Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 14 Heights And Distances are provided here with simple step-by-step explanations. These solutions for Heights And Distances are extremely popular among Class 10 students for Maths Heights And Distances Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 656:

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 657:

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 657:

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 657:

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 657:

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 657:

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 657:

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 657:

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 657:

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 657:

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 657:

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 658:

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 658:

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 658:

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 658:

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 658:

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 658:

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 658:

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 658:

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 659:

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 659:

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 659:

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 659:

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 659:

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 659:

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 659:

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 659:

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 659:

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 659:

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 660:

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 660:

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 660:

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 660:

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 660:

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 660:

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 660:

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 660:

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 661:

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 661:

Answer:




Let AB be the light house. Suppose C and D be the two positions of the boat.

Here, AB = 100 m.

Let the speed of the boat be v m/min.

So,

CD = v m/min × 2 min = 2v m          [Distance = Speed × Time]

In right ∆ABC,

tan60°=ABBC3=100 mBCBC=1003=10033m       .....1

In right ∆ABD,

tan30°=ABBD13=100 mBC+CD13=10010033+2v          From 1
2v+10033=10032v=1003-100332v=10031-132v=1003×23
v=100 × 1.7323=57.73 m/min

Thus, the speed boat is 57.73 m/min.

Page No 661:

Answer:


Let AB be the building and CD be the tower.

Draw AE ⊥ CD.



Suppose the height of the tower be h m.

Here, AB = 8 m

∠DAE = ∠ADX = 30º        (Alternate angles)

∠DBC = ∠BDX = 45º        (Alternate angles)

CE = AB = 8 m

∴ DE = CD − CE = (h − 8) m

Distance between the building and tower = BC

In right ∆BCD,

tan45°=CDBC1=hBCBC=h m

In right ∆AED,

tan30°=DEAE13=h-8h                   AE=BC3h-83=h
3h-h=833-1h=83h=833-1h=833+13-1×3+1
h=833+12         a-ba+b=a2-b2h=433+1h=12+43 m

∴ Height of the tower = 12+43 m

Also,

Distance between the tower and building = BC = 12+43 m                 (BC = h m)

Thus, the the height of the tower is 12+43 m and the distance between the tower and the building is 12+43 m.



Page No 671:

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 671:

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 671:

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 671:

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 671:

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 672:

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 672:

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 672:

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 672:

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 672:

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 672:

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 672:

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 672:

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 672:

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 673:

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 673:

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 673:

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 673:

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 673:

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.

Page No 673:

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.

Page No 673:

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

Page No 673:

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 673:

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.

Page No 673:

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



Page No 674:

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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