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RS Aggarwal 2015 Solutions for Class 10 Math 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 Math Heights And Distances Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2015 Book of class 10 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2015 Solutions. All RS Aggarwal 2015 Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 572:

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 572:

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 = 60o, OA = 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 573:

Question 3:

An observe, 1.6 m tall, is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°. Determine the height of the tower.

Answer:

Let XY be the horizontal ground and X be the position of the observer.
Let Y be the position of the tower such that XY is 45 m.
Let AY = h m be the height of the tower.
Now,
Height of the observer, CX = 1.6 m

From the right ∆ABC, we have:
 ABBC = tan 30o = 13
⇒ AB45 = 13 
⇒  AB = 453 = 451.732 = 25.98 m
∴ AY = AB + BY = 25.98 + 1.6 = 27.58 m          [∵ CX = BY]
⇒ h = 27.58 m

​Hence, the height of the tower is 27.58 m.

Page No 573:

Question 4:

The height of a tree is 10 m. It is bent by the wind in such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom did the tree get bent?

Answer:

Let AB be the tree broken at point C such that CB takes the position CD and ∠ADC = 60o.
If CB = CD = y m, then AC = 10 - y m.

In ∆ADC, we have:
ACCD=sin 60°⇒10 - yy = sin 60°= 32
⇒ 20 - 2y = 3y
⇒ y (3 + 2) =20 
⇒ y = 20(3 + 2) m
Now,
AC = 10 - y = 10 - 20(3 + 2) = (103 + 20) - 20(3 + 2) = 103(3 + 2) 

     = 10 × 1.732(1.732 + 2) = 17.323.73 = 4.64 m

Hence, the height from the bottom from where the tree got bent is 4.64 m.

Page No 573:

Question 5:

A 1.5-m-tall boy stands at a distance of 3 m from a lamp post and casts a shadow of 4.5 m on the ground. Find the height of the lamp post.

Answer:

Let AB be the lamppost, CD be the boy and CE be the shadow of CD.
Also, let ∠AEB = θ and the height of the lamppost be h.



From the right ∆​ECD, we have:
tan θ = CDCE = 1.54.5 = 13

Now, from the right ∆EAB, we have:
ABAE = tan θ = 13                             [∵ tan θ = 13]

⇒ h(AC + CE) = h(4.5 + 3) = 13

⇒ h = 7.53 = 2.5 m

∴ Height of the lamppost = AB = h = 2.5 m

Page No 573:

Question 6:

The angle of elevation of the top of a building from a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base from the point A.

Answer:

Let CD be the building. Then, the angle of elevation from the top of the building on the ground is such that ∠DAC = 30o.
After moving to point B, the angle of elevation changes such that ∠DBC = 45o and AB = 30 m.
Let:
CD = h m and BC = x m

In the right ∆DCB, we have:
hx = tan 45o = 1
⇒ h = x

Now, in the right ∆DCA, we have:
hx + 30= tan 30o = 13

On putting h = x in the above equation, we get:
hh + 30 =13 
⇒ 3h = h + 30
⇒ h (3 - 1) = 30
⇒ h = 30(3 - 1) =30 × (3 +1)(3 - 1) × (3 +1) = 30 (3 + 1)2  = 15 (3 + 1) =15 × 2.732 =40.98 m

Hence, the height of the building is 40.98 m.
Now,
 x = 40.98 m

∴ Distance of the base of the building from point A = AC = x + 30 = 40.98 + 30 = 70.98 m

Page No 573:

Question 7:

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 m towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.

Answer:

Let CD be the tower making an angle of elevation at point A on the ground such that ∠DAC = 30o.
On moving towards the foot of the tower at point B, AB = 20 m and ∠DBC = 60o.
Let:
CD = h m and BC = x m

In the right ∆DCA, we have:
hx + 20 = tan 30o= 13
⇒ 3h = x + 20                    ...(i)

Now, in the right ∆DCB, we have:
hx = tan 60o = 3
⇒ hx = 3
⇒ x = h3                           ...(ii)
 By using (ii) in (i), we get:
3h = h3+ 20
⇒ 3h = h + 203
⇒ 2h = 203
⇒ h = 103 = 10 ×1.732 = 17.32 m

∴ Height of the tower = 17.32 m
Now,
x = h3 = 1033 = 10 m
∴ Distance of the tower from point A = AC = x + 20 = 10 + 20 = 30 m

Page No 573:

Question 8:

From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find height of the tower, and the distance between the tower and the building.

Answer:

Let AB be the tower and CD be the building.
We have:
CD = 15 m, ∠ADE = 30o and ∠ACB = 60o
Let AB = h m and BC = x m such that ED =x m.

In the right ∆ABC, we have:
hx = tan 60o = 3
⇒ x = h3
Now, in the right ∆AED, we have:
h - 15x = tan 30o = 13 
On putting x= h3 in the above equation, we get:
h - 15h3 = 13
 
⇒ (h - 15)3h = 13
⇒ 3h - 45 = h
⇒ 2h = 45
⇒ h = 22.5 m

​Hence, the height of the tower is 22.5 m.
Now,
Distance between the tower and the building = x = h3 = 22.53 = 22.51.732 = 12.99 m

Page No 573:

Question 9:

From a window 15 m high above the ground, in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 30° and 45° respectively. Show that the height of the opposite house is 23.66 m.

Answer:

Let AB be the house and CD be the window.
We have:
CD = 15 m such that BE = 15 m, ∠ADE = 30o and ∠DBC = 45o.
Let AB = h m such that AE = (h - 15) m and let BC = x m such that DE = x m.

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

⇒ 15x = 1
⇒ x = 15 m

Now, in the right ∆ADE, we have:
AEDE = tan 30o = 13
⇒ h - 15x = 13
On putting x = 15 in the above equation, we get:
h - 1515 = 13
⇒ 3(h - 15) = 15
⇒ 3h - 153 = 15
⇒ 3h = 40.98
⇒ h = 40.983 = 40.981.732= 23.66 m
Hence proved.

Page No 573:

Question 10:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff. At a point on the plane, 30 metres away from the tower, an observe notices that the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the flagstaff and that of the tower.

Answer:

Let OX be the horizontal line, AC be the vertical tower and BC be the vertical flagstaff.
We now have:
OA = 30 m, ∠BOA = 60o and ∠COA = 45o
Let:
AC= h m and BC = x m



In the right ∆AOC, we have:
ACOA = tan 45o = 1

⇒ h30 = 1
⇒ h = 30 m

Now, in the right ∆AOB, we have:
ABOA = tan 60o = 3

⇒ h + x30 = 3
On putting h = 30 in the above equation, we get:
30 + x = 303
⇒ x = 303 - 30 = 21.96 m

We now have:
Height of the flagstaff = x = 21.96 m
Height of the tower = h = 30 m

Page No 573:

Question 11:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. From a point on the plane the angles of elevation of the bottom and the top of the flagstaff are 30° and 60°. Find the height of the tower.

