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Page No 635:

Question 1:

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower.

Answer:

Let AB be the tower standing vertically on the ground and O be the position of the observer.
We now have:
OA = 20 m , OAB= 90o and ∠AOB = 60o
Let:
AB = h m

Now, in the right ∆OAB, we have:
ABOA = tan 60o = 3
                                        
⇒ h20 = 3
                       
⇒ h = 203 = (20 × 1.732) = 34.64

Hence, the height of the pole is 34.64 m.                          

Page No 635:

Question 2:

A kite is flying at a height of 75 m from the level ground, attached to a string inclined at a 60° to the horizontal. Find the length of the string assuming that there is no slack in it.

Answer:

Let OX  be the horizontal ground and A be the position of the kite.
Also, let O be the position of the observer and OA be the thread.
Now, draw AB ⊥ OX.
We have:
BOA= 60oOA = 75 m and ∠OBA= 90o
Height of the kite from the ground = AB = 75 m
Length of the string, OA= x m

In the right ∆OBA, we have:
ABOA = sin 60o = 32
75x = 32
⇒ x = 75× 23 = 1501.732 = 86.6 m
Hence, the length of the string is 86.6 m.

Page No 635:

Question 3:

An observer 1.5 m tall is 30 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 60°.
Find the height of the chimney.                                                                                                            [CBSE 2013C]

Answer:



Let CE and AD be the heights of the observer and the chimney, respectively.

We have,
BD=CE=1.5 m, BC=DE=30 m and ACB=60°In ABC,tan60°=ABBC3=AD-BD30AD-1.5=303AD=303+1.5AD=30×1.732+1.5AD=51.96+1.5AD=53.46 m

So, the height of the chimney is 53.46 m (approx.).

Page No 635:

Question 4:

The angles of elevation of the top of a tower from two points at distances of 5 metres and 20 metres from the base of the tower and in the same straight line with it, are complementary. Find the height of the tower.                                                                               [CBSE 2014]

Answer:



Let the height of the tower be AB.

We have,
AC=5 m, AD=20 mLet the angle of elevation of the top of the tower i.e. ACB from point C be θ. Then,the angle of elevation of the top of the tower i.e. ADB from point D=90°-θNow, in ABC,tanθ=ABACtanθ=AB5          .....iAlso, in ABD,cot90°-θ=ADABtanθ=20AB           .....iiFrom i and ii, we getAB5=20ABAB2=100AB=100 AB=10 m

So, the height of the tower is 10 m.

Page No 635:

Question 5:

The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. [Use 3 = 1.732]                                  [CBSE 2014]

Answer:



Let BC and CD be the heights of the tower and the flagstaff, respectively.

We have,
AB=120 m, BAC=45°, BAD=60°Let CD=xIn ABC,tan45°=BCAB1=BC120BC=120 mNow, in ABD,tan60°=BDAB3=BC+CD120BC+CD=1203120+x=1203x=1203-120x=1203-1x=1201.732-1x=1200.732x=87.8487.8 m

So, the height of the flagstaff is 87.8 m.

Page No 635:

Question 6:

From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45°. Find (i) the height of the tower, (ii) the depth of the tank.

Answer:



Let BC be the tower and CD be the water tank.

We have,AB=40 m, BAC=30° and BAD=45°In ABD,tan45°=BDAB1=BD40BD=40 mNow, in ABC,tan30°=BCAB13=BC40BC=403BC=403×33BC=4033 mi The height of the tower, BC=4033=40×1.733=23.06723.1 mii The depth of the tank, CD=BD-BC=40-23.1=16.9 m



Page No 636:

Question 7:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 60°. Find the height of the tower.
[Use 3 = 1.732]                                                                                                                                                                     [CBSE 2011]

Answer:


Let AB be the tower and BC be the flagstaff.

We have,
BC=6 m, AOB=30° and AOC=60°Let AB=hIn AOB,tan30°=ABOA13=hOAOA=h3              .....iNow, in AOC,tan60°=ACOA3=AB+BCh3         Using i3h=h+63h-h=62h=6h=62h=3 m

So, the height of the tower is 3 m.

Page No 636:

Question 8:

A statue 1.46 m tall, stands 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.Use 3=1.73

Answer:

Let AC be the pedestal and BC be the statue such that BC = 1.46 m.
We have:
ADC = 45o and ∠ADB = 60o
Let:
AC = h m and AD = x m

In the right ∆ADC, we have:
 ACAD = tan 45o = 1

⇒ hx = 1
⇒ h = x 
Or,
x = h

Now, in the right ∆ADB, we have:
ABAD = tan 60o = 3

⇒ h +1.46x = 3

On putting x = h in the above equation, we get:
h + 1.46h = 3

⇒ h + 1.46 = 3h
⇒ h(3 - 1) = 1.46
⇒ h = 1.46(3 - 1) = 1.460.73 = 2m  

Hence, the height of the pedestal is 2 m.

Page No 636:

Question 9:

The angle of elevation of the top of an unfinished  tower at a distance of 75 m form its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°?

