Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 58

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Question 1
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A straight tree breaks due to wind and the broken parts bends without getting detached such that the top of the tree touches the ground making an angle of 30$$^o$$ with the ground. The distance from the part of the tree to the point when the top touches the ground is 10 m. The height of the tree is _____.

A
10($$\sqrt3$$ - 1) m

B
10($$\sqrt3$$ + 1) m

C
10($${\displaystyle{\sqrt3}{4}}$$ - 1) m

D
10($$\sqrt3$$ m

Solution

Refer figure, let AB be the tree broken at point C
such that CD takes position CD and $$\angle$$ADC = 30$$^o$$.
Then, $$AD = 10$$ m.
Let, $$AC = x$$ m and CD $$= y$$ m
Then the height of the tree $$= (x + y)$$ m

Now, $$\displaystyle{\frac{AC}{AD} = \tan30^o \Rightarrow \frac{x}{10} \Rightarrow x = \frac{10}{\sqrt3}}$$ m

And, $$\displaystyle{\frac{AD}{CD} = \cos30^o \Rightarrow \frac{\sqrt3}{2} = \frac{10}{y} \Rightarrow y = \frac{20}{\sqrt3}}$$ m

$$\therefore$$ Height of tree $$= x + y$$
= $$\displaystyle{\frac{10}{\sqrt3} + \frac{20}{\sqrt3}}$$ m

= $$\displaystyle{\frac{30}{\sqrt3}}$$m = 10$$\sqrt3$$m

So, option D is correct.
Question 2
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The elevation of the top of a hill at each of the three angular points X, Y, Z of a horizontal $$\displaystyle \Delta XYZ$$ is $$\displaystyle \alpha $$, then the height of the hill is

A
$$\displaystyle \frac{YZ.\tan \alpha. \mathrm{cosec} \:X}{2}$$

B
$$\displaystyle X Z. \tan \alpha .\mathrm{cosec}\: Y$$

C
$$\displaystyle \frac{XY.\cot \alpha }{2\sin Z}$$

D
$$\displaystyle \frac{XY.\tan \alpha }{2\sin Y}$$
Question 3
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A tree $$10(2+\sqrt 3)$$ metres high is broken by the wind at a height $$10\sqrt 3$$ metres from its root in such a way that top struck the ground at certain angle and horizontal distance from the root of the tree to the point where the top meets the ground is $$10m$$. Find the angle of elevation made by top of the tree with the ground.

A
$${80}^{o}$$

B
$${60}^{o}$$

C
$${40}^{o}$$

D
$${20}^{o}$$

Solution

Consider the given figure,
$$AC$$
$$=10(2+\sqrt{3})-10\sqrt{3}$$
$$=20m$$
$$=Hypotenuse.$$
Hence
$$\sin(\angle C)$$
$$=\dfrac{AB}{AC}$$

$$=\dfrac{10\sqrt{3}}{20}$$

$$=\dfrac{\sqrt{3}}{2}$$
$$=\sin60^{0}$$
Hence
$$\angle C= 60^{0}$$
Hence the angle of elevation made by top of the tree with the ground is $$60^{0}$$.
Question 4
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A man is watching from the top of a tower, a boat speeding away from the tower. The boat makes an angle of depression of $$\displaystyle 45^{0}$$ with the man's eye when at a distance of 60 meters from the tower. After 5 seconds the angle of depression becomes $$\displaystyle 30^{0}$$. What is the approximate speed of the boat assuming that it is running in still water?

A
$$32$$ kmph

B
$$36$$ kmph

C
$$38$$ kmph

D
$$40$$ kmph

E
$$42$$ kmph

Solution

Let $$AB$$ be the tower and $$C$$ and $$D$$ be the point of observation from boat.
Given $$ AC =60 \text{ m}$$ and $$\angle ADB=30^{0}$$ and $$ \angle ACB=45^{0}$$

Let height of tower is $$h \text{ m}$$
In $$\Delta BAC$$,
$$\tan 45^{0}=\cfrac{AB}{AC}$$
$$\Rightarrow 1=\cfrac{AB}{60}$$
$$\Rightarrow AB=60 \text{ m}$$

In $$\Delta BAD$$,
$$\tan 30^{o}=\cfrac{AB}{AD}$$
$$\Rightarrow \cfrac{1}{\sqrt{3}}=\cfrac{60}{AD}$$
$$\Rightarrow AD=60\sqrt{3} \text{ m}$$

$$\therefore CD=AD-AC=60\sqrt{3}-60=60(\sqrt{3}-1)$$

Then, speed $$=\cfrac{60(\sqrt{3}-1)}{5} \text{ m/s}$$
And approximate speed $$=\cfrac{12\times 0.73\times 3600}{1000}\approx 32 \text{ km/h}$$
Question 5
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The angle of elevation of the top of the tower from two points at distance $$a\ m$$ and $$b\ m$$ from the base and in the same straight line with it are complementary, then the height of the tower is