Answer:

Let OX be the horizontal line, AC be the vertical tower and BC be the vertical flagstaff such that BC = 5 m.
Let:
AC = h m and OA = x m



In the right ∆AOC, we have:
ACOA = tan 30o = 13

⇒ hx = 13
⇒ x = 3h

Now, in the right ∆AOB, we have:
ABOA = tan 60o = 3

⇒ (h + 5)x = 3

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

⇒ (h + 5)3h =3
⇒ h + 5 = 3h
⇒ 2h = 5
⇒ h = 52 = 2.5 m
​Hence, the height of the tower is 2.5 m.

Page No 573:

Question 12:

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.

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) = 1.460.732 = 1.99 m = 2 m

Hence, the height of the pedestal is 2 m.



Page No 574:

Question 13:

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 574:

Question 14:

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.

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 = 181.732 = 10.39

Thus, we have:
Height of the flagpole = 10.39 m
Height of the tower = 5.19 m

Page No 574:

Question 15:

From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill.

Answer:

Let AB be the hill and DE be the pillar. Draw CD ⊥ AB.
Thus, we have:
AB = 200 m, ∠BEA = 60o and  ∠BDC = 30o
Now, let AE = x m such that CD = x m and let DE = h m such that AC = h m.



In the right ∆AEB, we have:
ABAE = tan 60o = 3

⇒ 200x = 3
⇒ x = 2003 = 115.47 m
Now, in the right ∆BDC, we have:
BCCD = tan 30o = 13

⇒ (200 - h)x = 13

By putting x = 2003 in the above equation, we get:
(200 - h)3200 = 13
⇒ 600 - 3h = 200
⇒ 3h = 400
⇒ h = 4003 = 133.33 m

We now have:
Height of the pillar = 133.33 m
Distance of the pillar from the hill = 115.47 m

Page No 574:

Question 16:

From a window, 60 metres high above the ground, of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of the opposite house is 60(1+3) metres.

Answer:

Let AB be the house and DE be the window such that DE = 60 m. Draw CE ⊥ AB.
Thus, we have:
∠BEC = 60o and ∠EAD = 45o
Let AB = h m and CA = 60 m such that BC = (h - 60) m and AD = x m.

In the right ∆ADE, we have:
DEAD = tan 45o = 1

⇒ 60x = 1
⇒ x = 60 m
Now, in the right ∆BEC, we have:
BCEC = tan 60o = 3

⇒ (h - 60)x= 3
 
On putting x = 60 in the above equation, we get:
(h - 60)60 = 3

⇒ h - 60= 603
⇒ h = 603 + 60 = 60(3 + 1) m

Hence, the height of the opposite house is 60(3 + 1) m.

Page No 574:

Question 17:

The angle of elevation of a jet plane 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 plane is flying at a constant height of 15003 m, find the speed of the jet plane.

Answer:

Let C and C' be the two positions of the jet plane and BC and B'C' be its constant heights.
Thus, we have:
∠CAB=60° and ∠C'AB'=30°Let: AB=x and BB'=y



Now, in the right ∆ABC, we have:
tan 60°=BCAB⇒3=15003x⇒x=1500

Now, in the right ∆AB'C', we have:
tan 30°=B'C'AB'⇒13=15003x+y⇒x+y=1500×3=4500⇒y=4500-1500⇒y=3000

​BB' is 3000 m. This means that to cover a distance of 3000 m, the jet plane takes 15 seconds.
∴ Speed of the jet plane =3000 m15 sec=200 msec=200×185=720 kmh

Page No 574:

Question 18:

The angles of elevation and depression of the top and bottom of a lighthouse form the top of a building, 60 m high, are 30° and 60° respectively. Find (i) the difference between the heights of the lighthouse and the building, and (ii) the distance between the lighthouse and the building.

Answer:

Let AB be the lighthouse and DE be the building. Draw CE ∥ BD.
We have:
DE = 60 m, ∠AEC = 30o and ∠EBD = 60o
Let AB = h m such that BC = 60 m and AC = (h - 60) m.
Let BC = x.

In the right ∆EDB, we have:
EDBD = tan 60o = 3

⇒ 60x = 3
⇒ x = 603 = 34.64 m
Now, in the right ∆AEC, we have:
ACCE = tan 30o = 13

⇒ (h - 60)x = 13
On putting x = 603 in the above equation, we get:
(h - 60)360 = 13
⇒ (h - 60)3 = 60
⇒3h-180 = 60⇒3h=240⇒h=80

∴ Difference between the heights of the lighthouse and building = AC = (h - 60) = (80 - 60) = 20 m
Distance between the lighthouse and the building = x = 34.64 m

Page No 574:

Question 19:

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 574:

Question 20:

The angle of elevation of the top of a building form a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base form the point A.

Answer:

Let CD be the building and A be the point on the ground making an angle of elevation on top of the building.
Thus, we have:
∠DAB = 30o, ∠DBC = 45o and AB = 30 m
Let:
CD = h m and BC = x m

In the right ∆DBC, we have:
CDBC = tan 45o = 1
⇒ hx = 1
⇒ h = x  m 
Or,
x = h m
In the right ∆DAC, we have:
CDAC = tan 30o = 13

⇒ h(x + 30) = 13
⇒ 3h = x + 30
By putting x = h in the above equation, we get:
3h = h + 30
⇒ h(3 - 1) = 30
⇒ h = 30(3 - 1) = 30(3 - 1)×(3 + 1)(3 + 1) = 15 (3 + 1) = 40.98 m

∴ x = 40.98 m

Distance of the base of the building from point A = AC = (x + 30) = (40.98 + 30) = 70.98 m
Height of the building = h = 40.98 m

Page No 574:

Question 21:

The angle of elevation of the top of a tree from a point A on the ground is 60°. On walking 20 metres away from its base, to a point B, the angle of elevation changes to 30°.Find the height of the tree.

Answer:

Let CD be the tree and A be the point on the ground such that ∠DAC = 60o, AB = 20 m and ∠DBC = 30o.
Let:
CD = h m and AC = x m

In the right ∆DAC, we have:
CDAC = tan 60o = 3

⇒ hx = 3  (or)  x = h3
In the right ∆DBC, we have:
CDBC = tan 30o = 13

⇒ h(x + 20) = 13
⇒ 3h = x + 20

On putting x = h3, we get:
3h = h3 + 20
⇒ 3h = h + 203
⇒ 2h = 203
⇒ h = 2032 = 103 = 17.32 m
Hence, the height of the tree is 17.32 m.

Page No 574:

Question 22:

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 the foot of the tower is 60°. Find the height of the tower.

Answer:

Let AB be the tower and CD be the building. Draw CD⊥ AB.
We have:
CE = 7 m, ∠BCD = 60o and ∠CAE = 60o
Let AB = h m and  AE = x m such that CD = x m and BD = (h - 7) m.

In the right ∆​CAE, we have:
CEAE = tan 30o = 13

⇒ 7x = 13
⇒ x = 73
Now, in the right ∆BCD, we have:
BDCD = tan 60o = 3

⇒ (h - 7)x = 3
⇒ h - 7 = x3
By putting x = 73 in the above equation, we get:
h - 7 = 73×3 ⇒h-7=21⇒h=28

Hence, the height of the tower is 28 m.