Answer:

Let AB be the unfinished tower, AC be the raised tower and O be the point of observation.
We have:
OA = 75 m, ∠AOB = 30o and ∠AOC = 60o
Let AC = H m such that BC = (H-h) m.

In ∆AOB, we have:
ABOA= tan 30o=13

h75= 13

⇒ h = 753 m = 75×33×3= 253 m
In ∆AOC, we have:
ACOA= tan 60o =3

⇒ H75= 3
H = 753 m

∴ Required height = (H-h) = ( 753 - 253 ) = 503 m = 86.6 m

Page No 636:

Question 10:

On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top of bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it.Use 3=1.73

Answer:

Let OX be the horizontal plane, AD be the tower and CD be the vertical flagpole.
We have:
AB = 9 m, ∠DBA = 30o and ∠CBA = 60o
Let:
AD = h m and CD= x m



In the right ∆ABD, we have:
ADAB = tan 30o = 13
⇒ h9=13
⇒ h = 93 = 5.19 m
Now, in the right ∆ABC, we have:
ACBA = tan 60o = 3

⇒ h + x9 = 3
⇒ h + x = 93

By putting h = 93 in the above equation, we get:
93 + x= 93
⇒ x = 93 - 93
⇒ x = 27 - 93 = 183 = 181.73 = 10.4

Thus, we have:
Height of the flagpole = 10.4 m
Height of the tower = 5.19 m

Page No 636:

Question 11:

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on
the road, the angle of elevation of the top of one pole is 60° and the angle of depression from the top of another pole at P is 30°. Find the
height of each pole and distances of the point P from the poles.                                                                                                         [CBSE 2015]

Answer:


Let AB and CD be the equal poles; and BD be the width of the road.

We have,
AOB=60° and COD=60°In AOB,tan60°=ABBO3=ABBOBO=AB3Also, in COD,tan30°=CDDO13=CDDODO=3CDAs, BD=80BO+DO=80AB3+3CD=80AB3+3AB=80          Given: AB=CDAB13+3=80AB1+33=80AB43=80AB=8034AB=203 mAlso, BO=AB3=2033=20 mSo, DO=80-20=60 m

Hence, the height of each pole is 203 m and point P is at a distance of 20 m from left pole and 60 m from right pole.

Page No 636:

Question 12:

Two men are on opposite sides of a tower. they measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men. Take 3=1.732

Answer:

Let CD be the tower and A and B be the positions of the two men standing on the opposite sides. Thus, we have:
DAC = 30o, ∠DBC = 45o and CD= 50 m
Let AB = x m and BC= y m such that AC = (x - y) m.

In the right ∆DBC, we have:
CDBC = tan 45o = 1

⇒ 50y = 1
⇒ y = 50 m

In the right ∆ACD, we have:
CDAC = tan 30o = 13

⇒ 50(x - y) = 13
⇒ x - y = 503
On putting y = 50 in the above equation, we get:
x - 50 = 503
⇒ x = 50 + 503 = 50 (3 + 1) = 136.6 m

∴ Distance between the two men = AB = x = 136.6 m

Page No 636:

Question 13:

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. [Take 3= 1.73]                                                                          [CBSE 2011]

Answer:



Let PQ be the tower.

We have,

PQ=100 m, PAQ=30° and PBQ=45°In APQ,tan30°=PQAP13=100APAP=1003 mAls, in BPQ,tan45°=PQBP1=100BPBP=100 mNow, AB=AP+BP=1003+100=1003+1=100×1.73+1=100×2.73=273 m

So, the distance between the cars is 273 m.

Page No 636:

Question 14:

A straight highway leads to the foot of a tower. A man standing on the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds 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 form this point.

Answer:



Let PQ be the tower.

We have,PBQ=60° and PAQ=30°Let PQ=h, AB=x and BQ=yIn APQ,tan30°=PQAQ13=hx+yx+y=h3              .....iAlso, in BPQ,tan60°=PQBQ3=hyh=y3                     .....iiSubstituting h=y3 in i, we getx+y=3y3x+y=3y3y-y=x2y=xy=x2As, speed of the car from A to B=AB6=x6 units/secSo, the time taken to reach the foot of the tower i.e. Q from B=BQspeed=yx6=x2x6=62=3 sec

So, the time taken to reach the foot of the tower from the given point is 3 seconds.



Page No 637:

Question 15:

A TV tower stands vertically on a bank of canal. From a point on 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.

Answer:


Let PQ=h m be the height of the TV tower and BQ=x m be the width of the canal.

We have,

AB=20 m, PAQ=30°, PBQ=60°, BQ=x and PQ=hIn PBQ,tan60°=PQBQ3=hxh=x3           .....iAgain, in APQ,tan30°=PQAQ13=hAB+BQ13=x320+x         Using i3x=20+x3x-x=202x=20x=202x=10 mSubstituting x=10 in i, we geth=103 m

So, the height of the TV tower is 103 m and the width of the canal is 10 m.

Page No 637:

Question 16:

The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, then find the height of the building.                                                                          [CBSE 2013]

Answer:



Let AB be the building and PQ be the tower.

