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SSC > Height and Distance

Explore popular questions from Height and Distance for SSC. This collection covers Height and Distance previous year SSC questions hand picked by experienced teachers.

Q 1.

Correct2

Incorrect-0.5

A kite is flying at an inclination of {tex} 60 ^{\circ} {/tex} with the horizontal. If the length of the thread is 120 m, then the height of the kite is:

{tex} 60\sqrt{3} m {/tex}

B

60 m

C

{tex} \frac{60}{\sqrt{3}} m {/tex}

D

120 m

Explanation

Triangle ABC,

{tex} sin 60 ^{\circ} = \frac{h}{120} {/tex}

=> {tex} h = 120 \times \frac{\sqrt{3}}{2} {/tex}

= {tex} 60\sqrt{3} m {/tex}

Q 2.

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An observer standing on a 300 m high tower obeserver see two boat in the same direction, their angles of depression are {tex} 60 ^{\circ} {/tex} and {tex} 30 ^{\circ} {/tex} respectively. The distance between boats is:

A

173.2 m

346.4 m

C

25 m

D

72 In

Explanation



In {tex} \triangle BCD, \cot 60 ^{\circ} = \frac{BC}{300} {/tex}

=> {tex} BC = 300 \times \frac{1}{\sqrt{3}} {/tex}

In {tex} \triangle ACD, \cot 30 ^{\circ} = \frac{AC}{300} {/tex}

=> {tex} AC = 300\sqrt{3} {/tex}

Therefore, Distance between two boats = AB

= {tex} AC -BC = 300\sqrt{3} - \frac{300}{\sqrt{3}} {/tex}

= {tex} 300\frac{ \left(3-1\right) }{\sqrt{3}} = \frac{600 \times \sqrt{3}}{3} = 346.4 m {/tex}

Q 3.

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The angle of elevation of the top of an unfinished pillar at a point 150m from its base is {tex} 30 ^{\circ} {/tex}. If the angle of elevation at the same point is to be {tex} 45 ^{\circ} {/tex}, then the pillar has to be raised to a height of how many meters?

A

59.4m

B

61.4m

C

62.4m

63.4m

Explanation

Given, BC=150 m

In {tex} \angle ACB = 30 ^{\circ} \ and\ \angle DCB = 45 ^{\circ} {/tex}



Then, AD = ?

In {tex} \triangle ABC, \tan 30 ^{\circ} = \frac{AB}{BC} {/tex}

=> {tex} \frac{1}{\sqrt{3}} = \frac{AB}{150} {/tex}

{tex} \therefore\ AB= 86.6 m {/tex}

In {tex} \triangle DBC, \tan 45 ^{\circ} = \frac{DB}{BC} {/tex}

=> {tex} 1 = \frac{AD+AB}{BC} {/tex}

=> BC = AD + 86.6

=> 150 - 86.6 = AD

Therefore, AD = 63.4 m

Q 4.

Correct2

Incorrect-0.5

The angle of elevation of the top of an unfinished pillar at a point 150m from its base is {tex} 30 ^{\circ} {/tex}. If the angle of elevation at the same point is to be {tex} 45 ^{\circ} {/tex}, then the pillar has to be raised to a height of how many meters?

A

59.4m

B

61.4m

C

62.4m

63.4m

Explanation

Given, BC=150 m

In {tex} \angle ACB = 30 ^{\circ} \ and\ \angle DCB = 45 ^{\circ} {/tex}



Then, AD = ?

In {tex} \triangle ABC, \tan 30 ^{\circ} = \frac{AB}{BC} {/tex}

=> {tex} \frac{1}{\sqrt{3}} = \frac{AB}{150} {/tex}

{tex} \therefore\ AB= 86.6 m {/tex}

In {tex} \triangle DBC, \tan 45 ^{\circ} = \frac{DB}{BC} {/tex}

=> {tex} 1 = \frac{AD+AB}{BC} {/tex}

=> BC = AD + 86.6

=> 150 - 86.6 = AD

Therefore, AD = 63.4 m

Q 5.

