Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 59

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Question 1
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A tree breaks due to a storm and the broken part bends down such that the top of the tree touches the ground making an angle of $$30$$$$^o$$ with the ground at a distance of $$30$$ m. Then, the height of the tree is:

A
$$35$$$$\sqrt3$$

B
$$60$$$$\sqrt3$$

C
$$45$$$$\sqrt3$$

D
$$30$$$$\sqrt3$$

Solution

Let AB be the tree broken at point C such that
the broken part CB takes the position CO and touches the at O.
Then, OA = 30 m & $$\angle$$COA = 30$$^o$$
In $$\triangle$$OAC, tan 30$$^o$$ = $$\displaystyle{\frac{AC}{OA} \Rightarrow \frac{1}{\sqrt3} = \frac{AC}{30} \Rightarrow AC = \frac{30}{\sqrt3} = 10\sqrt3}$$
and,
cos 30$$^o$$ = $$\displaystyle{\frac{OA}{OC} \Rightarrow \frac{\sqrt3}{2} = \frac{30}{OC} \Rightarrow OC = \frac{60}{\sqrt3} \Rightarrow OC = 20\sqrt3}$$

Thus, the height of the tree $$= AC + CB$$
$$= AC + OC$$ .... ($$\because$$ BC = OC)
$$= 10$$$$\sqrt3$$ + 20$$\sqrt3$$
$$= 30$$$$\sqrt3$$ m
Question 2
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Alex observed that the angle of elevation to the top of $$800-foot$$ Mount Colin was $$23^{\circ}$$. To the nearest foot, how much closer to the base of Mount Colin must Alex move so that his angle of elevation is doubled?

A
$$200$$

B
$$400$$

C
$$489$$

D
$$1112$$

Solution

Let the initial position of Alex be C and final position be D.
Height of tower $$=800ft$$
Angle of elevation $$=\angle ACD$$
$$=23^o$$
In triangle $$\triangle ABC$$
$$\tan 23^o=\dfrac{AB}{BC}$$
$$\Rightarrow 0.424=\dfrac{800}{BC}$$
$$\Rightarrow BC=\dfrac{800}{0.424}=1886.8$$
In $$\triangle ABD$$
$$\tan D=\dfrac{AB}{BD}$$
$$\Rightarrow \tan 46 =\dfrac{800}{BD}$$
$$\Rightarrow BD=\dfrac{800}{\tan 46}=\dfrac{800}{1.035}$$
$$=772.95$$
Distance to be travelled $$DC=BC-BD$$
$$=1886.8-772.95$$
$$=1113.85 \approx 1112$$
Question 3
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A 15 m high eucalyptus tree standing erect on the ground breaks at the height of 5 m from it. The broken part bends such that the top of the tree touches the ground. Find the angle made by the broken part of the tree with the ground.

A
30$$^o$$

B
45$$^o$$

C
60$$^o$$

D
90$$^o$$

Solution

Based on the given information, we can draw the figure given above.
Here, $$AC$$ represents the tree when erect and $$BD$$ represents the broken part of the tree.
While $$BC$$ represents the remaining erect tree.
$$DC$$ is the distance between the top of the tree and the base of the tree.
$$\therefore\ \angle C = 90^o, BC = 5\ m,\ AC = 15\ m$$
$$BD = AB = AC - BC$$
$$ = 10\ m$$
We know that, $$\sin \theta=\dfrac{\text{Opposite Side}}{\text{Hypotenuse}}$$

Hence, $$\sin \theta=\displaystyle{\frac{BC}{BD} = \frac{5}{10}}$$
$$ = \dfrac{1}{2}$$

We know that $$\sin 30^o=\dfrac12$$.

Hence, $$\sin \theta =\sin 30^o$$

$$\therefore\ \theta=30^o$$

Hence, the broken part of the tree makes an angle of $$30^o$$ with the ground.
Question 4
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A tower $$AB$$ leans towards west marking an angle $$\alpha$$ with the vertical. The angular elevation of $$B$$, the top most point of the tower is $$\beta$$ as observed from a point $$C$$ due east of $$A$$ at a distance $$'d'$$ from $$A$$. If the angular elevation of $$B$$ from a point $$D$$ due east of $$C$$ at a distance $$2d$$ from $$C$$ is $$r$$, then $$2\tan \alpha$$ can be given as

A
$$3\cot \beta - 2\cot \gamma$$

B
$$3\cot \gamma - 2\cot \beta$$

C
$$3\cot \beta - \cot \gamma$$

D
$$\cot \beta - 3\cot \gamma$$

Solution

By $$m - n$$ theorem at $$C$$
$$(d + 2d) \cot \beta = d\cot \gamma - 2d\cot (90^{\circ} + \alpha)$$
$$3d\cot \beta = d\cot \gamma + 2d \tan \alpha$$
$$\Rightarrow 3\cot \beta = \cot \gamma + 2\tan \alpha$$
$$\therefore 2\tan \alpha = 3\cot \beta - \cot \gamma$$
Question 5
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The angle of elevation and the angle of depression are $$30^o$$ and $$30^o$$ respectively when seen from the top of the first building to the top and base of the second building. If the distance between the bases of two building is 12 m, then find the height of big building.