We have,

PQ=60 m, APB=30°, PAQ=60°In APQ,tan60°=PQAP3=60APAP=603AP=6033AP=203 mNow, in ABP,tan30°=ABAP13=AB203AB=2033 AB=20 m

So, the height of the building is 20 m.

Page No 637:

Question 17:

The horizontal distance between two towers is 60 metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 metres, find the height of the first tower.Use 3=1.732

Answer:

Let DE be the first tower and AB be the second tower.
Now, AB= 90 m and AD = 60 m such that CE = 60 m and ∠BEC = 30o.
Let DE = h m such that AC= h m and BC = (90 - h) m.

In the right ∆BCE, we have:
BCCE = tan 30o = 13
⇒ (90 - h)60 = 13
⇒ (90 - h)3 = 60
⇒ h3 = 903 - 60 
⇒ h = 90 - 603 = 90 - 34.64 = 55.36 m
∴ Height of the first tower = DE = h = 55.36 m

Page No 637:

Question 18:

The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 metres, then find the height of the chimney.

According to pollution control norms, the minimum height of a smoke emitting chimney should be 100 metres. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question?                                                   [CBSE 2014]

Answer:


Let PQ be the chimney and AB be the tower.


We have,

AB=40 m, APB=30° and PAQ=60°In ABP,tan30°=ABAP13=40APAP=403 mNow, in APQ,tan60°=PQAP3=PQ403 PQ=120 m

So, the height of the chimney is 120 m.

Hence, the height of the chimney meets the pollution norms.


In this question, management of air pollution has been shown.
a

Page No 637:

Question 19:

From the top of a 7-metre-high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. [Use 3 = 1.732]                                                                                                      [CBSE 2013C]

Answer:



Let AB be the 7-m high building and CD be the cable tower.

We have,

AB=7 m, CAE=60°, DAE=ADB=45°Also, DE=AB=7 mIn ABD,tan45°=ABBD1=7BDBD=7 mSo, AE=BD=7 mAlso, in ACE,tan60°=CEAE3=CE7CE=73 mNow, CD=CE+DE=73+7=73+1 m=71.732+1=72.732=19.12419.12 m

So, the height of the tower is 19.12 m.

Page No 637:

Question 20:

The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60°. Find the height of the tower and its distance from the point A.                                                                                                                                     [CBSE 2012]

Answer:


Let PQ be the tower.

We have,

AB=20 m, PAQ=30° and PBQ=60°Let BQ=x and PQ=hIn PBQ,tan60°=PQBQ3=hxh=x3               .....iAlso, in APQ,tan30°=PQAQ13=hAB+BQ13=x320+x          Using i20+x=3x3x-x=202x=20x=202x=10 mFrom i,h=103=10×1.732=17.32 mAlso, AQ=AB+BQ=20+10=30 m

So, the height of the tower is 17.32 m and its distance from the point A is 30 m.

Page No 637:

Question 21:

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.                                                                                                           [CBSE 2011]

Answer:


Let PQ be the tower.

We have,

AB=10 m, MAP=30° and PBQ=60°Also, MQ=AB=10 mLet BQ=x and PQ=hSo, AM=BQ=x and PM=PQ-MQ=h-10In BPQ,tan60°=PQBQ3=hxx=h3              .....iNow, in AMP,tan30°=PMAM13=h-10xh3-103=xh3-103=h3            Using i3h-30=h3h-h=302h=30h=302 h=15 m

So, the height of the tower is 15 m.

Page No 637:

Question 22:

The angles of depression of the top and bottom of a tower as seen from the top of a 603-m-high cliff are 45° and 60°, respectively. Find the height of the tower.                                                                                                                                                                  [CBSE 2012]

Answer:


Let AD be the tower and BC be the cliff.

We have,

BC=603 m, CDE=45° and BAC=60°Let AD=hBE=AD=hCE=BC-BE=603-hIn CDE,tan45°=CEDE1=603-hDEDE=603-hAB=DE=603-h              .....iNow, in ABC,tan60°=BCAB3=603603-h          Using i180-h3=603h3=180-603h=180-6033h=180-6033×33h=1803-1803h=1803-13 h=603-1        =601.732-1        =600.732Also, h=43.92 m

So, the height of the tower is 43.92 m.

Page No 637:

Question 23:

A man on the deck of a ship, 16 m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°. Calculate the distance of the cliff from the ship and height of the cliff.

Answer:

Let AB be the deck of the ship above the water level and DE be the cliff.
Now,
AB = 16 m such that CD = 16 m and ∠​BDA = 30o and ∠EBC= 60o.
If AD = x m and DE = h m, then CE = (h - 16) m.



