Some Applications of Trigonometry : Previous Year's Questions


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( 3 Marks Questions )
Q.1     A man standing on the deck of a ship,which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and height of the hill.

[AI 2006]

Sol.
1

In \Delta BDC
\tan 30^\circ {\rm{ }} = {{10} \over x}
 \Rightarrow {1 \over {\sqrt 3 }} = {{10} \over x} \Rightarrow x = 10\sqrt 3 = 10 \times 1.732 = 17.32 m
In \Delta AEC
\tan 60^\circ ={{AE} \over {CF}}
 \Rightarrow \sqrt 3 = {{AE} \over {10\sqrt 3 }} \Rightarrow AE = 30m
So, Height of the hill  = {\rm{ }}30{\rm{ }} + {\rm{ }}10{\rm{ }} = {\rm{ }}40 m
Distance of the hill from the ship = {\rm{ }}17.32 m .


Q.2     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 meters towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and distance of the tower from the point A. \left( {\sqrt 3= 1.732} \right).

[Delhi 2006C]

Sol.       Let height of the tower be h and BQ  = {\rm{ }}x

2

In \Delta PQB
\tan 60^\circ {\rm{ }} = {h \over x}
 \Rightarrow \sqrt 3 = {h \over x}
 \Rightarrow h = \sqrt 3 x………(1)
In \Delta PQA
\tan 30^\circ = {h \over {20 + x}}
 \Rightarrow {1 \over {\sqrt 3 }} = {h \over {20 + x}}
 \Rightarrow \sqrt 3 h = 20 + x
 \Rightarrow \sqrt {3 \times } \sqrt 3 x = 20 + x [Using equation (1)]
 \Rightarrow 3x - x = 20
 \Rightarrow 2x = 20
 \Rightarrow x = 10
From equation (1)
h = 10\sqrt 3 = 10 \times 1.732 = 17.32 m
So, Height of the tower  = {\rm{ }}17.32 m.
Distance of tower from point A  = {\rm{ }}20{\rm{ }} + {\rm{ }}10{\rm{ }} = {\rm{ }}30 m


Q.3 A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression of the point ‘A’ from the top of the tower is 45°. Find the height of the tower. \left( {\sqrt 3= 1.73} \right)

[AI 2007]

Sol.       Let PQ is the tower of height h, PR be pole and AQ = x 3

In \Delta PQA
\tan 45^\circ = {h \over x}
 \Rightarrow 1 = {h \over x}
 \Rightarrow h = x………… (1)

In \Delta RQA
\tan 60^\circ = {{5 + h} \over x}
 \Rightarrow \sqrt 3 = {{5 + h} \over x} \Rightarrow \sqrt 3 x = 5 + h
 \Rightarrow \sqrt 3 x = 5 + x [Using equation (1)]
 \Rightarrow x\left( {\sqrt 3 - 1} \right) = 5
1} \right)} \over {{{(\sqrt 3 )}^2} - {{(1)}^2}}} = {{5 \times 2.73} \over 2} = 6.825m" />
 \Rightarrow h = 6.825m [Using equation (1)]
So, height of the tower = {\rm{ }}6.825 m


Q.4     A statue 1.46 m tall stand on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. \left( {\sqrt 3= 1.73} \right)

[AI 2008]

Sol.     Let height of pedestal = h and distance between point of observation & pedestal = x

2

In \Delta CBD
\tan 45^\circ = {h \over x}
 \Rightarrow 1 = {h \over x}
 \Rightarrow h = x ……… (1)
In \Delta ABD
\tan 60^\circ = {{1.46 + h} \over x}
 \Rightarrow \sqrt 3 = {{1.46 + h} \over x}
 \Rightarrow 1.46 + h = \sqrt 3 x
 \Rightarrow 1.46 + h = \sqrt 3 h [Using equation (1)]
 \Rightarrow h\left( {\sqrt 3 - 1} \right) = 1.46
 \Rightarrow h = {{1.46} \over {\sqrt 3 - 1}} = 2m


Q.5     A person standing on the bank of a river observes that the angle of the elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and the width of the river. \left( {\sqrt 3= 1.732} \right)