Page No 575:

Question 23:

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

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

Page No 575:

Question 24:

From a point P on the ground, the angle of elevation of a 10-m-tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from  P is 45°. Find the length of the flagstaff and the distance of the building from the point P.

Answer:

Let AC be the building, P be the point on the ground and BC be the flagstaff.
Thus, we have:
AC = 10 m, ∠CPA = 30o and ∠BPA = 45o
Let:
BC = h m and AP = x m

In the right ∆CAP, we have:
ACAP = tan 30o = 13
⇒ 10x = 13

⇒ x = 103  = 17.32 m

In the right ∆BAP, we have:
ABAP = tan 45o = 1

⇒ (h + 10)x = 1
⇒ h + 10 = x

On putting x = 103 in the above equation, we get:
h + 10 = 103
⇒ h = 103 - 10 = 10 (3 - 1) = 7.32 m

Thus, we have:
Length of the flagstaff = BC = h = 7.32 m
Distance of the building from point P = AP = x = 17.32 m

Page No 575:

Question 25:

The angles of depression of the top and bottom of a 10 m high building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building and the distance between the two buildings.

Answer:

Let AB be the multistorey building and DE be the high building such that DE = 10 m. Draw CD ⊥ AB.
Thus, we have:
∠​BDC = 30o and ∠BEA= 45o
Let AB = h m such that BC = (h - 10) m and AE = x m such that CD = x m.

In the right ∆BEA, we have:
ABAE = tan 45o = 1

⇒ hx = 1
⇒ x = h 
Or,
h = x
Now, in the right ∆BDC, we have:
BCCD = tan 30o = 13

⇒ (h - 10)x = 13
⇒ (h - 10)3 = x
By putting x = h in the above equation, we get:
(h - 10)3 = h
⇒ h (3 - 1) = 103
⇒ h = 103(3 -1) × (3 + 1)(3 + 1) = 53 (3 +1) = 5 (3 + 3) = 23.66 m
We have:
Height of the multistorey building = AB = h = 23.66 m
Distance between the two buildings = AE = x = h = 23.66 m

Page No 575:

Question 26:

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 575:

Question 27:

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.

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 575:

Question 28:

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.

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 575:

Question 29:

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of each pole and the distances of the point form the poles.

Answer:

Let AB and CD be the two poles of equal heights such that AB = CD = h m.
∴ Distance between the two poles = AD = 80 m
We have:
∠BOA = 60o and ∠COD = 30o
Let AO = x m such that DO = (AD - AO) = (80 - x)  m.

​In the right ∆BOA, we have:
ABAO = tan 60o = 3

⇒ hx = 3
⇒ h = x3                   ...(i)

Now, in the right ∆COD, we have:
CDOD = tan 30o = 13
⇒ h(80 - x) = 13
⇒ h3 = 80 - x
⇒ h = (80 - x)3          ...(ii)
From (i) and (ii), we have:
x3 = (80 - x)3
⇒ 3x = 80 - x
⇒ 4x = 80
⇒ x = 804 = 20 m
∴ h = x3 = 203 = 34.64 m

We now have:
Height of each pole = AB = CD = h = 34.64 m
Distance of the pole AB from point O = x = 20 m
Distance of the pole CD from point O = (80 - x) = ( 80 - 20) = 60 m

Page No 575:

Question 30:

As observed form the top of a 75-m-tall lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:



Let AB be the lighthouse such that AB = 75 m and C and D be the positions of the two ships.
Thus, we have:
∠ACB = 45o and ∠ADB =30o
Let BD = x m and BC = y m such that CD = (x - y) m.
In the right ∆​ABC, we have:
ABBC = tan 45o = 1

⇒ 75y = 1
⇒ y = 75 m
 
In the right ∆ABD, we have:
ABBD = tan 30o = 13

⇒ 75x =13
= x = 753  m

∴ Distance between the two ships = CD = (x - y) = (753 - 75) = 75 (3 - 1) = 54.90 m

Page No 575:

Question 31:

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 576:

Question 32:

A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat to reach the shore.

Answer:

Let C be the cliff and A and D be the two positions of the boat.
We have:
∠BAC = 30o and ∠BDC = 60o
Let the speed of the boat be v metres per minute.
Also,
AD = Distance covered by the boat in six minutes
     = 6v m
Suppose the boat takes t minutes to reach B from D. 
∴ DB = vt m
Let:
BC = h m

In the right ∆DBC, we have:
BCDC = tan 60o = 3

⇒ hvt = 3
⇒ h = 3vt          ...(i)

From the right ∆ACB, we have:
BCAC = tan 30o = 13
⇒ h(6v + vt) = 13

⇒ h = (6v + vt)3  ...(ii)
From (i) and (ii), we have:
3vt = (6v + vt)3
⇒ 3vt = 6v + vt
⇒ 2vt = 6v
t = 6v2v = 3 minutes
Hence, the required time is three minutes.

Page No 576:

Question 33:

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 a 20 m high building, finds the angle of elevation of the same bird to be 45°. Boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl.

Answer:

Let O be the position of the bird, B be the position of the boy and FG be the roof on which G is the position of the girl.
Let:
OL ⊥ BF and GM ⊥ OL
Thus, we have:
BO = 100 m, ∠OBL = 30o, FG = 20 m and ∠OGM= 45o.
Let:
OG = x m

From the right ∆OLB, we have:
OLOB = sin 30o = 12
⇒ OL100 = 12
⇒ OL = 1002 = 50 m
∴ OM = (OL - LM) = (50 - 20) = 30 m

From the right ∆OMG, we have:
OMOG = sin 45o = 12
⇒ 30x = 12
⇒ x = 302 = 30 ×1.41 = 42.3 m

∴ Distance of the bird from the girl = OG = x = 42.3 m

Page No 576:

Question 34:

A 1.5-m-tall boy is standing at some distance from a 30-m-tall building. The angle of elevation from his eyes to the top of the building increase form 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

Let AB be the building and CD and EF be the two positions of the boy. Draw DFG ∥ CEA.
Thus, we have:
CD = EF = 1.5 m, AB = 30 m, ∠GDB = 30o and ∠GFB= 60o and GB= (AB - GA) = (30 - 1.5) = 28.5 m

Now, in ∆BDG, we have:
DGGB = cot 30o = 3

⇒ DG28.5= 3
⇒ DG= 28.53 = 28.53× 22 = 5732 m
In ∆BFG, we have:
FGGB= cot 60o =13
⇒ FG28.5= 13
⇒ FG = 28.53 =28.5 ×223 = 5723 m
∴ DF = (DG - FG) =(5732  - 5723) = 572(3 - 13) = 572(23×33) = 193 m = 32.87 m
Hence, the distance walked by the boy towards the building is 32.87 m.