In the right ∆BAD, we have:
ABAD = tan 30o = 13

⇒ 16x = 13

⇒ x = 163 = 27.68 m

In the right ∆EBC, we have:
ECBC = tan 60o = 3

⇒ (h - 16)x = 3
⇒ h - 16 = x3
⇒ h - 16 = 163 × 3 = 48        [∵ x = 163]
h = 48 + 16 = 64 m

∴ Distance of the cliff from the deck of the ship = AD = x = 27.68 m
And,
Height of the cliff = DE = h = 64 m



Page No 638:

Question 24:

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 tower PQ. [Take 3 = 1.73]                                                                                  [CBSE 2003C]

Answer:



We have,

XY=40 m, PXQ=60° and MYQ=45°Let PQ=hAlso, MP=XY=40 m, MQ=PQ-MP=h-40In MYQ,tan45°=MQMY1=h-40MYMY=h-40PX=MY=h-40        .....iNow, in MXQ,tan60°=PQPX3=hh-40             From ih3-403=hh3-h=403h3-1=403h=4033-1h=4033-1×3+13+1h=4033+13-1h=4033+12h=2033+1h=60+203h=60+20×1.73h=60+34.6 h=94.6 m

So, the height of the tower PQ is 94.6 m.

Page No 638:

Question 25:

The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, find the speed of the aeroplane.

Answer:



Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation.

We have,

PQ=BC=2500 m, PAQ=45° and BAC=30°In PAQ,tan45°=PQAQ1=2500AQAQ=2500 mAlso, in ABC,tan30°=BCAC13=2500ACAC=25003 mNow, QC=AC-AQ=25003-2500=25003-1 m=25001.732-1=25000.732=1830 mPB=QC=1830 mSo, the speed of the aeroplane=PB15=183015=122 m/s=122×36001000 km/h=439.2 km/h

So, the speed of the aeroplane is 122 m/s or 439.2 km/h.

Page No 638:

Question 26:

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30° On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres.

Answer:

Let AB be the tower.
We have:
CD= 150 m, ∠ACB = 30o and ∠ADB = 60o
Let:
AB = h m  and BD = x m

In the right ∆ABD, we have:
ABAD = tan 60o = 3

⇒ hx= 3
⇒ x = h3
Now, in the right ∆ACB, we have:
ABBC = tan 30o = 13

⇒ hx + 150 = 13
⇒ 3h = x + 150

On putting x = h3 in the above equation, we get:
3h = h3 + 150
⇒ 3h = h + 1503
⇒ 2h = 1503
⇒ h =15032 = 753 = 75 × 1.732 = 129.9 m
Hence, the height of the tower is 129.9 m.

Page No 638:

Question 27:

As observed form the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation.

Answer:

Let OA be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
OA= 100 m, ∠OBA= 30o and  ∠OCA = 60o



Let:
OC = x m and BC= y m
In the right ∆OAC, we have:
OAOC = tan 60o = 3

100x = 3
⇒ x = 1003 m
Now, in the right ∆OBA, we have:
OAOB = tan 30o = 13

⇒ 100x + y = 13
x + y = 1003

On putting x = 1003 in the above equation, we get:
y = 1003 - 1003 = 300 - 1003 = 2003 = 115.47 m

∴ Distance travelled by the ship during the period of observation = BC = y = 115.47 m

Page No 638:

Question 28:

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find width of the river.

Answer:



Let A and B be two points on the banks on the opposite side of the river and P be the point on the bridge at a height of 2.5 m.
Thus, we have:
DP = 2.5 m, ∠PAD= 30o and ∠PBD = 45o
In the right ∆APD, we have:
DPAD = tan 30o = 13
⇒ 2.5AD = 13
⇒ AD = 2.53 m
In the right ∆PDB, we have:
DPBD= tan 45o = 1
⇒ 2.5BD = 1
⇒ BD = 2.5 m

∴ Width of the river = AB = (AD + BD) = (2.53 + 2.5) = 6.83 m

Page No 638:

Question 29:

The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 metres.

Answer:

Let AB be the tower and C and D be two points such that AC = 4 m  and AD = 9 m.
Let:
AB = h m, ∠BCA = θ and ∠BDA = 90°-θ

In the right ∆BCA, we have:
tanθ= ABACtanθ= h4         ...(1)
In the right ∆BDA, we have:
tan90°-θ=ABADcot θ=h9                  tan90°-θ=cot θ1tan θ=h9       ...(2)     cot θ=1tan θ
Multiplying equations (1) and (2), we get:
tan θ×1tan θ=h4×h91=h236
⇒ 36 = h2
h = ±6

Height of a tower cannot be negative.
∴ Height of the tower = 6 m

Page No 638:

Question 30:

A ladder of length 6 metres 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 two walls of the room.                                                                                                                                                     [CBSE 2011]

Answer:


Let AB and CD be the two opposite walls of the room and the foot of the ladder be fixed at the point O on the ground.

We have,

AO=CO=6 m, AOB=60° and COD=45°In ABO,cos60°=BOAO12=BO6BO=62BO=3 mAlso, in CDO,cos45°=DOCO12=DO6DO=62DO=62×22DO=622DO=32 mNow, the distance between two walls of the room=BD=BO+DO=3+32=31+2=31+1.414=32.414=7.2427.24 m

So, the distance between two walls of the room is 7.24 m.

Page No 638:

Question 31:

From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.         [CBSE 2011]

Answer:



Let OP be the tower and points A and B be the positions of the cars.