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On walking 120 m towards a Chimney in a horizontal line through its base the angle of elevation of tip of the chimney changes from {tex} 30 ^{\circ} {/tex} to {tex} 45 ^{\circ} {/tex}. The height of the chimney is

A

120 m

B

{tex} 60 \left(\sqrt{3} - 1\right) {/tex} m

{tex} 60 \left(\sqrt{3} + 1\right) {/tex} m

D

None of these

Explanation

Let h be the height of the chimney

In {tex} \triangle BPC {/tex},

{tex} \tan 45 ^{\circ} =\frac{h}{x}= 1 h = x {/tex} ...(i)



{tex} \tan 30 ^{\circ} =\frac{h}{120+x}=\frac{1}{\sqrt{3}} {/tex}

=> {tex} \frac{h}{120+h}=\frac{1}{\sqrt{3}} {/tex}

=> {tex} \sqrt{3}h = 120 + h {/tex}

=> {tex} h = \frac{120}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{120 \left(\sqrt{3}+1\right) }{2} {/tex}

Therefore, Required height of chimney

{tex} \left(h\right) = 60 \left(\sqrt{3}+1\right) m {/tex}

Q 6.

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The angles of elevation of the top of an inaccessible tower from two points on the same straight line from the base of the tower are {tex} 30 ^{\circ} {/tex} and {tex} 60 ^{\circ} {/tex}, respectively. If the points are separated at a distance of 100 m then the height of the tower is close to

86.6 m

B

84.6 m

C

82.6 m

D

80.6 m

Explanation

Let h be the height tower.

Now, in {tex} \triangle ACD {/tex}

{tex} \tan60 ^{\circ} = \frac{h}{x} = \sqrt{3} {/tex}

=> {tex} x = \frac{h}{\sqrt{3}} {/tex}

and in {tex} \triangle ABD {/tex}



{tex} \tan 30 ^{\circ} = \frac{h}{100+x} = \frac{1}{\sqrt{3}} {/tex}

=> {tex} \left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)h=100 {/tex}

=> {tex} \frac{2}{\sqrt{3}}h = 100 {/tex}

=> {tex} h = 50\sqrt{3} {/tex}

=> {tex} h = 50 \times 1732 = 86.6 m {/tex}

So, required height is 86.6 m.

Q 7.

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As seen from the top and bottom of a building of height h m, the angles of elevation of the top of a tower of height {tex} \frac{\left(3 + \sqrt{3}\right)h}{2}m {/tex} are {tex} \alpha {/tex} and {tex} \beta {/tex} respectively. If {tex} \alpha {/tex} = {tex} 30 ^{\circ} {/tex}, then what is the value of {tex} \tan \beta {/tex}

1

B

{tex} \frac{1}{2} {/tex}

C

{tex} \frac{1}{3} {/tex}

D

None of these

Explanation

{tex} \alpha = 30 ^{\circ} {/tex}

In {tex} \triangle ABC, {/tex}

{tex} \tan \alpha = \tan 30 ^{\circ} = \frac{AC}{BC} = \frac{1}{\sqrt{3}} {/tex}

=> {tex} BC = \sqrt{3} AC = \sqrt{3} \left(AE - CE\right) {/tex}

= {tex} \sqrt{3} \left(AE - BD\right) {/tex}

= {tex} \sqrt{3} \left(\frac{3+\sqrt{3}}{2}-1\right)h {/tex}

= {tex} \frac{\sqrt{3}}{2} \left(1+\sqrt{3}\right)h {/tex}

Now, in {tex} \triangle ADE, {/tex}

{tex} \tan \beta = \frac{AE}{DE} {/tex}

=> {tex} \tan \beta = \frac{AE}{BC} {/tex}

= {tex} \frac{ \left(\frac{3+\sqrt{3}}{2}\right)h }{\frac{\sqrt{3}}{2} \left(1+\sqrt{3}\right)h } = \frac{\frac{\sqrt{3} \left(1+\sqrt{3}\right) }{2}h}{\frac{\sqrt{3} \left(1+\sqrt{3}\right) }{2}h} {/tex}

{tex} \therefore \tan \beta = 1 {/tex}

Q 8.

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As seen from the top and bottom of a building of height h m, the angles of elevation of the top of a tower of height {tex} \frac{\left(3 + \sqrt{3}\right)h}{2}m {/tex} are {tex} \alpha {/tex} and {tex} \beta {/tex} respectively. If {tex} \beta {/tex} = {tex} 30 ^{\circ} {/tex} and if {tex} \theta {/tex} is the angle of depression of the foot of the tower as seen from the top of the building, then what is {tex} \tan \theta {/tex} equal to?