[Delhi 2008]

Sol.     Let height of the tree = h, Width of the river = x
5

In \Delta ABC
\tan 60^\circ = {h \over x}
 \Rightarrow \sqrt 3 = {h \over x}
 \Rightarrow h = \sqrt 3 x…… (1)
In \Delta ABD
\tan 30^\circ = {h \over {40 + x}}
 \Rightarrow {1 \over {\sqrt 3 }} = {h \over {40 + x}}
 \Rightarrow \sqrt 3 h = 40 + x
\sqrt 3 \times \sqrt 3 x = 40 + x [Using equation (1)]
 \Rightarrow 3x - x = 40 \Rightarrow 2x = 40 \Rightarrow x = 20 m
On putting x=20 in(1),We get
h = \sqrt 3 \times 20 = 1.732 \times 20 = 534.64 m


Q.6     An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.

[AI 2009]

Sol.     Let A & D are two aeroplanes and AB  = {\rm{ }}3125 m

6

In \Delta ABC
\tan 60^\circ = {{3125} \over {BC}}
 \Rightarrow \sqrt 3 = {{3125} \over {BC}}
 \Rightarrow BC = {{3125} \over {\sqrt 3 }}
In \Delta DBC
\tan 30^\circ = {{BD} \over {BC}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{BD} \over {BC}}
 \Rightarrow{1\over {\sqrt 3}}={{BD}\over {{{3125} \over{\sqrt 3}}}} (Substituting value of BC)
 \Rightarrow BD = {{3125} \over 3}
AD = AB - BD = 3125 - {{3125} \over 3} = {{6250} \over 3} = 2083.33m


Q.7     The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high. Find the height of the building. \left( {\sqrt 3= 1.732} \right)

[Foreign 2009]

Sol.     Let AB be the building and PQ be the tower

7

In \Delta APQ
\tan 60^\circ = {{PQ} \over {AP}}
 \Rightarrow \sqrt 3 = {{50} \over {AP}}
 \Rightarrow AP = {{50} \over {\sqrt 3 }}...............(1)
In \Delta PAB
\tan 30^\circ = {{AB} \over {AP}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{AB} \over {{{50} \over {\sqrt 3 }}}}[Using (1)]
 \Rightarrow AB = {{50} \over {\sqrt 3 \times \sqrt 3 }} = {{50} \over 3} = 16.67 m


Q.8     A man on the deck of a ship, 12 m above water level, observes that the angle of elevation of the top of a cliff is 60° and the angle of depression of the base of the cliff is 30°. Find the distance of the cliff from the ship and the height of the cliff. {\rm{[Use }}\,\sqrt 3= 1.732]

[AI 2010]

Sol.     Let AB be the cliff and PQ be the ship

8

In \Delta PQB
\tan 30^\circ ={{PQ} \over {QB}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{12} \over x}
 \Rightarrow x = 12\sqrt 3 = 12 \times 1.732 = 20.784m
In \Delta AMP
\tan 60^\circ = {{AM} \over {PM}}
 \Rightarrow \sqrt 3 = {{AM} \over {12\sqrt 3 }}[Because PM=x]
 \Rightarrow AM = 36 m
Height of the cliff = AM + BM = {\rm{ }}36{\rm{ }} + {\rm{ }}12{\rm{ }} = {\rm{ }}48 m


Q.9     From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. {\rm{[Use }}\,\sqrt 3= 1.732]

[Delhi 2011]

Sol.     Let PQ be the tower and from A and B two man observe the tower.

9

In \Delta APQ
\tan 30^\circ = {{100} \over {AP}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{100} \over {AP}}
 \Rightarrow AP = 100\sqrt 3 m
In \Delta BPQ
\tan 45^\circ = {{100} \over {BP}}
 \Rightarrow 1 = {{100} \over {BP}}
 \Rightarrow BP = 100 m
AB = AP + BP = 100\sqrt 3 + 100 = 100 \times 1.732 + 100 = 273.2 m


Q.10     A ladder of length 6 m makes an angle of 45° with the floor while leaning against one wall of a room. If the foot of the ladder is kept fixed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of 60° with the floor. Find the distance between these two walls of the room.

