Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 38

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Question 1
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A pole $$6\ \text{m}$$ high casts a shadow of $$2$$$$\sqrt 3\text{ m}$$ on the ground. At that instance, the sun's elevation is :

A
$$30$$$$^{\circ}$$

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

C
$$60$$$$^{\circ}$$

D
$$90$$$$^{\circ}$$

Solution

Let the sun's elevation be $$\theta$$
In $$\triangle ABC, \angle A =90^{\circ}$$,

$$\therefore \tan \theta = \dfrac {AB}{AC} = \dfrac {6}{2\sqrt 3}$$

$$=\dfrac {2 \times 3}{2 \times \sqrt 3} = \sqrt 3$$

$$\therefore \theta = 60^{\circ}$$
Question 2
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From the base of a tower the angle of elevation of the top of the cliff is 60$$^o$$ and from the base of the cliff, the angle of elevation of the top a tower is 30$$^o$$. If the height of the tower is 50 m, then the height of the cliff is :

A
$$50\sqrt3$$ m

B
$$150$$ m

C
$$100$$m

D
$$50$$ m

Solution

Let AB be the cliff and CD be the tower.
Then, $$\angle$$ BDA = 60$$^o$$ and $$\angle$$CBD = 30$$^o$$.
Also, $$CD = 50$$ m.

In $$\triangle$$CDB,
tan 30$$^o$$ = $$\displaystyle{\frac{CD}{BD} \Rightarrow \frac{1}{\sqrt3} = \frac{50}{BD} \Rightarrow}$$ 50$$\sqrt3$$

In $$\triangle$$ABD,
tan 60$$^o$$ = $$\displaystyle{\frac{AB}{BD} \Rightarrow \sqrt3 = \frac{AB}{50\sqrt3} \Rightarrow}$$ AB = 150

$$\therefore$$ Height of the cliff is 150 m.
Question 3
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A boat is being rowed away from a cliff $$150$$ m high. From the top of the cliff, the angle of depression of the boat changes from 60$$^o$$ to 45$$^o$$ in 1 minute. If $$\sqrt3 = 1.73$$, the speed of the boat is _____

A
$$4.31$$ km/hr

B
$$3.81$$ km/hr

C
$$4.63$$ km/hr

D
$$3.91$$ km/hr

Solution

Referring to the figure, let AB be the cliff and C and D be the two positions of the boat.
Then, $$AB = 150$$ m, $$\angle ACB = 60^o$$ and $$\angle ADB = 45^o$$

Now, $$cot45^o $$ = $$\displaystyle{\frac{AD}{AB} \Rightarrow 1 = \frac{AD}{150} \Rightarrow }$$ AD = 150 m

and $$cot60^o$$ = $$\displaystyle{\frac{AC}{AB} \Rightarrow \frac{1}{\sqrt3} = \frac{AC}{150} \Rightarrow AC = \frac{150}{\sqrt3}}$$ m

$$\therefore$$ AC = $$\displaystyle{\frac{150 \times \sqrt3}{\sqrt3 \times \sqrt3}}$$ = 50$$\sqrt3$$ = 50 x 1.73 = 86.5 m.

$$\therefore CD = AD - AC = (150 - 86.5) = 63.5$$ m

$$\therefore$$ Speed of the boat = $$\displaystyle{\frac{63.5}{60}}$$ m/s

= $$\displaystyle{\left(\frac{63.5}{60} \times \frac{18}{5}\right)}$$ km/her

= $$3.81$$ km/hr

So, option B is correct.
Question 4
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Two points at distance x and y from the base point are on the same side of the line passing through the base pf a tower. The angle of elevation from these two points to the top of the tower are complementary. Then, the height of the tower is :

A
$$\sqrt{x}$$

B
$$\sqrt{y}$$

C
$$\sqrt{xy}$$

D
$$\sqrt{\displaystyle{\frac{x}{y}}}$$

Solution

Let the tower be AB and C and D be points on the same side
of the tower such that BD = x and BC = y
$$\therefore CD = (y - x)$$

