{DA}=\dfrac{9}{DA}\Rightarrow DA=9$$

Now in $$\triangle PDB$$
$$tan30^o=\dfrac{9}{DB}\Rightarrow DB=9\sqrt3$$

Also, we can see that $$\triangle DAB$$ is Right Angled Triangle
$$\therefore DA^2+AB^2=DB^2\Rightarrow 9^2+AB^2=(9\sqrt3)^2\Rightarrow AB^2=81\times2\Rightarrow AB=9\sqrt2$$

Now, Area of Rectangle $$=DA\times AB=9\sqrt2\times 9=81\sqrt2$$ $$m^2$$

Therefore, option (C) is the correct answer.