Page No 576:

Question 35:

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 579:

Question 1:

At a certain instant the ratio of the lengths of a pillar and its shadow are in the ratio 1:3. At that instant, the angle of elevation of the sum is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

Answer:

(a) 30°
Let AB be the pillar and AC be its shadow.
Let:
∠ACB = θ and AB:BC = 1:3

In the right ∆ABC, we have:
ABAC = tan θ

⇒k3k = tan θ⇒13= tan θ⇒tan 30°=tan θ
⇒ θ = 30o                               [∵ tan 30o = 13]
Hence, the angle of elevation of the sun is 30o.

Page No 579:

Question 2:

At a certain instant, the altitude of the sum is 60°. At that instant, the length of the shadow of a vertical tower is 100 m. The height of the tower is

(a) 503m
(b) 1003m
(c) 1003m
(d) 2003m

Answer:

(b) 1003m
Let AB be the tower and AC be its shadow.
We have:
∠ACB = 60o and AC = 100 m

Let:
AB = h m
In the right ∆BAC, we have:
ABAC = tan 60o = 3

⇒ h100 =3
⇒ h = 1003 m

Hence, the height of the tower is 1003 m.

Page No 579:

Question 3:

The height of a tower is 1003m. The angle of elevation of its top from a point 100 m away from its foot is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

Answer:

(c) 60°
Let AB be the tower and O be the point of observation.
We have:
AB = 1003 m and OB = 100 m

In ∆AOB, we have:
ABOB = tan θ

⇒ 1003100 = tan θ
⇒ 3=tan θ 
⇒ tan 60°=tan θ
⇒ θ = 60o

Hence, the angle of elevation is 60o.

Page No 579:

Question 4:

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 579:

Question 5:

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.

Page No 579:

Question 6:

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.

Page No 579:

Question 7:

The shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is

(a) 5(3+1)m
(b) 10(3-1)m
(c) 9 m
(d) 13 m

Answer:

(a) 5(3+1)m
Let AB be the tower and AC and AD  be its shadows.
Thus, we have:
∠ACB = 45o and ∠ADB = 30o
If AC = x m, then AD = ( x + 10) m.
Let:
AB = h m.

In ∆ACB, we have:
ACAB= cot 45o =1 

⇒ xh = 1
⇒ h = x                  ... (i)
Now, in ∆ADB, we have:
ADAB= cot 30o =3 
⇒ (x+10)h = 3
⇒ x+10 = 3h       ...(ii)
On putting the value of h from (i) in (ii), we get:
h+10 = 3h
⇒ (3-1)h = 10
⇒ h =10(3-1)
On multiplying the numerator and denominator by (3+1), we get:
h = 10(3-1)×(3+1)(3+1) = 10(3+1)3-1 = 10(3+1)2= 5(3+1) m

Hence, the height of the tower is 5(3+1) m.



Page No 580:

Question 8:

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.

Page No 580:

Question 9:

An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. The width of the river is

(a) 200+2003m
(b) 200-2003m
(c) 4003m
(d) 4003m

Answer:

(a) 200+2003m
Let A be the position of the aeroplane and XAY be the horizontal line.
Let BDC be the line on the ground such that AD ⊥ BC and AD = 200 m.
It observes an angle of depression such that ∠XAB = 45o and ∠YAC = 60o.
Also,
∠ABD = 45o and ∠ACD = 60o  (Alternate angles)

In ∆ABD, we have:
BDAD = cot 45o = 1

⇒ BD200 = 1
⇒ BD = 200 m
​Now, in ∆ACD, we have:
CDAD = cot 60o =13

⇒ CD200 = 13
⇒ CD = 2003 m
∴ Width of the river = BC = BD + CD = 200 + 2003 = 200 (1 + 13) m

Page No 580:

Question 10:

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.

Page No 580:

Question 11:

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

Page No 580:

Question 12:

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

Page No 580:

Question 13:

On the same side of a tower 300 m high, the angles of depression of two objects are 45° and 60° respectively. The distance between the poles is

(a) 127 m
(b) 117 m
(c) 217 m
(d) 473 m

Answer:

(a) 127 m
Let AB be the tower and C and D be the positions of the objects such that AB = 300  m, ∠ACB = 45o and ∠ADB = 60o.

Now, In ∆ACB , we have:
ACAB = cot 45o = 1
⇒ AC300 = 1
⇒ AC = 300 m

In ∆ADB, we have:
ADAB = cot 60o = 13
⇒ AD300 =13 
⇒ AD = 3003 = 300×33×3 = 30033 = 1003 m
∴ CD = (AC - AD )= (300 - 1003) = (300 - 100×1.732) = 300 - 173.2 = 126.8 ≈ 127 m

Hence, the distance between the two poles is 127 m.

Page No 580:

Question 14:

The heights of two poles are 80 m and 65 m. If the line joining their tops makes an angle of 45° with the horizontal, then the distance between the poles is

(a) 15 m
(b) 22.5 m
(c) 30 m
(d) 7.5 m
 

Answer:

(a) 15 m
Let AB and CD be the poles of heights of 80 m and 65 m, respectively.
Draw DE⊥ AB.
Now,
∠EDB = 45o
Also,
EB = AB - AE = AB - CD = 80 - 65 = 15 m

In ∆BED, we have:
EDEB = cot 45o = 1
⇒ ED15 = 1
⇒ ED = 15 m
∴ AC = ED  = 15 m

Hence, the distance between the poles is 15 m.

Page No 580:

Question 15:

From the foot of a tower, the angle of elevation of the top of a column is 60° and from the top of the tower, the angle of elevation is 30°. If the height of the tower is 25 m, the height of the column is

(a) 35 m
(b) 42.5 m
(c) 37.5 m
(d) 27.5 m

Answer:

(c) 37.5 m
Let AB be the tower and CD be the column.
Draw BE⊥ CD.
Given: 
AB = 25 m, ∠EBD = 30o and ∠CAD = 60o
Thus, we have;
AB = CE = 25 m.
Let ED = x m.


In the right ∆BED, we have:
BEED = cot 30o = 3

⇒ BEx = 3  ⇒ BE = 3x

In the right ∆ACD, we have:
ACCD = cot 60o = 13

⇒ AC25 + x  =13   ⇒ AC = (25 + x)3
Now, 
BE = AC
⇒ 3x = (25 + x) 3 
⇒ 3x = 25 + x
⇒ 2x = 25
⇒ x = 12.5 m
∴ Height of the column = CD = CE + ED = 25 + 12.5 = 37.5 m



Page No 581:

Question 16:

A straight tree breaks due to storm and the broken part bends so that the top of the trees touches the ground making an angle of 30° with the ground. the distance from the foot of the tree to the point, where the top touches the ground is 10 m. The height of the tree is

(a) 103m
(b) 1033m
(c) 10(3+1)m
(d) 10(3-1)m

Answer:

(a) 103m
Let AB be the tree broken at point C such that CB takes the position CD so that ∠ADC = 30o and AD = 10 m.