{tex} \frac{ \left(3 - \sqrt{3}\right) }{3\sqrt{3}} {/tex}

B

{tex} \frac{ \left(3 + \sqrt{3}\right) }{3\sqrt{3}} {/tex}

C

{tex} \frac{ \left(2 - \sqrt{3}\right) }{3\sqrt{3}} {/tex}

D

None of these

Explanation

Given that, {tex} \beta = 30 ^{\circ} {/tex}

In {tex} \triangle BDE, \tan \theta =\frac{BD}{DE}=\frac{h}{DE} {/tex}

=> {tex} \tan \theta =\frac{h}{\frac{3}{2} \left(1+\sqrt{3}\right)h } {/tex}

= {tex} \frac{2}{3}\frac{ \left(\sqrt{3}-1\right) }{\left(\sqrt{3}+1\right) \left(\sqrt{3}-1\right) } = \frac{2 \left(\sqrt{3}-1\right) }{3.2} {/tex}

= {tex} \frac{ \left(\sqrt{3}-1\right) }{3} \times \frac{\sqrt{^{3}}}{\sqrt{3}} = \frac{ \left(3-\sqrt{3}\right) }{3\sqrt{3}} {/tex}

Q 9.

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The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is {tex} 60 ^{\circ} {/tex}, and the angle of elevation of the top of the second tower from the foot of the first tower is {tex} 30 ^{\circ} {/tex}. The distance between the two towers is n times the height of the shorter tower. What is n equal to?

A

{tex} \sqrt{2} {/tex}

{tex} \sqrt{3} {/tex}

C

{tex} \frac{1}{2} {/tex}

D

{tex} \frac{1}{3} {/tex}

Explanation

Let h be the height of shorter tower. Then, the distance between the two towers is given by nh m.



In {tex} \triangle BCD, \tan 30 ^{\circ} = \frac{h}{nh} {/tex}

{tex} \frac{1}{\sqrt{3}}=\frac{1}{n} => n=\sqrt{3} {/tex}

Q 10.

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A spherical balloon of radius r subtends angle {tex} 60 ^{\circ} {/tex} at the eye of an observer. If the angle of elevation of its centre is {tex} 60 ^{\circ} {/tex} and h is the height of the centre of the balloon, then which one of the following is correct?

A

h=r

B

{tex} h = \sqrt{2}r {/tex}

{tex} h = \sqrt{3}r {/tex}

D

h=2r

Explanation


In {tex} \triangle ABO {/tex},



{tex} \sin 60 ^{\circ} = \frac{OB}{AO} {/tex}

=> {tex} AO = \frac{OB}{\sin 60 ^{\circ} } {/tex}

Now, in {tex} \triangle AOC, {/tex}

{tex} \sin \frac{60 ^{\circ} }{2} = \frac{OC}{AO} {/tex}

=> {tex} AO = \frac{OC}{\sin 30 ^{\circ} } {/tex}

From Eqs. (i) and (ii), we get

{tex} \frac{OB}{\sin 60 ^{\circ} }=\frac{OC}{\sin 30 ^{\circ} }3 {/tex}

=> {tex} \frac{h}{\frac{\sqrt{3}}{2}}=\frac{r}{\frac{1}{2}} {/tex}

Therefore, {tex} h = \sqrt{3}r {/tex}

Q 11.

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At a point on a horizontal line through the base of a monument the angle of elevation of the top of the monument is found to be such that its tangent is {tex} \frac{1}{5} {/tex}. On walking 138 metres towards the monument the secant of the angle of elevation is found to be {tex} \frac{\sqrt{193}}{12} {/tex}. The height of the monument (in metre) is

42

B

49

C

35

D

56

Explanation


{tex} \textbf{Shortcut approach} {/tex}
{tex} \textbf{Ist Case:} {/tex}
{tex} tan \theta =\frac{AB}{BC}=\frac{Perpendicular}{Base}=\frac{1}{5} {/tex}
{tex} \textbf{IInd Case:} {/tex}
{tex} Sec \alpha =\frac{AD}{BD}=\frac{Hypo}{Base} {/tex}
{tex} \frac{\sqrt{193}}{12} {/tex}
{tex} In\ \triangle ABD {/tex}
{tex} Hypo=\sqrt{193} {/tex}
{tex} Base=12 {/tex}
Then perpendicular = 7
(By pythagoras theorem)
In both case perpendicular will be equal
{tex} tan \theta =\frac{1 \times 7}{5 \times 7}=\frac{7\leftarrow Perpen}{35\leftarrow Base} {/tex}

{tex} AB = 42 m {/tex}

Q 12.