We have,

AB=100 m, OAP=60° and OBP=45°Let OP=hIn AOP,tan60°=OPOA3=hOAOA=h3Also, in BOP,tan45°=OPOB1=hOBOB=hNow, OB-OA=100h-h3=100h3-h3=100h3-13=100h=10033-1×3+13+1h=10033+13-1h=1003+32h=503+1.732h=504.732 h=236.6 m

So, the height of the tower is 236.6 m.

Disclaimer: The answer given in the texbook is incorrect. The same has been rectified above.

Page No 638:

Question 32:

An electrician has to repair an electric fault on a pole of height 4 metres. He needs to reach a point 1 metre below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use, which when inclined at an angle of 60° to the horizontal would enable him to reach the required position? [Use 3 = 1.73]

Answer:


Let AC be the pole and BD be the ladder.

We have,

AC=4 m, AB=1 m and BDC=60°And, BC=AC-AB=4-1=3 mIn BDC,sin60°=BCBD32=3BDBD=3×23BD=23BD=2×1.73 BD=3.46 m

So, he should use 3.46 m long ladder to reach the required position.



Page No 639:

Question 33:

From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60°, respectively. Find
(i) the horizontal distance between AB and CD,
(ii) the height of the lamp post,
(iii) the difference between the heights of the building and the lamp post.                                                                                    [CBSE 2009]

Answer:


We have,

AB=60 m, ACE=30° and ADB=60°Let BD=CE=x and CD=BE=yAE=AB-BE=60-yIn ACE,tan30°=AECE13=60-yxx=603-y3         .....iAlso, in ABD,tan60°=ABBD3=60xx=603x=603×33x=6033x=203Substituting x=203 in i, we get203=603-y3y3=603-203y3=403y=4033y=40 mi the horizontal distance between AB and CD=BD=x=203=20×1.732=34.64 mii the height of the lamp post=CD=y=40 miii the difference between the heights of the building and the lamp post=AB-CD=60-40=20 m

Page No 639:

Question 34:

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.          [CBSE 2017]

Answer:

Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º  and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters



Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters.
 In DAB, we havetan45°=ABAD1=hvth=vt       .....(i)InCAB, we havetan30°=ABAC13=hvt+12v3h=vt+12v       .....(ii)
Substituting the value of h from equation (i) in equation (ii), we get
3t=t+12t=123-1=123-1×3+13+1=63+1 min

Page No 639:

Question 35:

An aeroplane is flying at a height of 300 m above the ground. Flying at this height the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° ad 60° respectively. Find the width of the river. [Use 3=1.732]           [CBSE 2017]

Answer:

Let CD be the height of the aeroplane above the river at some instant. Suppose A and B be two points on both banks of the river in opposite directions.


Height of the aeroplane above the river, CD = 300 m
Now,
CAD = ADX = 60º       (Alternate angles)
CBD = BDY = 45º        (Alternate angles)
In right ∆ACD,
tan60°=CDAC3=300ACAC=3003=1003 m
In right ∆BCD,
tan45°=CDBC1=300BCBC=300 m
∴ Width of the river, AB
= BC + AC
=300+1003=300+100×1.73 =300+173=473 m
Thus, the width of the river is 473 m.

Page No 639:

Question 36:

From a point on the ground the angles of elevation of the bottom and top of a communication tower fixed on the top of a 20-m-high building are 45° and 60° respectively. Find the height of the tower. [Take 3=1.732]             [CBSE 2017]

Answer:


Let BC be the 20 m high building and AB be the communication tower of height h fixed on top of the building. Let D be a point on ground such that CD = x m and angles of elevation made from this point to top and bottom of tower are 45° and 60°.
In BCD: tan 45°=BCCD=20x
1=20xx=20 m.

Also, in ACD: tan 60°=ACCD=20+hx
3=20+hx3=20+h2020+h=203h=203-20h=203-1=201.73-1=20×0.73h=14.64 m.

Page No 639:

Question 37:

From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45° and 30° respectively. Find the height of the hill.            [CBSE 2017]

Answer:

Let PQ be the hill of height h km. Let R and S be two consecutive kilometre stones, so the distance between them is 1 km.
Let QR = x km.

In PQR,    tan45°=PQQR1=hxh=x     ...i

In PQS,    tan30°=PQQS13=hx+13h=x+1     ...ii
From equation (i) and (ii) we get,
    3h=h+1h3-1=1h=13-1=3+13-13+1h=3+12=2.732=1.365 km
Hence, the height of the hill is 1.365 km.

Page No 639:

Question 38:

If at some time of the day the ratio of the height of a vertically standing pole to the length of its shadow on the ground is 3:1 then find the angle of elevation of the sun at that time.             [CBSE 2017]

Answer:


Let AB be the vertically standing pole of height h units and CB be the length of its shadow of s units.
Since, the ratio of length of pole and its shadow at some time of day is given to be 3:1.
ABBC=31.
 In ABC: 
tanθ=ABBC=31=3tanθ=tan 60°θ=60°.



Page No 646:

Question 1:

Choose the correct answer of the following question:

If the height of a vertical pole is equal to the length of its shadow on the ground, the angle of elevation of the sun is

(a) 0°                         (b) 30°                         (c) 45°                         (d) 60°                                                                            [CBSE 2014]

Answer:



Let AB represents the vertical pole and BC represents the shadow on the ground and θ represents angle of elevation the sun.