Let:
AC = x m  and CB = CD = y m. So, the height of the tree is AB = (x+y) m.
Now, in ∆ADC, we have:
ACAD= tan 30o = 13

⇒ x10= 13
⇒ x = 103  m
Again,
ADCD = cos 30o = 32

⇒ 10y=32
⇒ y = 203 m
∴ Height of the tree = (x+y) = (103+203) = 30×33×3= 3033=103 m

Page No 581:

Question 17:

An observer standing 50 m away from a building notices that the angles of elevation of the top and bottom of a flagstaff on the building are 60° and 45° respectively. The height of the flagstaff is

(a) 503m
(b) 50(3+1)m
(c) 50(3-1)m
(d) 503m

Answer:

(c) 50(3-1)m
Let AB be the tower, BC be the flagstaff and O be the position of the observer. Thus, we have:
OA = 50 m, ∠AOB = 45o and ∠AOC = 60o

Now, in ∆AOB, we have:
ABOA = tan 45o= 1

⇒ AB50 = 1
⇒ AB = 50 m
In ∆AOC, we have:
ACOA= tan 60o= 3

⇒ AC50= 3
⇒ AC = 503 m

∴ Height of the flagstaff = BC = ( AC-AB)=(503-50)= 50(3-1) m

Page No 581:

Question 18:

A boat is being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of the boat changed from 60° to 45° in 1 minute.
If 3=1.73, the speed of the boat is
(a) 4.31 km/hr
(b) 3.81 km/hr
(c) 4.63 km/hr
(d) 3.91 km/hr

Answer:

(b) 3.81 km/h
Let AB be the cliff and C and D be the two positions of the ships. Thus, we have:
AB= 150 m, ∠ACB = 60o and ∠ADB = 45o

Now, in ∆ADB, we have:
ADAB = cot 45o =1 

⇒ AD150 = 1
⇒ AD = 150 m
In ∆ACB, we have:
ACAB=cot 60o = 13

⇒ AC150 = 13
⇒ AC = 1503 = 150×33×3 = 15033=503 m
Now,
AC = 50×1.73 = 86.5 m
∴ CD = (AD-AC) =(150 - 86.5) = 63.5 m
Speed of the boat = DistanceTime=63.560 m/s = (63.560×185) km/h = 3.81 km/h

Page No 581:

Question 19:

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 581:

Question 20:

If the length of shadow a pole on a level ground is twice the length of the pole, the angle of  elevation of the sun is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

Answer:

(d) None of these

Let AB be the pole and AC be its shadow such that AC= 2 AB.
Let:
∠ACB = θ

In ∆ACB, we have:

ABAC = tan θ

⇒ AB2 AB= tan θ

⇒ tan θ =12
It is different from 30o, 45o and 60o.

Page No 581:

Question 21:

A pole 6 m high casts a shadow 23m long on the ground. At that instant, the sun's elevation is

(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(c) 60°
Let AB be the pole and AC be its shadow.
Let ∠ACB = θ, AB = 6 m and AC = 23 m.

In ∆ACB, we have:

ABAC=tan θ
 
⇒ 623= tan θ

⇒ tan θ = 3
⇒ θ = 60o

Hence, the angle of elevation of the sun is 60°.

Page No 581:

Question 22:

Two men standing on opposite sides of a flagstaff measure the angles of the top of the flagstaff as 30° and 60°. If the height of the flagstaff is 18 m, the distance between the men is
(a) 24 m
(b) 243m
(c) 243m
(d) 31.2 m

Answer:

(b) 243m
Let AB be the flagstaff and C and D be the positions of the two men. Thus, we have:
AB = 18 m,∠ACB = 30o and ∠ADB = 60o

In ∆ACB, we have:

ACAB=cot 30o=3

⇒ AC18=3
⇒ AC= 183 m

In ∆ADB, we have:

ADAB= cot 60o=13

⇒ AD18=13

⇒ AD = 183 m
∴ CD = ( AC+AD)= (183+183) = 723×33 = 7233 = 243 m

Hence, the distance between the two men is 243 m.

Page No 581:

Question 23:

The angled of elevation of an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60°. The height of the aeroplane from the ground is

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

Answer:

(d) 12(3+3)km
Let AB be the position of the aeroplane and B and C be the two stones (1 km apart) such that BC = 1 km.
Let AD be the perpendicular meeting BC produced at D. Thus, we have:
∠ABD = 45o and ∠ACD = 60o
Let:
AD = h km and CD = x km


In ∆ABD, we have:

ADBD=tan 45o=1

⇒ h(x+1)= 1
⇒ h = x+1    (or)      x = h-1         ...(i)

In ∆ACD, we have:

ADCD= tan 60o= 3
 
⇒ hx=3
⇒ h = 3x    (or) x = h3              ...(ii)
Putting the value of x from (i) in (ii), we get:
h-1 =h3
⇒ 3(h-1) = h
⇒ h (3-1) = 3
⇒ h = 3(3-1) = 3(3-1)×(3+1)(3+1)= 3+32 km
Hence, the height of the aeroplane from the ground is 12(3+3) km.



Page No 582:

Question 24:

Match the following columns:

Column I Column II
(a) The length of a shadow of a
tower is 3 times the height of
the tower. The angle of
elevation of the sun is........ .		 (p) 40 m
(b) The angle of depression of the
top of a tower at a point 40 m
from its base 45°. The height
of the tower is ........ .		 (q) 60°
(c) The angle of elevation of
the top of a tower from a
point 15 m away from its base is
30°. The height of the tower is
.......... .		 (r) 30°
(d) At a point 14 m away from the
base of a 143m high pillar, the
angle of elevation of its top is
........ .		 (s) 53m

Answer:

(a) Let the length of the shadow be y m. Then, according to the given condition, y = 3h m.
   Let θ be the angle of elevation. Then, AB = h m and BC = y = h3 m.

In ∆​ABC, we have:
ABBC = tanθ
⇒ tan θ = hy = hh3 = 13
⇒ θ = 30o
∴ (a) →(r)

(b) Let AB be the tower. Then, BC = 40 m and ∠ACB = 45o.
Let:
AB = h m

In ∆ABC, we have:
ABBC = tan 45o = 1
⇒ h40 = 1   ⇒ h = 40 m
∴ (b) → (p)

(c) Let AB be the tower. Then, BC = 15 m and ∠ACB = 30o.
Let:
AB = h m

In ∆ABC, we have:
ABBC = tan 30o = 13
⇒ h15 = 13
⇒  h = 153× 33 = 1533 = 53 m
∴ (c) → (s)

(d)  Let AB be the pillar and θ be the angle of elevation. Thus, we have:
AB = 143 m and BC = 14 m

In ∆ABC, we have:
ABBC = tan θ
⇒ tan θ = 14314 = 3
⇒ θ = 60o
∴ (d) → (q)

Page No 582:

Question 25:

Match the following columns:

Column I Column II
(a) From the top of a 10 m high
building, the angle of elevation
of the top of a tower is 60° and
the angle of depression of its
foot is 45°. The height of the
tower is ....... .		 (p) 12.7 m
(b) A tower is 30 m high. Its
shadow is x metres shorter
when the sun's altitude is 45°
than when it is 30°. Then,
x = .......... .		 (q) 27.3 m
(c) From the top of a cliff 30 m
high, the angles of depression
of the top and bottom of a
tower are 30° and 45°
respectively. The height of the
tower is ......... .		 (r) 69.2 m
(d) Two men on either side of a
temple 30 m high observe the
angles of elevation of the top of
the temple to be 30° and 60°
respectively. The distance
between the two men is ........ .		 (s) 21.9 m