Also, let $$\angle ADB = \theta$$ then $$\angle ACB = 90 - \theta$$

From $$\triangle$$ADB, we have tan $$\theta$$ = $$\displaystyle{\frac{AB}{DB} = \frac{AB}{x}}$$ .....1

From $$\triangle ABC$$, we have $$\tan (90 - \theta$$) = $$\displaystyle{\frac{AB}{BC} = \frac{AB}{y}}$$

$$\therefore \cot \theta = \displaystyle{\frac{AB}{y}}$$ ..... 2($$\because$$ tan(90 - $$\theta$$) = cot $$\theta$$)

From 1 and 2
$$\tan \theta \times cot \theta =1$$

$$\therefore$$$$\displaystyle{\frac{AB}{x} \times \frac{AB}{y} = 1 \Rightarrow AB^2 = xy \Rightarrow AB = \sqrt{xy}}$$

$$\therefore$$ The height of the tower is $$\sqrt{xy}$$
Question 5
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The angle of elevation of the top of a tower at a distance 30 m from its foot on a horizontal plane is found to be 30$$^o$$. The height of the tower is _____

A
$$17.3$$ m

B
$$57.96$$ m

C
$$17.8$$ m

D
$$173$$ m

Solution

Let AB be the tower and O be the point of observation,
then OA = 30 m and $$\angle$$AOB = 30$$^o$$

$$\therefore$$ In $$\triangle$$OAB, tan 30$$^o$$ =

$$\displaystyle{\frac{AB}{OA} \Rightarrow \frac{1}{\sqrt3} = \frac{AB}{30} \Rightarrow AB = \frac{30}{\sqrt3}}$$

$$\therefore$$ AB = $$\displaystyle{\frac{30 \times \sqrt3}{\sqrt3 \times \sqrt3}}$$ = 10$$\sqrt3$$ = 10 x 1.73 = 17.3

$$\therefore$$ Height of the tower is 17.3 m.
Question 6
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From the top of a tower, the angle of depression of a man standing 40 m away from the tower is 45. Then the height of the tower is:

A
40$$\sqrt3$$ m

B
$$40$$ m

C
$$20$$ m

D
$$45$$ m

Solution

Referring figure, let AB be the tower and C be the position of the man.
$$\therefore BC = 40$$ m and $$\angle ACB = 45$$ $$^o$$ ...(alternate angles)
In $$\triangle$$ ABC
$$\tan 45^o$$ = $$\displaystyle{\frac{AB}{BC}}$$

$$\therefore$$ 1 = $$\displaystyle{\frac{AB}{40}}$$

$$\therefore$$ The length of the tower (height of tower) is 40 m
Question 7
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A ladder is leaning against a wall such that its upper end touches the wall of the height of 3 m and the ladder is inclined at an angle having measure of 30$$^o$$ with the ground. The length of the ladder is:

A
$$3$$ m

B
$$6$$ m

C
$$9$$ m

D
$$12$$ m

Solution

In the figure, AC = ladder, AB = wall = 3m, $$\angle$$ACB = 30$$^o$$

In $$\triangle$$ABC, sin 30$$^o$$ = $$\displaystyle{\frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{3}{AC} \Rightarrow}$$ AC = 6 m.

$$\therefore$$ The length of ladder is 6 m.
Question 8
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From the top of a hill $$h$$ metres high the angles of depressions of the top and the bottom of a pillar are $$\alpha$$ and $$ \beta $$ respectively. The height (in metres) of the pillar is