[Foreign 2011]

Sol.      Let AB & CD be two opposite walls & the ladder is fixed at point O.

10

In \Delta ABO
Cos60^\circ = {{BO} \over 6}
 \Rightarrow {1 \over 2} = {{BO} \over 6}
 \Rightarrow BO = 3 m

In \Delta CDO
Cos{45^\circ }= {{DO} \over 6}
 \Rightarrow {1 \over {\sqrt 2 }} = {{DO} \over 6}
 \Rightarrow DO = {6 \over {\sqrt 2 }} = 3\sqrt 2 m
Distance between two walls = BO + DO = 3 + 3 \times 1.41 = 3 + 4.23 = 7.23 m


Q.11     The horizontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. {\rm{[Use }}\,\sqrt 3= 1.732]

[Delhi 2013]

Sol.      Let AB & CD be two poles

11

Let AB = x
Therefore MD = AB = x
And CM = CD – MD  = {\rm{ }}24{\rm{ }}-{\rm{ }}x
Now,AM = BD = 15 m
In \Delta AMC
\tan 30^\circ = {{CM} \over {AM}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{24 - x} \over {15}}
 \Rightarrow 24 - x = {{15} \over {\sqrt 3 }}
 \Rightarrow 24 - x = 5\sqrt 3
 \Rightarrow 24 - x = 5 \times 1.732

 \Rightarrow 24 - x = 8.66
 \Rightarrow x = 24 - 8.66 = 15.34 m


 ( 5 Marks Questions )

 Q.1   The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of the tower PQ and the distance XQ.

[Foreign 2006]

Sol.
12

Let height of the tower (PQ) = h
PM = YX = {\rm{ }}40
And MQ = PQ – PM = {\rm{ }}h{\rm{ }}-{\rm{ }}40
Let PX = MY = x
In \Delta QPX
\tan 60^\circ = {h \over x}
 \Rightarrow \sqrt 3 = {h \over x}
 \Rightarrow h = x\sqrt 3 …….. (1)
In \Delta QMY
\tan 45^\circ = {{h - 40} \over x}
 \Rightarrow 1 = {{h - 40} \over x}
 \Rightarrow x = h - 40 \Rightarrow h = x + 40………….(2)
From equation (1) & (2)
h = {h \over {\sqrt 3 }} + 40
 \Rightarrow h - {h \over {\sqrt 3 }} = 40
 \Rightarrow h\left( {{{\sqrt 3 - 1} \over {\sqrt 3 }}} \right) = 40
 \Rightarrow h = 40 \times {{\sqrt 3 } \over {\sqrt 3 - 1}}
 \Rightarrow h = 40 \times {{\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}
 \Rightarrow h = {{40\left( {3 + \sqrt 3 } \right)} \over {{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}
 \Rightarrow h = {{40\left( {3 + 1.732} \right)} \over 2} = 20\left( {4.732} \right) = 94.64m...........(3)
In\,\,\Delta PQX
\sin 60^\circ = {{PQ} \over {XQ}}
 \Rightarrow {{\sqrt 3 } \over 2} = {h \over {XQ}}
 \Rightarrow XQ = {2 \over {\sqrt 3 }}h
 \Rightarrow XQ = {2 \over {\sqrt 3 }} \times 94.64[Using (3)]
 \Rightarrow XQ = {{2 \times 94.64 \times \sqrt 3 } \over {\sqrt 3 \times \sqrt 3 }} = {{2 \times 94.64 \times \sqrt 3 } \over 3}
 \Rightarrow XQ = 109.28m


Q.2     A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45° how soon after this, will the car reach the tower ?

[AI 2006C]

Sol.