Answer:

(a) Let AB be the building and CD be the tower. Draw BE ⊥ CD.
Now, we have:
AB = 10 m , ∠ACB = 45o and ∠​DBE = 60o

From the right ∆BCA, we have:
  ACAB = cot 45o = 1
⇒ AC10 = 1   ⇒ AC = 10 m
From the right ∆DBE, we have:
DEBE = tan 60o = 3
⇒ DE10 = 3        [∵ BE = AC]
⇒  DE = 103 = 10×1.73 = 17.3 m
Height of the tower = CD = (DE + CE) = (17.3 + 10) = 27.3 m
∴ (a) - (q)

(b) Let AB be the tower such that AB = 30 m. Thus, we have:
     ∠BCA = 45o and ∠BDA = 30o

From the right ∆BAC, we have:
ACAB = cot 45o = 1
⇒             AC30 = 1
⇒ AC = 30 m
From the right ∆BAD, we have:
ADAB = cot 30o = 3
⇒     AD30 = 3
⇒ AD = 303 m

Now, 
x = ( AD - AC) = (303 - 30) = 30 (3 -1) = 30×(1.73 - 1) = 30×0.73 = 21.9 m
∴ (b) - (s)

(c) Let AB be the cliff such that AB = 30 m and CD be the tower. Thus, we have:
     ∠BDE = 30o and ∠BCA = 45o
     If CD = AE = x m, then BE = (AB - AE) = (30 - x) m.

From the right ∆BAC, we have:
     ABAC = tan 45o = 1
 ⇒               30AC = 1
 ⇒  AC = 30 m
From the right ∆BED, we have:
BEDE = tan 30o = 13
 ⇒   (30 - x)DE = 13
⇒ DE = (30 - x)3 m

We know:
DE = AC = 30 m
∴ 30 = (30 - x) 3
         ⇒ 30 = 303 - x3
         ⇒ x3 = 30 (3 - 1) 
         ⇒ x = 30 (3 -1)3 = 30 (3 - 1) 33 = 103 (3 - 1) = 17.3×0.73 = 12.7 m
Hence, the height of the tower is 12.7 m.
∴ (c) - (p)

(d) Let AB be the temple and C and D be the points of the observer. Thus, we have:
      AB = 30 m , ∠BCA = 30o and ∠BDA = 60o

From the right ∆BAC, we have:
ABAC = tan 30o = 13
 ⇒ 30AC = 13
 ⇒ AC = 303 m

From the right ∆BAD, we have:
ABAD = tan 60o = 3
 ⇒          30AD = 3
 ⇒    AD = 303 = 30×33×3 = 103 m
Distance between the two men = (AC + AD) = (303 + 103) = 403 = 40×1.73 =69.2 m
∴ (d) - (r)



Page No 583:

Question 26:

Assertion (A)
If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then h1 : h2 = 3 : 1.

Reason (R)
Figure
In the given figure, we have: h1x=tan60° and h2x=tan30°

(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

There are two towers AB and CD of heights h1 and h2, respectively. Let AC be the distance between the two towers and O be the midpoint of AC.
Thus, we have:
AO = CO = x m
Now, in ∆AOB, we have:
ABAO = tan 600 = 3
⇒ h1x = 3  ⇒ h1 = x3
Again, in ∆DOC, we have:
DCOC = tan 30o = 13
⇒ h2x = 13 ⇒ h2 = x3
Now,
h1h2 = x3x3 = 31
∴ h1:h2 = 3:1

Hence, both assertion (A) and reason (R) are true and (R) is the correct explanation of (A).



Page No 584:

Question 27:

Assertion (A)
A person standing on the bank of a river observes that the angle of 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°. Then, the height of the tree is 203m.

Reason (R)
The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving a metres towards the tower, the angle of elevation is β. Then, the height of the tower is
α tan α tan β(tan β-tan α).

(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

Let AB be the tree on the opposite side of the bank of the river and C be the position of the man standing on the bank of the river.
Thus, we have:
CD = 40 m, ∠ACB = 30o and ∠ADB = 60o
If AB = h m and BC = x m, then BD = (BC - CD) = (x - 40) m.

Then, in the right ∆ABC, we have:
ABBC = tan 30o = 13 
⇒ hx = 13 
⇒ x = h3
In the right ∆ABD, we have:
ABBD = tan 60o = 3 
 ⇒ h(x - 40) = 3  
 ⇒ h = 3 (x - 40)  
 ⇒ h = 3 (h3 - 40)    [∵ x = h3 ]
 ⇒ h = 3h - 403  ⇒ 2h = 403  ⇒ h = 4032 = 203 m
Hence, assertion (A) is true.

Assertion (A):
Let:
a = 40 m, α = 30o and β = 60o
Given:
h = a tan α tan βtan β - tan α
Putting the values of a, α and β in the above equation, we get:
h = 40 tan 30o tan 60o  tan 60o - tan 30o = 40 × 13× 33 - 13 = 4032 = 203m
Hence, reason (R) is true.

Both assertion (A) and reason (R) are true, and (R) is the correct reason of assertion (A).



Page No 591:

Question 1:

The angle of elevation of a jet plane form a point P on the ground is 60°. After a flight of 15 second, the angle of elevation changes to 30°. If the jet plane is flying at a constant of 15003m, then the jet plane is
(a) 360 km/hr
(b) 720 km/hr
(c) 540 km/hr
(d) 270 km/hr

Answer:

(b) 720 km/h
Let C be the position of the jet plane and P be the point on the ground such that ∠CPD = 60o, ∠C'PD' = 30o and CD = 15003 m.
Let DD' = x m and PD = y m.

In the right ∆CPD, we have:
CDPD= tan 60o = 3
⇒ 15003y = 3
⇒ y = 1500 m
Now, in the right ∆C'PD', we have:
C'D'D'P = tan 30o = 13
⇒ 15003(x + y) = 13
⇒ x + y = 15003 × 3 = 4500
Putting the value of y = 1500 in the above equation, we get:
x + 1500 = 4500 ⇒ x = 4500 - 1500 = 3000 m
∴ Distance  between the two points = DD' =x = 3000 m
Speed = DistanceTime = 300015 = 200 m/s = 200 × 185 = 720 km/h



Page No 592:

Question 2:

A man standing on the bank of a river observes that the angle of 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°. The width of the river is
(a) 20 m
(b) 22 m
(c) 24 m
(d) 25 m

Answer:

(a) 20 m
Let AB be the tree and D be the position of the man on the opposite side of the tree. Thus, we have:
∠​BDA = 60o, ∠BCA = 30o and CD = 40 m
If AB = h m and BC = x m, then BD = (x - 40) m.