A
$$\dfrac { h\tan { \beta } -h\tan { \alpha } }{ \tan { \beta } } $$

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

C
$$\dfrac { h\tan { \beta } +h\tan { \alpha } }{ \tan { \beta } } $$

D
$$\dfrac { h\tan { \beta } +h\tan { \alpha } }{ \tan { \alpha } } $$

Solution

Let $$AB$$ be a hill whose height is $$h$$ metres and $$CD$$ be a pillar of height $$h$$ metres.
In $$\Delta EDB$$,
$$\tan { \alpha } =\dfrac { h-{ h }^{ \prime } }{ ED } $$ ......(i)
and in $$\Delta ACB$$,
$$\tan { \beta } =\dfrac { h }{ AC } =\dfrac { h }{ ED } $$ .......(ii)
Divide equations (i) and (ii) to eliminate $$ED$$:
$$\dfrac{\tan { \alpha }}{\tan \beta} =\dfrac { h-{ h }^{ \prime } }{ h } $$

$$\quad h.\dfrac { \tan { \alpha } }{ \tan { \beta } } =h-{ h }^{ \prime }$$

$$\implies \quad { h }^{ \prime }=\dfrac { h\tan { \beta } -h\tan { \alpha } }{ \tan { \beta } } $$
Question 9
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$$P$$ is a point on the segment joining the feet of two vertical poles of heights $$a$$ and $$b$$. The angles of elevation of the tops of the poles from $$P$$ are $$45^0$$ each. Then, the square of the distance between the tops of the poles is

A
$$\displaystyle \frac {a^2\, +\, b^2}{2}$$

B
$$a^2\, +\, b^2$$

C
$$2(a^2\, +\, b^2)$$

D
$$4(a^2\, +\, b^2)$$

Solution

Let $$AD$$ be the pole of height $$a$$ and $$BC$$ be the pole of height $$B$$

In $$\Delta$$ $$APD$$, we have
$$\tan\, 45^o\, =\, \displaystyle \frac {AD}{AP}$$
$$\Rightarrow 1\, =\, \displaystyle \frac {a}{AP}$$
$$\, \Rightarrow\, AP\, =\, a$$

And in $$\Delta$$ $$BPC$$, we have,
$$\tan 45^o$$ $$ =\, \displaystyle \frac {BC}{PB}\, $$
$$\Rightarrow 1\, =\, \displaystyle \frac {b}{BP}$$
$$\, \Rightarrow\, BP\, =\, b$$

$$\therefore$$ $$DE=AB $$
$$= AP+BP$$
$$=a + b$$

and $$CE =BC-BE= b - a$$

In $$\Delta$$ $$DEC$$, applying pythagoras theorem,
$$DC^2=DE^2\, +\, EC^2$$
$$=\, (a\, +\, b)^2\, +\, (b\, -\, a)^2$$
$$=a^2+2ab+b^2+a^2-2ab+b^2$$
$$=\, 2(a^2\, +\, b^2)$$
Question 10
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A person observes the top of a tower from a point $$A$$ on the ground. The elevation of the tower from this point is $$60^{\circ}$$. He moves $$60\ m$$ in the direction perpendicular to the line joining $$A$$ and base of the tower. The angle of elevation of the tower from the point is $$45^{\circ}$$. Then the height of the tower (in meters) is

A
$$60\sqrt {\displaystyle \frac{3}{2} }$$

B
$$60\sqrt {2 }$$

C
$$60\sqrt {3 }$$

D
$$60\sqrt {\displaystyle \frac{2}{3} }$$

Solution

Let the base of tower be $$B$$ and the top be $$T$$
Initial position is $$A$$. Let the new position be $$C$$
$$BT=\text{height of the tower}=h$$
$$\angle TAB=60^{\circ }$$ and $$\angle TCB=45^{\circ }$$
$$tan60^{\circ }=\dfrac{h}{AB}\\\Rightarrow AB=\dfrac{h}{\sqrt{3}}$$

Now in $$\Delta TCB$$

$$tan45^{\circ }=\dfrac{h}{CB}\Rightarrow CB=h$$

Applying Pythagoras theorem in $$\triangle CAB,$$
$$CB^{2}=CA^{2}+AB^{2}$$
$$\Rightarrow h^{2}=60^{2}+(\dfrac{h}{\sqrt{3}})^{2}\Rightarrow h=60\sqrt{\dfrac{3}{2}}m$$
So, Option (A) is correct