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The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot: of the tower to a point B, the angle of elevation increases to 60°. The height of the tower is

A

{tex} \sqrt{3}m {/tex}

B

{tex} 5\sqrt{3}m {/tex}

{tex} 10\sqrt{3}m {/tex}

D

{tex} 20\sqrt{3}m {/tex}

Explanation



AB = 'h' metre

{tex} In\ \triangle ABC {/tex}

{tex} tan30 ^{\circ}=\frac{AB}{BC} {/tex}

{tex} \Rightarrow \frac{h}{(d+20)} {/tex}

{tex} \frac{1}{\sqrt{3}}=\frac{h}{(d+20)} {/tex}

{tex} \sqrt{3}h=d+20........(i) {/tex}

{tex} In\ \triangle ABD {/tex}

{tex} tan60 ^{\circ}=\frac{AB}{BD}=\frac{h}{d} {/tex}

{tex} \sqrt{3}=\frac{h}{d} {/tex}

{tex} h=\sqrt{3}d {/tex}

{tex} d=\frac{h}{\sqrt{3}}.........(ii) {/tex}

Put the value of d in equation (i)

{tex} \sqrt{3}h=\frac{h}{\sqrt{3}}+20 {/tex}

{tex} \sqrt{3}h=\frac{h+20\sqrt{3}}{\sqrt{3}} {/tex}

{tex} 3h=h+20\sqrt{3} {/tex}

{tex} 2h=20\sqrt{3} {/tex}

{tex} h=\frac{20\sqrt{3}}{2} {/tex}

{tex} h=10\sqrt{3}\small\ meter {/tex}

Q 13.

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Incorrect-0.5

The top of two poles of height 24m and 36m are connected by a wire. If the wire makes an angle of 60° with the horizontal, then the length of the wire is

A

{tex} 6m {/tex}

{tex} 8\sqrt{3}m {/tex}

C

{tex} 8m {/tex}

D

{tex} 6\sqrt{3}m {/tex}

Explanation



AC = wire

AB and CD are two poles

{tex} In\triangle AEC {/tex}

{tex} sin660 ^{\circ} =\frac{AE}{AC} {/tex}

{tex} \Rightarrow \frac{\sqrt{3}}{2}=\frac{12}{AC} {/tex}

(AE = AB - CD = 36 - 24 = 12 m)

{tex} AC=\frac{24}{\sqrt{3}} {/tex}

{tex} =8\sqrt{3} {/tex}

Q 14.

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From a tower 125 metres high the angle of depression of two objects, which are in horizontal line through the base of the tower are 45° and 30° and they are on the same side of the tower. The distance (in metres) between the objects is

A

{tex} 125\sqrt{3} {/tex}

{tex} 125(\sqrt{3}-1) {/tex}

C

{tex} 125/(\sqrt{3}-1) {/tex}

D

{tex} 125(\sqrt{3}+1) {/tex}

Explanation



AB = Tower

{tex} In\triangle ABC\ tan45 ^{\circ} =\frac{AB}{BC} {/tex}

{tex} 1=\frac{AB}{BC}=AB:BC=1:1......(i) {/tex}

{tex} In \triangle ABD=tan30 ^{\circ}=\frac{AB}{BD} {/tex}

{tex} =AB:BD=1:\sqrt{3}.......(ii) {/tex}

Now,



CD = BD - BC

{tex} =(\sqrt{3}-1)units {/tex}

{tex} AB=1unit=125\ metre {/tex}

{tex} CD=(\sqrt{3}-1)units =125(\sqrt{3}-1)metre {/tex}

Q 15.

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From a point P on the ground the angle of elevation of the top of a 10m tall building is 30° A flag is hoisted at the top of var building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff (Take {tex} \sqrt{3}=1.732 {/tex})

A

{tex} 10(\sqrt{3}+2)m {/tex}

B

{tex} 10(\sqrt{30}+1)m {/tex}

C

{tex} 10\sqrt{3}m {/tex}

{tex} 7.32m {/tex}

Explanation



AB - building = 10 m

{tex} In\triangle ABP {/tex}

{tex} tan30 ^{\circ} =\frac{AB}{BP} {/tex}

{tex} \frac{1}{\sqrt{3}}=\frac{AB}{BP}=AB:BP=1:\sqrt{3}.....(i) {/tex}

{tex} In \triangle FBP {/tex}

{tex} tan45 ^{\circ} =\frac{FB}{BP} {/tex}

{tex} 1=\frac{FB}{BP}=FB:BP=1:1.......(ii) {/tex}

now,

{tex} AB\ \ :\ \ BP\ \ :\ \ FB {/tex}

{tex} 1\ \ :\ \ \sqrt{3} {/tex}



FB = 17.32 m

FA = FB - AB

=17.32 - 10

=7.32 metre

Q 16.