In ∆ABC, we have:

ABBC = tan 30o = 13
⇒ hx = 13   ⇒ x = h3     ...(i)
In ∆ADB, we have:

ABBD = tan 60o = 3
⇒ h(x - 40) = 3  
⇒ x - 40 = h3 
⇒ x = 40 + h3                  ...(ii)
From (i) and (ii), we have:
h3 = 40 + h3
⇒ 3h = 403 + h
⇒ 2h = 403 ⇒ h = 203 m
From (i), we get:
x = h3 = 203×3 = 60 m
​∴ Width of the river = BD = (BC - CD) = (x - 40) = (60 - 40) = 20 m

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Question 3:

As observed from the top of a 75 m tall lighthouse, the angle of depression of two ships is 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, then the distance between the ships is
(a) 75(3+1)m
(b) 75(3-1)m
(c) 25(3+1)m
(d) 25(3-1)m

Answer:

(b) 75(3-1)m
Let OB be the lighthouse. From O, the angle of depression is observed.
We have:
∠OAB = 45o , ∠OCB = 30o and OB = 75 m
Let BC = y m and AB = x m such that AC = (BC - AB) = (y - x) m.

In ∆OBA, we have:

OBAB = tan 45o = 1

⇒75x = 1 ⇒ x = 75 m

In ∆OBC, we have:

OBBC = tan 30o = 13

⇒ 75y = 13  ⇒ y = 753 m
∴ Distance between the two ships = AC = ( y - x) = (753 - 75) = 75 (3 - 1) m

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Question 4:

If the angles of elevation of a tower from two points as distances a and b, where a > b from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is
(a) a+b
(b)a-b
(c) ab
(d) ab

Answer:

(d) ab
Let AB be the tower and C and D be the two points. Thus, we have:
BC= a, BD = b, ∠ADB = 60o and ∠ACB = 30o
Let:
AB = h m

In right ∆​ABC, we have:
tan 30°=ABBC⇒13=ha⇒a3=h       ...(1)

In right ∆ABD, we have:
tan 60°=hb⇒3=hb⇒b3=h     ...(2)
On multiplying (1) and (2), we get:
a3×b3=h2⇒ab=h2⇒h=ab
Hence, the height of the tower is ab.

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Question 5:

A ladder 20 m long touches a wall at the height of 10 m. Find the angle made by the ladder with the horizontal.

Answer:

Let AC be the ladder such that AC = 20 m. It touches the wall at a height AB = 10 m.
Let θ be the angle.

In ∆ABC, we have:
sin θ = ABAC = 1020 = 12

⇒ θ = 30o

Hence, the angle made by the ladder with the horizontal is 30°.

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Question 6:

The angle of elevation of the top of a pillar at a distance of 18 m from its foot is 30°. Find the height of the pillar.

Answer:

Let AB be the pillar, C be the foot of the person such that BC = 18 m. Thus, we have:
∠ACB = 30o

From right ∆ABC, we have:
ABBC = tan 30o = 13
⇒ h18 = 13
⇒ h = 183× 33 = 1833 = 63 m
∴ Height of the tower = AB = h = 63 m

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Question 7:

Find the angle of elevation of the sum when the length of of the shadow of a pole is 3 times the height of the pole.

Answer:

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.

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Question 8:

A Minar A casts a shadow 150 m long at the same time when a Minar B casts a shadow 120 m long on the ground. If the height of the Minar B is 80 m, find the length of Minar A.

Answer:

Let AB and DE be the Minar A and Minar B, respectively. Let BC and EF be the shadows of Minar A and Minar B.
Thus, we have:
BC = 150 m, EF = 120 m and DE = 80 m
Let AB = h and θ be the angle of elevation of both Minars.

From ∆DEF, we have:
tan θ = DEEF = 80120 = 23

From ∆ABC, we have:
tan θ = ABBC = h150

⇒23 = h150                                              [ tan θ = 23]

⇒ h = 150×23 = 100 m
∴ Length of Minar A = AB = h = 100 m

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Question 9:

If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then show that h1 : h2 = 3 : 1.

Answer:

Let AB and CD be the two towers subtending angles 60o and 30o, respectively.
Now,
∠APB = 30o and  ∠DPC = 60o
Let P be the midpoint of the line segment joining the two towers.
Let:
AB = h1, CD = h2 and BP = CP = x

From ∆DCP, we have:
DCPC = tan 30o = 13
⇒h2x = 13
⇒ x = h23                    ...(i)

From ∆ABP, we have:
ABBP = tan 60o = 3

⇒ h1x = 3
⇒ x = h13                    ...(ii)
From (i) and (ii), we get:
h13 = h23

⇒ h1h2 = 31
∴ h1:h2 = 3:1

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Question 10:

Find the height of a tower whose shadow is 10 m long when the sun's altitude is 45°.

Answer:

Let AB be the tower such that AB = h m and BC be its shadow such that BC = 10 m and ∠ACB = 45o.

From ∆ABC, we have:
ABBC = tan 45o = 1
⇒ h10 = 1
⇒ h = 10 m

Hence, the height of the tower is 10 m.

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Question 11:

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 the height of the lamp post.

Answer:

Let AB be the building and CD be the vertical lamppost.
Here,
AB = 60 m, ∠ACB = 60o and ∠ADE = 30o
Let CD = h m such that BE = h m. Thus, AE = (AB - BE) = (60 - h) m and BC = DE = x m.

From ∆ABC, we have:
ABBC = tan 60o = 3
 
⇒ 60x = 3
⇒ x = 603                   ...(i)

From ∆AED, we have:
AEDE = tan 30o = 13

⇒ (60 - h)x = 13

⇒ x = (60 - h) 3       ...(ii)

From (i) and (ii), we have:
603 = (60 - h) 3
⇒ 60 = 3 (60 - h)
⇒ 60 = 180 - 3h
⇒ 3h = 180 - 60 = 120
⇒ h = 1203 = 40 m.
∴ Height of the lamppost = CD = h = 40 m



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Question 12:

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.

Answer:

Let B be the aeroplane flying vertically below the aeroplane A at a height of 3125 m, that is, BC = 3125 m.
The angles of elevation of both the aeroplanes from the ground are such that ∠BDC = 30o and ∠ADC = 60o.
Let:
AB = b m and CD = x m

From the right ∆BCD, we have:
BCCD = tan 30o = 13
⇒ 3125x = 13
⇒ x = 31253

From the right  ∆ACD, we have:
ACCD = tan 60o = 3

⇒ (b+ 3125)x = 3

⇒ b + 3125 = x3
Putting the value of x = 31253, we get:
 b+ 3125 = 31253 × 3 = 9375
⇒ b = 9375 - 3125 = 6250 m

∴ Distance between the two aeroplanes = AB = b = 6250 m   

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Question 13:

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 O on the ground is 60° and the angle of depression of the point O from the top of the tower is 45°. Show that the height of the tower is 52(3+1)m.