Let the tower be PQ and from point A, its angle of elevation is 30° after 12 min from point B its angle of elevation is 45°.
Let the height of the tower be h & the distance between BQ = x

3
Let the speed of the car be y m/min
AB = 12 \times y = 12y [Because Distance = Speed \times Time]
AQ = AB + BQ = 12y + x
In \Delta BQP
\tan 45^\circ = {{PQ} \over {BQ}}
 \Rightarrow 1 = {h \over x}
 \Rightarrow x = h……….. (1)
In \Delta AQP
\tan 30^\circ = {{PQ} \over {AQ}}
 \Rightarrow {1 \over {\sqrt 3 }} = {h \over {12y + x}}
 \Rightarrow h\sqrt 3 = 12y + x…………. (2)
From equation (1) & (2)
x\sqrt 3= 12y + x
 \Rightarrow x\left( {\sqrt 3 - 1} \right) = 12y
 \Rightarrow x = {{12y} \over {\sqrt 3 - 1}} = {{12} \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 - 1}}y
 \Rightarrow x = {{12\left( {1.732 + 1} \right)} \over {{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}y = {{12 \times 2.732} \over 2}y = 16.392y
Time = {{Dis\tan ce} \over {Speed}} = {{16.392y} \over y} = 16.392 min


Q.3     A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of 20 m high building, finds the angle of elevation of the same bird to be 45°. Both the boy and the girl are on opposite sides of the bird. Find the distance of bird from the girl.

[Delhi 2007]

Sol. hemant

Let CD = X
In \Delta ABC
In \sin 30^\circ = {{AC} \over {AB}}
 \Rightarrow {1 \over 2} = {{AC} \over {100}}
 \Rightarrow AC = 50\,m
Now, in \Delta AFE
\sin 45^\circ = {{AF} \over {AE}}
 \Rightarrow {1 \over {\sqrt 2 }} = {{30} \over {AE}}
 \Rightarrow AE = 30\sqrt 2 = 30 \times 1.41
 \Rightarrow AE = 42.42\,m
Therefore, distance of bird from the girl = 42.42 m.


Q.4     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 720 km/hr, find the constant height. \left( {\sqrt 3= 1.732} \right)

[AI 2008]

Sol.

Speed of jet fighter  = 720km/h = 720 \times {5 \over {18}}m/\sec = 200\,sec
Distance covered in 15 sec = 200 × 15 = 3000 m
 \Rightarrow PB{\rm{ }} = {\rm{ }}3000{\rm{ }}m

14QC = PB = 3000 m
AC = AQ + QC = y + 3000
In \Delta AQP
\tan {60^\circ } = {{PQ} \over {AQ}}
 \Rightarrow \sqrt 3 = {h \over y} \Rightarrow h = \sqrt 3 y ………… (1)
In \Delta BCA
\tan 30^\circ = {{BC} \over {AC}}
 \Rightarrow {1 \over {\sqrt 3 }} = {h \over {y + 3000}}
 \Rightarrow h = {{y + 3000} \over {\sqrt 3 }} ………. (2)
From equation (1) & (2)
\sqrt 3 y = {{y + 3000} \over {\sqrt 3 }}
 \Rightarrow 3y = y + 3000
 \Rightarrow 2y = y + 3000
 \Rightarrow y = 1500
On putting y = 1500 in (1),We get
h = y\sqrt 3= 15000\sqrt 3= 1500 \times 1.732= 2598 m


 Q.5   The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of 3600 h = y\sqrt 3= 15000\sqrt 3= 1500 \times 1.732 m, find the speed in km / hr of the plane.

[Foreign 2008]

Sol.     Let OS = x And SQ = y

15

In \Delta OSR
\tan 60^\circ = {{RS} \over {OS}}
 \Rightarrow \sqrt 3 = {{3600\sqrt 3 } \over x}
 \Rightarrow x = 3600m............(1)
In \Delta OQP
\tan 30^\circ = {{PQ} \over {OQ}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{3600\sqrt 3 } \over {x + y}}
 \Rightarrow x + y = 3600 \times 3..............(2)
On putting x = 3600 in (2) ,We get
3600 + y = 10800
 \Rightarrow y{\rm{ }} = {\rm{ }}10800{\rm{ }}-{\rm{ }}3600{\rm{ }} = {\rm{ }}7200{\rm{ }}m
PR = QS = 7200m
Speed = {{Dis\tan ce} \over {Time}}\, = {{7200} \over {30}} = 240\,m/\sec
 \Rightarrow Speed = 240 \times {{18} \over 5} = 864 km/hr