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Incorrect-0.5

The angle of elevation of ladder leaning against a house is 60° and the foot of the ladder is 6.5 metres from the house. The length of the ladder is

A

{tex} \frac{13}{\sqrt{3}} {/tex}

13 meters

C

15 meters

D

3.25 metres

Explanation



AC = Ladder

BC = 6.5 metres

{tex} In \triangle ABC {/tex}

{tex} Cose60 ^{\circ}=\frac{BC}{AC} {/tex}

{tex} AC = 13 m {/tex}

Q 17.

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Incorrect-0.5

A vertical pole and a vertical tower are standing on the same level ground. Height of the pole is 10 metres. From the top of the pole the angle of elevation of the top of the tower and angle of depression of the foot 0f the towervare 60° and 30° respectively. The height of the tower is

A

20 m

B

30 m

40 m

D

50 m

Explanation



AB = pole CE = tower

AB = 10 metre

{tex} In\triangle ABE {/tex}

{tex} tan30 ^{\circ} =\frac{AB}{BE} {/tex}

{tex} \frac{1}{\sqrt{3}}=\frac{AB}{BE}=AB:BE=1:\sqrt{3}.....(i) {/tex}

{tex} In\triangle ACD {/tex}

{tex} tan60 ^{\circ} =\frac{CD}{AD} {/tex}

{tex} \frac{\sqrt{3}}{1}=\frac{CD}{AD}=CD:AD=\sqrt{3}:1.....(ii) {/tex}

AD = BE and AB = DE

Now,



CE = CD + DE

= 30 + 10 = 40 metre

Q 18.

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Incorrect-0.5

A 1.6 m tall observer is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in metres is (Take {tex} \sqrt{3} {/tex} =1.732)

A

25.98

B

26.58

27.58

D

27.98

Explanation



Let height of observer = AB

{tex} \sqrt{3} {/tex} unit = 45 m

1 unit = {tex} \frac{45}{\sqrt{3}}={/tex}25.98 m

So,

ED = 25.98 m

So,

Height of tower = ED+CD

= 25.98 + 1.6

= 27.58 m

Q 19.

Correct2

Incorrect-0.5

The angle of elevation of the top of a tower from a point on the ground is 30° and moving 70 meters towards the tower it becomes 60°. The height of the tower is

A

{tex} 10\small\ meter {/tex}

B

{tex} \frac{10}{\sqrt{3}}\ meter {/tex}

C

{tex} 10\sqrt{3}\ meter {/tex}

{tex} 35\sqrt{3}\ meter {/tex}

Explanation



{tex} In\triangle ACD {/tex}

{tex} \angle ACB= \angle CAD+ \angle ADC {/tex}

{tex} 60 ^{\circ} = \angle CAD+30 ^{\circ} {/tex}

{tex} \angle CAD = 30 ^{\circ} {/tex}

So,

AC = CD

AC = 70m

{tex} cosec60 ^{\circ} =\frac{AC}{AB} {/tex}

{tex} \frac{2}{\sqrt{3}}=\frac{70}{AB} {/tex}

{tex} AB=35\sqrt{3}m {/tex}

Q 20.

Correct2

Incorrect-0.5

From the peak of a hill which is 300m high, the angle of depression of two sides of a bridge lying on a ground are 45° and 30° (both ends of the bridge are on the same side 30f the hill). Then the length of the bridge is

{tex} 300(\sqrt{3}-1)m {/tex}

B

{tex} 300(\sqrt{3}+1)m {/tex}

C

{tex} 300\sqrt{3}m {/tex}

D

{tex} \frac{300}{\sqrt{3}}m {/tex}

Explanation



AB = height of peak = 300 m

CD = length of Bridge

{tex} In \triangle ABC {/tex}

{tex} tan45 ^{\circ} =\frac{AB}{BC} {/tex}

{tex} 1=\frac{AB}{BC}=AB:BC=1:1 {/tex}

{tex} In\triangle ABD {/tex}

{tex} tan30 ^{\circ} =\frac{AB}{BD} {/tex}

{tex} \frac{1}{\sqrt{3}}=\frac{AB}{BD}\Rightarrow AB:BD=1:\sqrt{3} {/tex}

Now,



{tex} CD = BD - BC {/tex}

{tex} CD \Rightarrow \sqrt{3}-1{/tex}

AB = 1 unit = 300 metre

{tex} (\sqrt{3}-1)units = 300(\sqrt{3}-1)metre {/tex}

Q 21.