In ABC,tanθ=ABBCtanθ=xx            As, the height of the pole, AB=the length of the shadow, BC=xtanθ=1tanθ=tan45° θ=45°


Hence, the correct answer is option (c).

Page No 646:

Question 2:

Choose the correct answer of the following question:

If the height of a vertical pole is 3 times the length of its shadow on the ground, then the angle of elevation of the sun at that time is

(a) 30°                    (b) 45°                     (c) 60°                     (d) 75°                                                                                [CBSE 2012, 14]

Answer:


Here, AO be the pole; BO be its shadow and θ be the angle of elevation of the sun.

Let BO=xThen, AO=x3In AOB,tanθ=AOBOtanθ=x3xtanθ=3tanθ=tan60° θ=60°

Hence, the correct answer is option (c).



Page No 647:

Question 3:

If the length of the shadow of a tower is 3 times its height then the angle of elevation of the sun is

(a) 45°
(b) 30°
(c) 60°
(d) 90°

Answer:

(b) 30°

Let AB be the pole and BC be its shadow.



Let AB = h and BC = x  such that x = 3 h (given) and θ be the angle of elevation.
From ∆ABC, we have:
ABBC = tan θ

⇒ hx = h3h = tan θ
⇒ tan θ = 13
⇒ θ = 30o

Hence, the angle of elevation is 30o.

Page No 647:

Question 4:

Choose the correct answer of the following question:

If a pole 12 m high casts a shadow 43 m long on the ground, then the sun's elevation is

(a) 60°                    (b) 45°                    (c) 30°                    (d) 90°                                                      [CBSE 2013C]

Answer:



Let AB be the pole, BC be its shadow and θ be the sun's elevation.

We have,AB=12 m and BC=43 m In ABC,tanθ=ABBCtanθ=1243tanθ=33tanθ=33×33tanθ=333tanθ=3tanθ=tan60° θ=60°

Hence, the correct answer is option (a).

Page No 647:

Question 5:

Choose the correct answer of the following question:

The shadow of a 5-m-long stick is 2 m long. At the same time, the length of the shadow of a 12.5-m-high tree is

(a) 3 m                    (b) 3.5m                   (c) 4.5 m                    (d) 5 m                                                                      [CBSE 2011]

Answer:



Let AB be a stick and BC be its shadow; and PQ be the tree and QR be its shadow.

We have,

AB=5 m, BC=2 m, PQ=12.5 mIn ABC,tanθ=ABBCtanθ=52        .....iNow, in PQR,tanθ=PQQR52=12.5QR             Using iQR=12.5×25=255 QR=5 m

Hence, the correct answer is option (d).

Page No 647:

Question 6:

Choose the correct answer of the following question:

A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder is

(a) 43 m                   (b) 43 m                   (c) 22 m                   (d) 4 m                                        [CBSE 2014]

Answer:


Let AB be the wall and AC be the ladder.

We have,


BC=2 m and ACB=60°In ABC,cos60°=BCAC12=2AC AC=4 m

Hence, the correct answer is option (d).

Page No 647:

Question 7:

Choose the correct answer of the following question:

A ladder 15 m long makes an angle of 60° with the wall. Find the height of the point, where the ladder touches the wall.

(a) 153 m

(b) 1532m

(c) 152 m

(d) 15 m

Answer:


Let AB be the wall and AC be the ladder.

We have,

AC=15 m and BAC=60°In ABC,cos60°=ABAC12=AB15 AB=152 m

Hence, the correct answer is option (c).

Page No 647:

Question 8:

Choose the correct answer of the following question:

From a point on the ground, 30 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The height of the
tower is

(a) 30 m                    (b) 103 m                    (c) 10 m                    (d) 303 m                                                                       [CBSE 2014]

Answer:



Let AB be the tower and point C be the point of observation on the ground.

We have,

BC=30 m and ACB=30°In ABC,tan30°=ABBC13=AB30AB=303AB=303×33AB=3033 AB=103 m

Hence, the correct answer is option (b).

Page No 647:

Question 9:

Choose the correct answer of the following question:

The angle of depression of a car parked on the road from the top of a 150-m-high tower is 30°. The distance of the car from the tower is

(a) 503 m                   (b) 1503 m                   (c) 1502 m                     (d) 75 m                                                                [CBSE 2014]

Answer:


Let AB be the tower and point C be the position of the car.

We have,

AB=150 m and ACB=30°In ABC,tan30°=ABBC13=150BC BC=1503 m

Hence, the correct answer is option (b).

Page No 647:

Question 10:

Choose the correct answer of the following question:

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 there 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°                                                                                                      [CBSE 2012]

Answer:


Let point A be the position of the kite and AC be its string.

We have,

AB=30 m and AC=60 mLet ACB=θIn ABC,sinθ=ABACsinθ=3060sinθ=12sinθ=sin30° θ=30°

Hence, the correct answer is option (b).