A
$$\sqrt{a - b}$$ m

B
$$\sqrt{a + b}$$ m

C
$$\sqrt{ab}$$ m

D
$$\sqrt{a/b}$$ m

Solution

Let AB be the tower and C and D be the points of observations on AC.
If $$\angle$$ACB = $$\theta$$ then $$\angle$$ADB = 90$$^o$$ - $$\theta$$ and $$AB = h$$ m

Also, $$AC = a, AD = b$$ and $$CD = (a - b)$$

$$\therefore$$ tan$$\theta$$ = $$\displaystyle{\frac{AB}{AC} = \frac{h}{a}}$$, $$tan(90^o - \theta) = \displaystyle{\frac{AB}{CD} = \frac{h}{b} \Rightarrow \frac{h}{b}}$$

Also, tan$$\theta$$ . Cot $$\theta$$ = 1
$$\therefore$$ $$\displaystyle{\frac{h}{a} \times \frac{h}{b}}$$ = 1

$$\therefore h^2 = ab$$
$$\therefore h = \sqrt{ab}$$ m
$$\therefore$$ Height of the tower is $$\sqrt{ab}$$ m.

So, option C is correct.
Question 6
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$$AB$$ is a vertical pole. The end $$A$$ lies on the ground. $$C$$ is the midpoint of $$AB. P$$ is a point on the level ground such that the portion $$BC$$ subtends an angle $$\displaystyle \phi $$ at $$P$$. If $$AP = nAB$$ then the value of $$\displaystyle \cot \phi $$ is:

A
$$\displaystyle \frac{2n^{2}+1}{n}$$

B
$$\displaystyle \frac{n}{2n^{2}+1}$$

C
$$\displaystyle \frac{2n^{2}+1}{2n}$$

D
$$\displaystyle \frac{2n}{2n^{2}+1}$$

Solution

Let $$BC=AC=x$$ and $$\angle APC = \theta$$
$$\because C$$ is the mid point
$$\Rightarrow AB=2x$$
$$\Rightarrow AP=nAB=2nx$$

Now, In $$\triangle APC,$$

$$\Rightarrow \tan \theta =\dfrac{AC}{PA}=\dfrac{x}{2nx}$$

$$\Rightarrow \tan \theta =\dfrac{1}{2n}\longrightarrow (1)$$

In $$\triangle BPA,$$

$$\Rightarrow \tan { \left( \theta +\phi \right) } =\dfrac { AB }{ AP } =\dfrac { 2x }{ 2nx } =\dfrac { 1 }{ n } $$

$$\Rightarrow \dfrac { \tan { \theta } +\tan { \phi } }{ 1-\tan { \theta } \tan { \phi } } =\dfrac { 1 }{ n } $$

From $$(1),$$ using $$\tan \theta =\dfrac{1}{2n}$$

$$\Rightarrow \dfrac { \dfrac { 1 }{ 2n } +\tan { \phi } }{ 1-\dfrac { 1 }{ 2n } \tan { \phi } } =\dfrac { 1 }{ n } $$

$$\Rightarrow n\left( \dfrac { 1 }{ 2n } +\tan { \phi } \right) =1-\dfrac { 1 }{ 2n } \tan { \phi } $$

$$\Rightarrow \dfrac { 1 }{ 2 } +n\tan { \phi } =1-\dfrac { 1 }{ 2n } \tan { \phi } $$

$$\Rightarrow \tan { \phi } \left( n+\dfrac { 1 }{ 2n } \right) =1-\dfrac { 1 }{ 2 } $$

$$\Rightarrow \tan { \phi } \left( \dfrac { { 2n }^{ 2 }+1 }{ 2n } \right) =\dfrac { 1 }{ 2 } $$

$$\Rightarrow \tan { \phi } =\dfrac { 2n }{ 2\left( { 2n }^{ 2 }+1 \right) } =\dfrac { n }{ { 2n }^{ 2 }+1 } $$

$$\Rightarrow \cot { \phi } =\dfrac { 1 }{ \tan { \phi } } =\dfrac { { 2n }^{ 2 }+1 }{ n } $$

Hence, the answer is $$\dfrac { { 2n }^{ 2 }+1 }{ n }. $$
Question 7
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Four point $$A, B, C, D$$ are in a plane so that $$B$$ is in the line joining $$A$$ and $$C$$ also $$B$$ is due north of $$D$$ and $$D$$ is due west of $$C$$. $$BD = 2$$, $$\displaystyle \angle BDA = 45^{\circ}$$ $$\displaystyle \angle BCD = 30^{\circ}$$. Then $$AD$$ equal to:

A
$$\displaystyle \sqrt{3}$$

B
$$\displaystyle 2\sqrt{3}$$

C
$$\displaystyle 3\sqrt{3}$$

D
None of these

Solution

According to Given Conditions we can make the Above Triangle

In $$\triangle ABD$$
$$\dfrac{AB}{BD}=tan45^o\Rightarrow \dfrac{AB}{2}=1\Rightarrow AB=2$$

$$AD^2=AB^2+BD^2\Rightarrow AD=\sqrt{2^2+2^2}=2\sqrt2$$

Therefore, Answer is $$2\sqrt2$$
Question 8
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On the ground level, the angle of elevation of a tower is $$ 30$$$$^o$$. On moving $$20$$ m nearer, the angle of elevation is $$60$$$$^o$$. The height of the tower is _____.