A
$$16 \sqrt {3} m$$

B
$$12 \sqrt {3} m$$

C
$$14 \sqrt {3} m$$

D
$$20 \sqrt {3} m$$

Solution

Given that the angle of elevation is $$30^{0}$$ and the angle of depression is $$30^{0}$$ when seen from the top of first building to the top and base of second building.
Let the height of first building be $$x$$ and the height of second building be $$y$$
From angle of depression ,we have $$tan(30^{0})=\frac{12}{x}$$
$$\Rightarrow x=12 cot(30^{0})=12 \sqrt3$$
From angle of elevation , we have $$tan(30^{0})=\frac{y-x}{12}$$
$$\Rightarrow y-x=12 tan(30^{0})=4 \sqrt3$$
$$\Rightarrow y=x+ 4 \sqrt3=16 \sqrt3$$
Question 6
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The angle of elevation of the top of a vertical tower from two points at distances $$a$$ and $$b$$ $$(a>b)$$ from the base and in the same line with it, are complimentary. If $$\theta$$ is the angle subtended at the top of the tower by the line joining these points, then $$\sin{\theta}$$ is equal to

A
$$\cfrac { a+b }{ a-b } $$

B
$$\cfrac { a-b }{ a+b } $$

C
$$\cfrac { (a-b)b }{ a+b } $$

D
$$\cfrac { a-b }{ (a+b)b } $$

Solution

$$ In\quad \triangle CAD,\\ \tan { \phi } =\frac { AC }{ b } \quad \_ \_ \_ (1) $$
$$ In\quad \triangle CAB,\\ \tan { (90-\phi ) } =\dfrac { AC }{ a } \quad \\ \Rightarrow \cot { \phi } =\frac { AC }{ a } \quad \_ \_ \_ \_ (2) $$
From (1) and (2),
$$\Rightarrow \dfrac { AC }{ b } =\dfrac { a }{ AC } $$
$$ \Rightarrow{ AC }^{ 2 }=ab $$
$${ \Rightarrow AC }=\sqrt { ab } $$
$$ \sin { \theta } =-\sin { (90-2\phi )=-\cos { 2\phi } } $$
$$ =-1+2\sin ^{ 2 }{ \phi } $$
$$ =-1+\frac { 2a }{ a+b } $$
As $$\sin { \phi } =\sqrt { \dfrac { ab }{ ab+{ b }^{ 2 } } }$$ and $$ \sin ^{ 2 }{ \phi } =\frac { a }{ a+b } $$ we get $$\sin \theta = \dfrac{a-b}{a+b}$$
Question 7
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Elevation angle of the top of the mirror from the foot of the tower of height $$h$$ is $$\alpha$$ and the tower subtend an angle $$\beta$$ at the top of the mirror. Then, height of mirror is

A
$$\dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) - \cot \alpha}$$

B
$$\dfrac {h\tan (\alpha - \beta)}{\tan (\alpha - \beta) - \tan \alpha}$$

C
$$\dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) + \cot \alpha}$$

D
None of the above

Solution

Let $$AB$$ be the tower of height $$h$$ and $$PQ$$ be the mirror of height $$x.$$
$$\therefore \angle PBQ = \alpha$$ and $$\angle APB = \beta$$

In $$\triangle APB,$$
$$\angle PBA = 90^o-\alpha$$
$$\angle PAB = \angle PAC+90^o$$

So, by using angle sum property,
$$\angle APB+\angle PBA+\angle PAB=180^o$$
$$\Rightarrow \beta+90^o-\alpha+\angle PAC+90^o=180^o$$
$$\Rightarrow \beta-\alpha+\angle PAC+180^o=180^o$$
$$\Rightarrow \angle PAC=\alpha-\beta$$

In right angle triangle $$PBQ,$$
$$\dfrac {x}{BQ} = \tan \alpha ..... (i)$$

And, in right angle triangle $$PCA,$$
$$\dfrac {x - h}{AC} = \tan (\alpha - \beta)$$
$$\Rightarrow \dfrac {x - h}{BQ} = \tan (\alpha - \beta) .... (ii)$$

Now divide equation $$(ii)$$ by $$(i),$$
$$\dfrac {x - h}{x} = \dfrac {\tan (\alpha - \beta)}{\tan \alpha}$$
$$\Rightarrow 1 -\dfrac {h}{x} = \dfrac {\cot \alpha}{\cot (\alpha - \beta)}$$
$$\Rightarrow x = \dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) - \cot \alpha}$$

Hence, option $$A$$ is correct.
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 $$p$$ and $$q$$, respectively. The height of hill is ?