Q.6    A straight highway leads to the foot of the tower. A man standing at the top of the tower observes a car at angle of depression of 30°, which is approaching the foot of the tower with a uniform speed, 6 second later the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

[Foreign 2008]

Sol.
16

Let height of tower PQ be h
A is initial point & B is final point
In \Delta PQA
\tan 30^\circ = {{PQ} \over {QA}}
 \Rightarrow {1 \over {\sqrt 3 }} = {h \over {x + y}}
 \Rightarrow h = {1 \over {\sqrt 3 }}\left( {x + y} \right)…………. (1)
In \Delta PQB
\tan 60^\circ = {{PQ} \over {QB}}
 \Rightarrow \sqrt 3 = {h \over y} \Rightarrow h = \sqrt 3 y………….(2)
From equation (1) and (2)
\sqrt 3 \,y = {1 \over {\sqrt 3 }}\left( {x + y} \right)
 \Rightarrow 3y = x + y
 \Rightarrow 2y = x
 \Rightarrow y = {x \over 2}
Therefore,Time taken = {6 \over 2} = 3\, seconds


Q.7     A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°. Find the height of the tower. \left( {\sqrt 3= 1.732} \right)

[Delhi 2009]

Sol.     Let height of the tower be h

17

In \Delta PQS
\tan 60^\circ = {{PQ} \over {SQ}}
 \Rightarrow \sqrt 3 = {{5 + h} \over x}
 \Rightarrow \sqrt 3 x = 5 + h ………(1)
In \Delta RQS
\tan 45^\circ = {{RQ} \over {SQ}}
 \Rightarrow 1 = {h \over x}
 \Rightarrow h = x ………… (2)
From equation (1) & (2)
5 + h = \sqrt 3 \,h
 \Rightarrow \sqrt 3 h - h = 5
\left( {\sqrt 3- 1} \right)h = 5
 \Rightarrow h = {5 \over {\sqrt 3 - 1}}
 \Rightarrow h = {5 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}
 \Rightarrow h = {{5\left( {\sqrt 3 + 1} \right)} \over {{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}
 \Rightarrow h = {5 \over 2}\left( {\sqrt 3 + 1} \right) = {5 \over 2}\left( {1.732 + 1} \right)
 \Rightarrow h = {5 \over 2} \times 2.732 = 6.83m


Q.8     From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° respectively. Find
(i) The horizontal distance between the building and the lamp post.
(ii) The height of the lamp post, \sqrt 3= 1.732

[AI 2009]

Sol.     Let PQ be building and AB be lamp post.

18

In \Delta PQB
\tan 60^\circ = {{PQ} \over {BQ}}
 \Rightarrow \sqrt 3 = {{60} \over {BQ}}
 \Rightarrow BQ = {{60} \over {\sqrt 3 }} = {{60 \times \sqrt 3 } \over 3} = 20\sqrt 3 = 20 \times 1.732 = 34.64m
In \Delta PMA
\tan 30^\circ = {{PM} \over {AM}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{PM} \over {20\sqrt 3 }}
 \Rightarrow PM{\rm{ }} = {\rm{ }}20{\rm{ }}m
MQ = PQ – PM = 60 – 20 = 40 m
AB = MQ = 40 m


Q.9      A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at that instants is 60°. After some time the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the internal.

[AI 2009]

Sol.     Let initially balloon is at R and its final position is at P.