A
$$20$$ m

B
$$15$$ m

C
$$10$$$$\sqrt3$$ m

D
$$10$$ m

Solution

From the given figure, AB be the tower, C and D
be the points of observation on AC such that
$$\angle BCD = 30^o$$, $$\angle BDA = 60^o$$ and $$CD = 20$$ m

Also, Let $$DA = x$$ m
Now,
tan 60$$^o$$ = $$\displaystyle{\frac{AB}{AD} \Rightarrow \frac{AB}{x} = \sqrt3 \Rightarrow AB = x\sqrt3}$$ .....1

tan 30$$^o$$ = $$\displaystyle{\frac{AB}{AC} \Rightarrow \frac{1}{\sqrt3} = \frac{AB}{20 + x} \Rightarrow AB = \frac{20 + x}{\sqrt3}}$$ .....2

$$\therefore$$ From 1 and 2,

x$$\sqrt3$$ = $$\displaystyle{\frac{20 + x}{\sqrt3} \Rightarrow 3x = 20 + x \Rightarrow x = 10 m}$$
Height=AB=$$x\sqrt3=10\sqrt3$$

So, option D is correct.
Question 9
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If the sun ray's inclination increases from $$\displaystyle 45^{\circ}$$ to $$\displaystyle 60^{\circ}$$ the length of the shadow of a tower decreases by $$50\ m$$. Find the height of the tower.

A
$$\displaystyle 25\left ( \sqrt{3} +1\right )$$

B
$$\displaystyle 25\left ( 3+\sqrt{3} \right )$$

C
$$\displaystyle 50\left ( \sqrt{3} +1\right )$$

D
$$\displaystyle 30\left ( 3+\sqrt{3} \right )$$

Solution

REF image
H is the height of the tower and BC and BD are the length of the shadow.

In $$\triangle ABC$$
$$\tan60^o=\dfrac{H}{x}\Rightarrow x=H\cot60^o=\dfrac{H}{\sqrt3}.......(1)$$

In $$\triangle ABD$$
$$\tan45^o=\dfrac{H}{x+50}\Rightarrow x+50=H\Rightarrow x=H-50.........(2)$$

From (1) and (2) we get,
$$\dfrac{H}{\sqrt3}=H-50 \Rightarrow H=\sqrt3H-50\sqrt3\Rightarrow H=\dfrac{50\sqrt3}{\sqrt3-1}=\dfrac{50\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=25(3+\sqrt3)$$

Therefore, Answer is $$25(3+\sqrt3)$$
Question 10
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The angles of elevation of an aeroplane flying vertically above the ground from two consecutive milestones(1 km) apart are 45$$^o$$ and 60$$^o$$. The height of the aeroplane from the ground is _____

A
$$\displaystyle{\frac{1}{2}}$$(3 + $$\sqrt3$$) km

B
$$\displaystyle{\frac{1}{2}}$$($$\sqrt3$$ + 1) km

C
(3 + $$\sqrt3$$) km

D
($$\sqrt3$$ + 1) km

Solution

Let A be the position of aeroplane and
B and C be the two milestones
$$\therefore$$ BC = 1 km.
Let AD = perpendicular meeting BC at D.
$$\angle$$ABD = 45$$^o$$ and $$\angle$$ACD = 60$$^o$$
let AD = h km, CD = x km
Now, tan45$$^o$$ = $$\displaystyle{\frac{h}{x + 1} \Rightarrow 1 = \frac{h}{x + 1} \Rightarrow}$$ x = h- 1

Also, tan60$$^o$$ = $$\displaystyle{\frac{h}{x} \Rightarrow \sqrt3 = \frac{h}{x} \Rightarrow x= \frac{h}{\sqrt3}}$$

$$\therefore$$$$\displaystyle{\frac{h}{\sqrt3}}$$ = h - 1 $$\therefore$$ h = $$\displaystyle{\frac{\sqrt3}{\sqrt3 - 1} = \frac{1}{2}(3 + \sqrt3)}$$ km

$$\therefore$$ The height of the aeroplane is $$\displaystyle{\frac{1}{2}}$$(3 + $$\sqrt3$$) km