Page No 647:

Question 11:

Choose the correct answer of the following question:

From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the
tower. The height of the tower is

(a) 20 m                    (b) 40 m                    (c) 60 m                    (d) 80 m                                                                                        [CBSE 2013C]

Answer:



Let AB be the cliff and CD be the tower.

We have,

AB=20 mAlso, CE=AB=20 mLet ACB=CAE=DAE=θIn ABC,tanθ=ABBCtanθ=20BCtanθ=20AE          As, BC=AEAE=20tanθ             .....iAlso, in ADE,tanθ=DEAEtanθ=DE20tanθ           Using itanθ=DE×tanθ20DE=20×tanθtanθDE=20 mNow, CD=DE+CE=20+20 CD=40 m

Hence, the correct answer is option (b).

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

Page No 647:

Question 12:

Choose the correct answer of the following question:

If a l.5-m-tall girl stands at a distance of 3 m from a lamp post and casts a shadow of length 4.5 m on the ground, then the height of the lamp post is

(a) 1.5 m                      (b) 2 m                      (c) 2.5 m                      (d) 2.8 m

Answer:


Let AB be the lamp post; CD be the girl and DE be her shadow.

We have,

CD=1.5 m, AD=3 m, DE=4.5 mLet E=θIn CDE,tanθ=CDDEtanθ=1.54.5tanθ=13            .....iNow, in ABE,tanθ=ABAE13=ABAD+DE             Using i13=AB3+4.5AB=7.53 AB=2.5 m

Hence, the correct answer is option (c).



Page No 648:

Question 13:

Choose the correct answer of the following question:

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30° than when it was
45°. The height of the tower is

(a) 23x m                   (b) 32x m                    (c) 3-1x m                    (d) 3+1x m

Answer:


Let CD = h be the height of the tower.

We have,

AB=2x, DAC=30° and DBC=45°In BCD,tan45°=CDBC1=hBCBC=hNow, in ACD,tan30°=CDAC13=hAB+BC13=h2x+h2x+h=h3h3-h=2xh3-1=2xh=2x3-1×3+13+1h=2x3+13-1h=2x3+12 h=x3+1 m

Hence, the correct answer is option (d).

Page No 648:

Question 14:

Choose the correct answer of the following question:

The lengths of a vertical rod and its shadow are in the ratio 1:3. The angle of elevation of the sun is

(a) 30°                    (b) 45°                     (c) 60°                     (d) 90°

Answer:


Let AB be the rod and BC be its shadow; and θ be the angle of elevation of the sun.

We have,AB:BC=1:3Let AB=xThen, BC=x3In ABC,tanθ=ABBCtanθ=xx3tanθ=13tanθ=tan30° θ=30°

Hence, the correct answer is option (a).

Page No 648:

Question 15:

Choose the correct answer of the following question:

A pole casts a shadow of length 23 m on the ground when the sun's elevation is 60°. The height of the pole is

(a) 43 m                    (b) 6 m                    (c) 12 m                    (d) 3 m                                                                     [CBSE 2015]

Answer:


Let AB be the pole and BC be its shadow.

We have,

BC=23 m and ACB=60°In ABC,tan60°=ABBC3=AB23 AB=6 m

Hence, the correct answer is option (b).

Page No 648:

Question 16:

Choose the correct answer of the following question:

In the given figure, a tower AB is 20 m high and BC, its shadow on the ground is 203 m long. The sun's altitude is

(a) 30°                   (b) 45°                   (c) 60°                   (d) none of these                                                                         [CBSE 2015]

Answer:


Let the sun's altitude be θ.

We have,

AB=20 m and BC=203 mIn ABC,tanθ=ABBCtanθ=20203tanθ=13tanθ=tan30° θ=30°

Hence, the correct answer is option (a).

Page No 648:

Question 17:

Choose the correct answer of the following question:

The tops of two towers of heights x and y, standing on a level ground subtend angles of 30° and 60°, respectively at the centre of the line joining their feet. Then, x : y is

(a) 1 : 2                  (b) 2 : 1                  (c) 1 : 3                  (d) 3 : 1                                                                                       [CBSE 2015]

Answer:



Let AB and CD be the two towers such that AB = and CD = y.

We have,

AEB=30°, CED=60° and BE=DEIn ABE,tan30°=ABBE13=xBEBE=x3Also, in CDE,tan60°=CDDE3=yDEDE=y3As, BE=DEx3=y3xy=13×3xy=13 x:y=1:3

Hence, the correct answer is option (c).

Page No 648:

Question 18:

The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30°. The height of the tower is

(a) 30 m
(b) 103m
(c) 20 m
(d) 102m

Answer:

(b) 103m
Let AB be the tower and O be the point of observation.
Also,
AOB = 30o and OB = 30 m
Let:
AB = h m

In ∆AOB, we have:
ABOB = tan 30o = 13

h30= 13
h = 303×33 = 3033 = 103 m

Hence, the height of the tower is 103 m.