19

PQ{\rm{ }} = {\rm{ }}88.2{\rm{ }}m
AB{\rm{ }} = {\rm{ }}1.2{\rm{ }}m
TQ{\rm{ }} = {\rm{ }}AB{\rm{ }} = {\rm{ }}1.2{\rm{ }}m
PT{\rm{ }} = {\rm{ }}PQ{\rm{ }}-{\rm{ }}TQ{\rm{ }} = {\rm{ }}88.2{\rm{ }}-{\rm{ }}1.2{\rm{ }} = {\rm{ }}87{\rm{ }}m
RS{\rm{ }} = {\rm{ }}PT{\rm{ }} = {\rm{ }}87{\rm{ }}m
In \Delta RSA
\tan 60^\circ = {{RS} \over {AS}}
 \Rightarrow \sqrt 3 = {{87} \over {AS}} \Rightarrow AS = {{87} \over {\sqrt 3 }}m
In \Delta PTA
\tan 30^\circ = {{PT} \over {AT}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{87} \over {AT}} \Rightarrow AT = 87\sqrt 3
ST = AT - AS = 87\sqrt 3 - {{87} \over {\sqrt 3 }} = {{174} \over {\sqrt 3 }} = 58\sqrt 3 m
RP = ST  = 58\sqrt 3 \,\,m
Therefore ,Distance covered by balloon  = 58\sqrt 3 \,\,m


Q.10     A tower stand vertically on a bank of a canal from a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°, from another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

[Foreign 2009]

Sol.       Let PQ be the tower and BQ be x width of canal

20

Suppose PQ = h
And BQ = x
In \Delta PQB
\tan 60^\circ = {{PQ} \over {BQ}}
 \Rightarrow \sqrt 3 = {h \over x} \Rightarrow h = \sqrt 3 x ……… (1)
In \Delta PQA
\tan 30^\circ = {{PQ} \over {AQ}}
 \Rightarrow {1 \over {\sqrt 3 }} = {h \over {20 + x}}
 \Rightarrow h = {{20 + x} \over {\sqrt 3 }} …………. (2)
From equation (1) & (2)
x\sqrt 3 \times \sqrt 3 = 20 + x
 \Rightarrow 3x - x = 20{\rm{ }} \Rightarrow 2x{\rm{ }} = {\rm{ 2}}0 \Rightarrow x = 10
On putting x=10 in (1),We get,
h = 10\sqrt 3= 10 \times 1.73 = 17.3\,m
Therefore, height of tower = 17.3 m
And Width of canal = 10 m


Q.11     The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud from the surface of the lake.

[AI 2010]

Sol.

Let the height of the cloud C from the lake be hm. A is position of the point 60 m above the lake. D is the reflection of the cloud in lake.

21

Let AE = x and CF = h
 \Rightarrow CE = h – 60
And DE = 60 + h
In \Delta AEC
\tan 30^\circ = {{CE} \over {AE}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{CE} \over {AE}} \Rightarrow AE = \left( {h - 60} \right)\sqrt 3 …………… (1)
In \Delta AED
\tan 60^\circ = {{ED} \over {AE}}
 \Rightarrow\sqrt 3 = {{60 + h}\over {AE}}\Rightarrow AE ={{h + 60}\over{\sqrt 3 }} ……… (2)
From equation (1) & (2)
\left( {h - 60} \right)\sqrt 3= {{h + 60} \over {\sqrt 3 }}
 \Rightarrow 3\left( {h - 60} \right) = h + 60
 \Rightarrow 2h = 60 + 180 \Rightarrow h = {{240} \over 2} = 120m
Therefore, height of the cloud = 120 m


Q.12     The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.

[AI 2012]

Sol.        Let height of the tower be h

22

In \Delta PQB
\tan 60^\circ = {{PQ} \over {BQ}}
 \Rightarrow \sqrt 3 = {h \over x}
 \Rightarrow h = x\sqrt 3 ……….. (1)
In \Delta PMA
\tan 30^\circ = {{PM} \over {AM}}
 \Rightarrow {1 \over {\sqrt 3 }} = {{h - 10} \over x}
 \Rightarrow x = \left( {h - 10} \right)\sqrt 3 …………… (2)
From equation (1) & (2)
h = \left( {h - 10} \right)\sqrt 3 \times \sqrt 3
 \Rightarrow h = 3h - 30 \Rightarrow 2h = 30 \Rightarrow h = {{30} \over 2} = 15m


Q.13    The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find
(i) the difference between the heights of the light-house and the building.
(ii) the distance between the light-house and the building.

