Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 38

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Question 1
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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

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Some Applications of Trigonometry test - 38

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Some Applications of Trigonometry test - 38

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Question 1

303030∘^{\circ}∘

454545∘^{\circ}∘

606060∘^{\circ}∘

909090∘^{\circ}∘

Solution

Question 2

50350\sqrt3503​ m

150150150 m

100100100m

505050 m

Solution

Question 3

4.314.314.31 km/hr

3.813.813.81 km/hr

4.634.634.63 km/hr

3.913.913.91 km/hr

Solution

Question 4

x\sqrt{x}x​

y\sqrt{y}y​

xy\sqrt{xy}xy​

xy\sqrt{\displaystyle{\frac{x}{y}}}yx​​

Solution

Question 5

17.317.317.3 m

57.9657.9657.96 m

17.817.817.8 m

173173173 m

Solution

Question 6

403\sqrt33​ m

404040 m

202020 m

454545 m

Solution

Question 7

333 m

666 m

999 m

121212 m

Solution

Question 8

htan⁡β−htan⁡αtan⁡β\dfrac { h\tan { \beta } -h\tan { \alpha } }{ \tan { \beta } } tanβhtanβ−htanα​

htan⁡β−htan⁡αtan⁡α\dfrac { h\tan { \beta } -h\tan { \alpha } }{ \tan { \alpha } } tanαhtanβ−htanα​

htan⁡β+htan⁡αtan⁡β\dfrac { h\tan { \beta } +h\tan { \alpha } }{ \tan { \beta } } tanβhtanβ+htanα​

htan⁡β+htan⁡αtan⁡α\dfrac { h\tan { \beta } +h\tan { \alpha } }{ \tan { \alpha } } tanαhtanβ+htanα​

Solution

Question 9

a2 + b22\displaystyle \frac {a^2\, +\, b^2}{2}2a2+b2​

a2 + b2a^2\, +\, b^2a2+b2

2(a2 + b2)2(a^2\, +\, b^2)2(a2+b2)

4(a2 + b2)4(a^2\, +\, b^2)4(a2+b2)

Solution

Question 10

603260\sqrt {\displaystyle \frac{3}{2} }6023​​

60260\sqrt {2 }602​

60360\sqrt {3 }603​

602360\sqrt {\displaystyle \frac{2}{3} }6032​​

Solution

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$30$ $^{\circ}$

$45$ $^{\circ}$

$60$ $^{\circ}$

$90$ $^{\circ}$

$50\sqrt3$ m

$150$ m

$100$ m

$50$ m

$4.31$ km/hr

$3.81$ km/hr

$4.63$ km/hr

$3.91$ km/hr

$\sqrt{x}$

$\sqrt{y}$

$\sqrt{xy}$

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

$17.3$ m

$57.96$ m

$17.8$ m

$173$ m

40 $\sqrt3$ m

$40$ m

$20$ m

$45$ m

$3$ m

$6$ m

$9$ m

$12$ m

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

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

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

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

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

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

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

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

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

$60\sqrt {2 }$

$60\sqrt {3 }$

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