Page No 648:

Question 19:

The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. If these is no slack in the string, the height of the kite from the ground is

(a) 503m
(b) 1003m
(c) 502m
(d) 100 m

Answer:

(a) 503m
Let AB be the string of the kite and AX be the horizontal line.
If BC ⊥ AX, then AB = 100 m and ∠BAC = 60o.
Let:
BC = h m

In the right ∆ACB, we have:
BCAB =sin 60o  = 32

h100 = 32
h = 10032 = 503 m
Hence, the height of the kite is 503 m.

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Question 20:

If the angles of elevation of the top of a tower form tow points at distances a and b from the base and in the same straight line with it are complementary, then the height of the tower is

(a) ab
(b) ab
(c) a+b
(d) a-b

Answer:

(b) ab
Let AB be the tower and C and D be the points of observation on AC.
Let:
ACB = θ, ∠ADB = 90-θ and AB = h m
Thus, we have:
AC = a, AD = b and CD = a - b

Now, in the right ∆ABC, we have:
tanθ = ABAC ⇒ ha = tanθ             ...(i)
In the right ∆ABD, we have:
tan(90-θ) = ABAD  ⇒ cotθ = hb    ...(ii)

On multiplying (i) and (ii), we have:
tanθ×cotθ = ha×hb
ha×hb = 1                     [ ∵ tanθ = 1cotθ]
⇒ h2 = ab
⇒  h = ab m

Hence, the height of the tower is ab m.

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Question 21:

On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is

(a) 10 m
(b) 103m
(c) 15 m
(d) 53m

Answer:

(b) 103m
Let AB be the tower and C and D be the points of observation such that ∠BCD = 30o, ∠BDA = 60o, CD = 20 m  and AD = x m.

Now, in ∆ADB, we have:
ABAD = tan 60o = 3

ABx = 3    
⇒  AB = 3x

In ∆ACB, we have:
ABAC= tan 30o = 13
AB20 + x = 13  ⇒ AB = 20 + x3
∴ 3x = 20 + x3 
⇒ 3x = 20 + x
⇒ 2x = 20 ⇒ x = 10
∴ Height of the tower AB = 3x = 103 m

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Question 22:

In a rectangle, the angle between a diagonal and a side is 30° and the length of this diagonal is 8 cm. the area of the rectangle is

(a) 16 cm2
(b) 163cm2
(c) 163cm2
(d) 83cm2

Answer:

(c) 163cm2
Let ABCD be the rectangle in which ∠BAC = 30o and AC = 8 cm.

In ∆BAC, we have:

ABAC= cos 30o = 32

AB8= 32
⇒ AB= 832= 43 m
Again,

BCAC= sin 30o = 12

BC8= 12
BC =82= 4 m
∴ Area of the rectangle = (AB×BC) = (43×4) = 163 cm2



Page No 649:

Question 23:

From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°. The height of the hill is

(a) 12(3-1)km
(b) 12(3+1)km
(c) (3-1)km
(d) (3+1)km

Answer:

(b) 12(3+1)km
Let AB be the hill making angles of depression at points C and D such that ∠ADB =45o, ∠ACB = 30o and CD = 1 km.
Let:
AB = h km and AD = x km

In ∆ADB, we have:
ABAD = tan 45o =1 

⇒ hx  = 1  ⇒ h = x        ...(i)
In ∆ACB, we have:
ABAC = tan 30o = 13

hx+1 = 13             ...(ii)
On putting the value of h taken from (i) in (ii), we get:
hh+1= 13
3h = h+1
(3-1)h = 1
h = 1(3-1)
On multiplying the numerator and denominator of the above equation by (3+1), we get:
h = 1(3-1)×(3+1)(3+1) = (3+1)3-1= (3+1)2 = 12(3+1) km

Hence, the height of the hill is 12(3+1) km.

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Question 24:

If the elevation of the sun changes form 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, is

(a) 7.5 m
(b) 15 m
(c) 103m
(d) 53m

Answer:

(c) 103m
Let AB be the pole and AC and AD be its shadows.
We have:
ACB = 30o, ∠ADB = 60o and AB = 15 m

In ∆ACB, we have:
ACAB = cot 30o = 3
⇒ AC15 = 3  ⇒ AC = 153 m

Now, in ∆ADB, we have:
ADAB = cot 60o = 13
⇒ AD15 = 13  ⇒ AD = 153  = 15× 33×3  = 1533 = 53 m

∴ Difference between the lengths of the shadows = AC - AD = 153 - 53 = 103 m

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Question 25:

An observer 1.5 m tall 28.5 away from a tower and the angle of elevation of the top of the tower form the eye of the observer is 45°. The height of the tower is

(a) 27 m
(b) 30 m
(c) 28.5 m
(d) none of these

Answer:

(b) 30 m
Let AB be the observer and CD be the tower.

Draw BE ⊥ CD, Let CD = h metres. Then,
AB = 1.5 m , BE = AC = 28.5 m and ∠EBD = 45o.
DE = (CD - EC) = (CD - AB) = (h - 1.5) m.
In right  ∆BED, we have:
DEBE = tan 45o = 1

(h - 1.5)28.5= 1

h - 1.5 = 28.5
⇒ h = 28.5 + 1.5 = 30 m
Hence the height of the tower is 30 m.



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