[AI 2012]

Sol.      Let PQ be the light house and AB be the building

23

MQ = AB = 60 m
Suppose BQ = y
And PM = x
In \Delta PMA
\tan 30^\circ = {{PM} \over {AM}}
 \Rightarrow {1 \over {\sqrt 3 }} = {x \over y} [Because AM = BQ]
 \Rightarrow y = x\sqrt 3 …………. (1)
In \Delta ABQ
\tan 60^\circ = {{AB} \over {BQ}}
 \Rightarrow \sqrt 3 = {{60} \over y}
 \Rightarrow y = {{60} \over {\sqrt 3 }} = 20\sqrt 3
From equation (1) x = {y \over {\sqrt 3 }} = {{20\sqrt 3 } \over {\sqrt 3 }} = 20\,m
Therefore,
(i) The difference between the height of the light house and the building = 20 m
(ii) The distance between the light house and the building  = 20\sqrt 3 m


Q.14     From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 30°. Determine the height of the tower.

[Foreign 2013]

Sol.       Let AB be the building & PQ be the cable tower

24

In \Delta ABQ
\tan 30^\circ = {{AB} \over {BQ}}
 \Rightarrow {1 \over {\sqrt 3 }} = {7 \over x}
 \Rightarrow x = 7\sqrt 3
In \Delta AMP
\tan 60^\circ = {{PM} \over {AM}}
 \Rightarrow \sqrt 3 = {{h - 7} \over {7\sqrt 3 }} [Because AM = x]
 \Rightarrow h - 7 = 21 \Rightarrow h = 21 + 7 = 28m


 ( Multiple Choice Questions )

 Q.1 At some time of the day, the length of the shadow of a tower is equal to its height. Then the Sun’s altitude at that time is - (a) 30° (b) 60° (c) 90° (d) 45°

[Foreign 2011]

Sol. 4

\tan \theta= {{AB} \over {BC}}
 \Rightarrow \tan \theta = {h \over h} [Because AB = BC]
 \Rightarrow \tan \theta= 1
 \Rightarrow \tan \theta= \tan 45^\circ
 \Rightarrow \theta= 45^\circ


Q.2 A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that here is no slack in the string, the angle of elevation of the kite at the ground is -
(a) 45°       
(b) 30°      (c) 60°     (d) 90°

[AI 2012]

Sol. 5

\sin \theta= {{AB} \over {AC}}
 \Rightarrow \sin \theta= {{30} \over {60}}
 \Rightarrow \sin \theta= {1 \over 2}
 \Rightarrow \sin \theta= \sin 30^\circ
 \Rightarrow \theta= 30^\circ


Q.3     The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30°. The distance of the car from the base of the tower (in m) is :
(a) 25\sqrt 3 (b) 50\sqrt 3 (c) 75\sqrt 3 (d) 150 [Delhi 2013]

Sol.       Let AB be the tower and the car be at C

27

\tan 30^\circ= {{75} \over x}
 \Rightarrow x = 75\sqrt 3 m


Q.4    A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is
(a) 15\sqrt 3 \,m
(b) {{15\sqrt 3 \,} \over 2}m
(c) {{15} \over 2}m
(d) 15 m

[AI 2013]

Sol.     Let AB be the wall and AC be the ladder

28

\sin 60^\circ= {{AB} \over {AC}}
 \Rightarrow {{\sqrt 3 } \over 2} = {{AB} \over {15}}
 \Rightarrow AB = {{15\sqrt 3 } \over 2}m


Q.5     If at some time, the length of the shadow of a tower is \sqrt 3 times its height, then the angle of elevation of the Sun, at that time, is
(a) 15°     (b) 30°     (c) 45°     (d) 60°

[Foreign 2013]

Sol.     Let tower be h and angle of elevation from point C be Q.

6

\tan \theta= {{AB} \over {BC}}
 \Rightarrow tan\theta= {h \over {\sqrt 3 h}}
 \Rightarrow tan\theta= {1 \over {\sqrt 3 }}
 \Rightarrow tan\theta= \tan 30
 \Rightarrow \theta= 30^\circ