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

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

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

D
None of these

Solution

Let $$AD = h$$ is the height of the building and $$EB$$ is height of the hill.
and $$\tan q = \dfrac {h + x}{y}$$
and $$\tan p = \dfrac {x}{y}$$
$$\Rightarrow y = x\cot p$$
Now, $$\tan q = \dfrac {h + x}{x\cot p}$$
$$\Rightarrow x \cot p = (h + x) \cot q$$
$$\Rightarrow x = \dfrac {h\cot q}{\cot p - \cot q}$$
$$\therefore h + x = h + \dfrac {h\cot q}{\cot p - \cot q}$$
$$= \dfrac {h\cot p}{\cot p - \cot q}$$
Question 9
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The angle of elevation of a jet fighter from a point $$A$$ on the ground is $$60$$. After a flight of $$15$$ seconds, the angle of elevation changes to $$30$$. If the jet is flying at a speed of $$720km/hr$$, find the constant height in m.

A
$$2598 m$$

B
$$2690 m$$

C
$$2355 m$$

D
$$2405 m$$

Solution

Let $$A$$ be the point of observation on the ground and $$B$$ and $$C$$ be two positions of the jet.
Let $$BL=CM=h$$metres.
Let $$AL=x$$ metres.
Speed of the jet$$=720$$ km\hr$$=720 \times \cfrac{5}{18}=200$$ m\s.
Time taken to cover the distance $$BC$$ is $$15$$ sec.
Distance $$BC=LM=200 \times 15$$ m=$$3000$$m
In $$\triangle ALB$$,
$$\cot 60^{\circ}=\cfrac{AL}{BL}=\cfrac{1}{\sqrt 3}=\cfrac{x}{h}$$
$$\therefore x=\cfrac{h}{\sqrt 3}$$-------------------------1
In $$\triangle AMC$$,
$$\cot 30^{\circ}=\cfrac{AM}{MC}$$
$$\Rightarrow \sqrt 3=\cfrac{AL+LM}{MC}$$
$$\Rightarrow \sqrt 3=\cfrac{x+3000}{h}$$
$$\Rightarrow h\sqrt 3-3000=x$$--------------------------2
From Equation 1 & 2,
$$\cfrac{h}{\sqrt 3}=h\sqrt 3-3000$$
$$\Rightarrow h=3h-\sqrt 3\times 3000$$
$$\Rightarrow 2h=5196$$
$$\Rightarrow h=2598m$$
Question 10
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The angles of depression of the top and the bottom of a 7 m tall building from the top of a tower are $$45^0$$ and $$60^0$$ respectively.Find the height of the tower in metres.

A
$$7(3+\sqrt{3})$$

B
$$\dfrac{7}{2}(3-\sqrt{3})$$

C
$$\dfrac{7}{2}(3+\sqrt{3})$$

D
$$7(3-\sqrt{3})$$

Solution

Height of the building $$AB=7\:m$$

Let height of the tower be $$PC$$ and the distance between the building and the tower be $$AC$$.

Angle of depression to top of building is $$\angle {QPB}=45^o$$

Angle of depression to bottom of building is $$\angle {QPA}=60^o$$

We have to find the height of the tower $$i.e., PC$$.

Draw $$BD$$ parallel to $$AC \: and\: PQ$$

Since $$PQ || BD, \angle{PBD}=\angle{QPB}\Rightarrow \angle{PBD}=45^o$$

Since $$PQ || AC, \angle{PAC}=\angle{QPA}\Rightarrow \angle{PAC}=60^o$$

Now, $$AC,BD$$ are parallel lines $$\Rightarrow AC=BD$$

Also, $$AB,CD$$ are parallel lines $$\Rightarrow AB=CD$$. $$\therefore CD=7\:m$$

Since $$PC \:perpendicular \:to\:AC$$, $$\angle{PDB}=\angle{PCA}=90^o$$

In right triangle $$PBD, tan\:B=\dfrac{PD}{BD}$$

$$\Rightarrow Cot\:45=\dfrac{BD}{PD}$$

$$\Rightarrow 1=\dfrac{BD}{PD}$$

$$\Rightarrow BD=PD \quad ...(1)$$

In right triangle $$PAC, tan\:A=\dfrac{PC}{AC}$$

$$\Rightarrow cot\:60=\dfrac{AC}{PC}$$

$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{AC}{PC}$$

$$\Rightarrow AC=\dfrac{PC}{\sqrt{3}} $$

$$\Rightarrow BD=\dfrac{PC}{\sqrt{3}} (\because AC=BD) \quad ...(2)$$

From $$(1), (2)$$ we get,

$$PD=\dfrac{PC}{\sqrt{3}}$$

$$\Rightarrow PD=\dfrac{PD+DC}{\sqrt{3}}$$

$$\Rightarrow \sqrt{3}PD=PD+DC$$

$$\Rightarrow (\sqrt{3}-1)PD=7$$

$$\Rightarrow PD=\dfrac{7}{\sqrt{3}-1}$$

Rationalising we get $$PD=\dfrac{7(\sqrt{3}+1)}{3-1}=\dfrac{7(\sqrt{3}+1)}{2}$$

Now, $$PC=PD+DC$$

$$=\dfrac{7(\sqrt{3}+1)}{2}+7$$

$$=\dfrac{7\sqrt{3}+7+14}{2}$$

$$=\dfrac{7\sqrt{3}+21}{2}$$

$$PC=\dfrac{7(\sqrt{3}+3)}{2}$$

Hence the height of the tower is $$\dfrac{7(\sqrt{3}+3)}{2}\: m$$