Answer:

Let AB be the tower and BC be the pole such that BC = 5 m.
∠​BDA = 45o and ∠CDA = 60o
Let:
AB = h m and AD = x m

In right ∆​BAD, we have:
ABAD = tan 45o = 1

⇒ hx = 1
⇒ x = h  
Or,
h = x

In right ∆CAD, we have:
ACAD = tan 60o = 3

⇒ (h + 5)x = 3

Putting the value of x = h in the above equation, we have:
(h + 5)h = 3

⇒ h + 5 = h3
⇒ h (3 - 1) = 5
⇒ h = 5(3 - 1) = 5(3 - 1)×(3 + 1)(3 + 1) =52(3 + 1) m

∴ Height of the tower = AB = h = 52(3 + 1) m

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Question 14:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5. From a point on the plane, the angles of elevation of the bottom and top of the flagstaff are respectively 30° and 60°. Show that the height of the tower is 2.5 m.

Answer:

Let AC be the vertical tower and BC be the vertical flagstaff such that BC = 5 m.
∠CDA = 30o and ∠BDA = 60o
Let:
AC = h m and AD = x m

In right ∆​CDA, we have:
ACAD = tan 30o = 13

⇒ hx = 13

⇒ x = h3

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

⇒ (h + 5)x = 3

Putting the value x = h3 in above equation, we get:
(h + 5)h3 = 3

⇒ h + 5 = 3h
⇒ 2h = 5
⇒ h= 52 = 2.5 m
∴ ​Height of the tower = AC = h = 2.5 m

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Question 15:

The angle of elevation of the top of a hill at the foot of the tower is 60° and the angle of elevation of the top of the tower form the foot of the hill is 30°. If the tower is 50 m high, find the height of the hill.

Answer:

Let AB be the hill and CD be the tower.
Now,
CD = 50 m, ∠ACB = 60o and ∠DBC = 30o
Let:
AB = h m and  BC = x m

​From right ∆DBC, we have :
CDBC = tan 30o = 13

⇒ 50x = 13

⇒ x = 503 m

From ∆ABC, we have:
ABBC = tan 60o = 3

⇒ hx = 3
⇒ h = x3 = 503×3 = 150 m             [∵ x = 503 ]

Height of the hill = AB = h = 150 m

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Question 16:

A tree breaks due to storm and the broken part bends so that the top of tree touches the ground, making an angle of 30° with the ground. If the distance between the foot of the tree to the point, where the top touches the ground is 8 m, show that the height of the tree is 83m.

Answer:

Let AB be the tree broken at point C so that CB takes the position CD such that ∠CDA = 30o and AD = 8 m.
Let:
AC = x m and CB = CD = y m

From right ∆CAD, we have:
ACAD = tan 30o = 13
 
⇒ x8 = 13

⇒ x = 83
 
Again, we have:
cos 30o = ADCD = 8y

⇒ 8y = 32

⇒ y = 163
Height of the tree = AB = (x + y) = (83 + 163) = 243 = 243× 33 = 83 m

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Question 17:

The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving metres towards the tower, the angle of elevation is β. Prove that the height of the tower is a tan α tan β(tan β-tan α).

Answer:

Let AB be the tower and C be the point on the ground such that ∠ACB = α.
On moving towards the tower at point D, we have CD = a and ∠ADB = β.
If AB = h and BC = x, then BD = (BC - CD) = ( x - a).

From ∆ACB, we have:
ABBC = hx = tan α

⇒ x = htan α                    ...(i)

From ∆ADB, we have:
ABBD = h(x - a) = tan β

⇒ x - a = htan β

⇒ x = a + htan β              ...(ii)
From (i) and (ii), we have:
htan β = htan α - a

⇒ htan α - htan β = a

⇒h tan β - h tan αtan α tan β = a

⇒ h = a tan α tan βtan β - tan α
Hence, the height of the tower is a tan α tan βtan β- tan α.

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Question 18:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angle of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is h tan α(tan β-tan α).

Answer:

Let BD be the tower and AD be the vertical flagstaff such that AD = h. Thus, we have:
∠ACB = β and  ∠DCB = α
Let:
BD = H and BC = x

In ∆DBC, we have:
BDBC = tan α

⇒Hx = tan α
⇒ x = Htan α 
Or,
H = x tan α                       ...(i)

In ∆ABC, we have:
ABBC = tan β

⇒ (H +h)x = tan β
⇒ x = (H + h)tan β                 ...(ii)
From (i) and (ii), we get:

Htan α = (H + h)tan β

⇒ H tan β = (H + h) tan α
⇒ H tan β = H tan α + h tan α
⇒ H ( tan β - tan α) = h tan α
⇒ H = h tan α(tan β - tan α)
Hence, the height of the tower is h tan α( tan β - tan α).

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Question 19:

The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection is β. Prove that the distance of the cloud from the point of observation is 2h sec α(tan β-tan α).

Answer:

Let AB be the surface of the lake and P be the point vertically above A such that AP = h m.
Let C be the position of the cloud and D be its reflection on the lake.
Let the height of the cloud be H metres. Thus, we have:
BC = BD = H m
Draw PQ⊥CD.
Thus, we have:
∠CPQ = α and ∠QPD = β
AP = BQ = h m
CQ = (BC - BQ) = (H - h) m and DQ = (H + h) m


From the right ∆CQP, we have:
CQPQ = tan α
⇒ (H - h)PQ = tan α
⇒ PQ = (H - h)tan α                  ...(i)
From the right ∆QPD, we have:
DQPQ =tan β
⇒ (H + h)PQ = tan β
⇒ PQ = (H + h)tan β                 ...(ii)

From (i) and (ii), we get:
(H - h)tan α = (H + h)tan β
 ⇒ H tan β - h tan α = H tan α + h tan β
 ⇒ H (tan β - tan α) = h (tan α + tan β)
 ⇒ H = h (tan α + tan β)(tan β - tan α)   ...(iii)
Using the value of H in (i), we get:
PQ = (H - h)tan α = Htan α - htan α = h(tan α + tan β)tan α (tan β - tan α) - htan α

                                                  = h tan α + h tan β - h tan β + h tan αtan α ( tan β - tan α) = 2h tan αtan α (tan β - tan α) = 2h(tan β - tan α)    ...(iv)

Now, to find PC, we have:
cos α = PQPC
⇒ PC= PQcos α = PQ sec α
Putting the value of PQ from (iv), we get:
PC = 2h sec α(tan β - tan α)

Distance of the cloud = PC = 2h sec α(tan β - tan α)



Page No 594:

Question 20:

A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Show that the speed of the boat is 10033 metres per minute.

Answer:

Let AB be the light house and C and D be the positions of the two ships. Thus, we have:
∠ACB = 30°
∠ADB = 60o
AB = 100 m
Let:
BC = x m and CD = y m
Thus, we have:
BD = (BC - CD) = (x - y) m

In ∆ABC, we have:

ABBC = tan 30o = 13 ⇒ 100x = 13⇒x = 1003

In ∆ABD, we have:

ABBD = tan 60o = 3⇒100(x - y) = 3
⇒   (x - y) = 1003 
∴ y = x - 1003 = 1003 - 1003 = 300 - 1003 = 2003 m
​Distance travelled in two minutes = CD = y = 2003 = 20033 m
Speed = DistanceTime = 200332 = 10033 metres per minute

Hence proved.



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