Correct2

Incorrect-0.5

From the top and bottom of a straight hill, the angle of depression and elevation of the top of a pillar of 10 m. height are observed to be 60° and 30° respectively. The height (in metres) of the hill is

A

30

B

80

C

60

40

Explanation

According to the question



From figure, AB = CD

height of pillar BC = AD = 10 m

In triangle ABC,

{tex} AB = BC\ cot 30 ^{\circ} {/tex}

{tex} AB = 10\sqrt{3} {/tex}

also {tex} AB=CD=10\sqrt{3}m {/tex}

In triangle CDE,

{tex} DE=DC\ tan60 ^{\circ} {/tex}

{tex} DE = 10\sqrt{3}\ \sqrt{3}=30 {/tex}

Height of the hill = AD + DE = 10 + 30 = 40m

Q 22.

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Incorrect-0.5

The distance between two parallel poles is {tex} 40\sqrt{3}m. {/tex} The angle of depression of the top of the second pole when seen from the top of first pole is 30°. What will be the height of second tower if the first pole is 100m long?

A

{tex} 50\sqrt{3}m {/tex}

B

{tex} 80m {/tex}

C

{tex} 35\sqrt{3}m {/tex}

{tex} 60m {/tex}

Explanation



{tex} tan30 ^{\circ} =\frac{x}{40\sqrt{3}} {/tex}

{tex} \frac{1}{\sqrt{3}}=\frac{x}{40\sqrt{3}} {/tex}

{tex} x=40 {/tex}

So, y = 100 - 40 = 60

Q 23.

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Incorrect-0.5

An earthing wire connected to the top of an electricity pole has its other end inside the ground. The foot of the wire is 1.5 m away from the pole and the wire is making an angle of 60° with the level of the ground. Determine the height of pole.

{tex} \frac{3\sqrt{3}}{2} {/tex}

B

{tex} 3\ m {/tex}

C

{tex} \sqrt{3}\ m {/tex}

D

{tex} \frac{\sqrt{3}}{2}\ m {/tex}

Explanation

Let height of Pole = h mtr.



{tex} tan60 ^{\circ} =\frac{h}{1.5} {/tex}

{tex} h=1.5 \times tan60 ^{\circ} {/tex}

{tex} =1.5 \times \sqrt{3} {/tex}

{tex} =\frac{3\sqrt{3}}{2}mtr {/tex}

Q 24.

Correct2

Incorrect-0.5

A person observes that the angle of elevation at the top of a pole of height 5 meter is 30°. Then the distance of the person from the pole is:

{tex} 5\sqrt{3}\small\ meter {/tex}

B

{tex} \frac{5}{\sqrt{3}}\small\ meter {/tex}

C

{tex} \sqrt{3}\small\ meter {/tex}

D

{tex} 10\sqrt{3}\small\ meter {/tex}

Explanation



1 unit = 5m

{tex} \sqrt{3}\ unit=5\sqrt{3}m {/tex}

Q 25.

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Incorrect-0.5

The cliff of a mountain is 180 m high, and the angles of depression of two ships on the either side of cliff are 30° and 60°. What is the distance between the two ships?

A

{tex} 400\ m {/tex}

B

{tex} 400\sqrt{3}\ m {/tex}

{tex} 415.68\ m {/tex}

D

{tex} 398.6\ m {/tex}

Explanation



{tex} AD : CD = \sqrt{3}:1.........(i) {/tex}



{tex} AD:BD=1:\sqrt{3}........(ii) {/tex}

From Eq. (i) & (ii) to make equal ratio

{tex} CD:AD:BD=1:\sqrt{3}:3 {/tex}

BC = DC + BD

BC = 1 + 3 = 4 units

{tex} \sqrt{3}\ units------180m {/tex}

{tex} 1\ unit------\frac{180}{\sqrt{3}} {/tex}

Now, Distance between the two ships

{tex} 4\ units-----\frac{180}{\sqrt{3}} \times 4 {/tex}

{tex} =\frac{180 \times 4 \times \sqrt{3}}{3}=240\sqrt{3}m {/tex}

{tex} 415.68m {/tex}