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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 22 - Heights and Distances

Selina Textbook Solutions Chapter 22 - Heights and Distances

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 22 - Heights and Distances.

All Selina textbook questions of Chapter 22 - Heights and Distances solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 22 - Heights and Distances Page/Excercise 22(A)

Solution 1

Let the length of the shadow of the tree be x m.

Height of the tree = m

If is the angle of elevation of the sun, then

Solution 2

Let the height of the tower be h m.

Given that angle of elevation is 60o

So, height of the tower is 277.12 m.

Solution 3

Let the height upto which the ladder reaches be h m.

Given that angle of elevation is 68o

So, the ladder reaches upto a height of 5.94 m.

Solution 4

Let one person A be at a distance x and the second person B be at a distance of y from the foot of the tower.

Given that angle of elevation of A is 30o

The angle of elevation of B is 38o

So, distance between A and B is x + y = 150.6 m

Solution 5

Let the length of the rope be x m.

Now,

So, the length of the rope is 120m.

Solution 6

Let the height of the tower be h m.

(i) Here

So, height of the tower is 21.6 m.

(ii) Here

So, height of the tower is 36.24 m.

Solution 7

Let the height of the tree after breaking be h m.

Here

Now, length of the tree broken by the wind =

So, height of the tree before it was broken is (15 + 21.21) m = 36.21 m.

Solution 8

Let AB be the unfinished tower and C be the top of the tower when finished. Let P be a point 80 m from the foot A.

In BAP,

In CAP,

Therefore, the tower must be raised by (138.56 - 46.19)m = 92.37 m

Solution 9

Let the length of the tower be h m.

(i) Here

Hence the length of the tower is 25.98 m.

(ii) Let the length of the shadow be x m.

(a) Here,

Hence the length of the shadow is 25.98 m

(b) Here,

Hence the length of the shadow is 15 m.

Solution 10

Let AB be the ladder and ABP = 32o24'.

When rotated, let the ladder be AC and CAQ = 32o24'.

Hence, width of the road = (16.08 + 25.32) = 41.4 m

Solution 11

Let P be the foot of the cliff on level ground.

Then, ACP = 48o and BCP = 57o

Hence, distance between the climbers = AB = BP - AP = 17.17 m

Solution 12

Let AB be the man and PQ be the flag-pole.

Given, AR = 9 m.

Also, PAR = 28o and QAR = 13o

Hence, height of the pole = PR + RQ = 6.867 m

Solution 13

Let AB be the cliff and C be the buoy.

Given, AB = 92 m.

Also, ACB = 20o

Hence, the buoy is at a distance of 253 m from the foot of the cliff.

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 22 - Heights and Distances Page/Excercise 22(B)

Solution 1

Solution 2

Let AB be the tree of height h m.

Let the two points be C and D such that CD = 20 m, ADB = 30o and ACB = 60o

Hence, height of the tree is 17.32 m.

Solution 3

Let AB be the building of height h m.

Let the two points be C and D such that CD = 40 m, ADB = 30o and ACB = 45o

Hence, height of the building is 54.64 m.

Solution 4

Let AB be the lighthouse.

Let the two ships be C and D such that ADB = 36o and ACB = 48o

(i) If the ships are on the same side of the light house,

then distance between the two ships = BD - BC = 48 m

(ii) If the ships are on the opposite sides of the light house,

then distance between the two ships = BD + BC = 228 m

Solution 5

Let AB and CD be the two towers of height h m.

Let P be a point in the roadway BD such that BD = 150 m, APB = 60o and CPD = 30o

Hence, height of the pillars is 64.95 m.

The point is from the first pillar.

That is the position of the point is from the first pillar.

The position of the point is 37.5 m from the first pillar.

Solution 6

Solution 7

Let AB be the tower of height h m.

Let the two points be C and D such that CD = 30 m, ADE = 45o and ACB = 60o

Hence, height of the tower is 70.98 m

(ii)

The horizontal distance from the points of observation is BC = 40.98 m

Solution 8

Let AB be the cliff and CD be the tower.

Here AB = 60 m, ADE = 30o and ACB = 60o

Hence, height of the tower is 40 m.

Solution 9

Let AB be the cliff and C and D be the two positions of the boat such that ADE = 30o and ACB = 60o

Let speed of the boat be x metre per minute and let the boat reach the shore after t minutes more.

Therefore, CD = 3x m ; BC = tx m

Hence, the boat takes an extra 1.5 minutes to reach the shore.

And, if the height of cliff is 500 m, the speed of the boat is 3.21 m/sec

Solution 10

Let AB be the lighthouse and C and D be the two positions of the boat such that AB = 150 m, ADB = 45o and ACB = 60o

Let speed of the boat be x metre per minute.

Therefore, CD = 2x m ;

Hence, the speed of the boat is 0.53 m/sec

Solution 11

Let AB be the tree of height 'h' m and BC be the width of the river. Let D be the point on the opposite bank of tree such that CD = 40 m. Here ADB = 30o and ACB = 60o

Let speed of the boat be x metre per minute.

Hence, height of the tree is 34.64 m and width of the river is 20 m.

Solution 12

Let AB and CD be the two towers

The height of the first tower is AB = 160 m

The horizontal distance between the two towers is

BD = 75 m

And the angle of depression of the first tower as seen from the top of the second tower is ACE = 45o.

Hence, height of the other tower is 85 m

Solution 13

Let AB be the tower and C and D are two points such that CD = 2y m, ADB = 45o and ACB = 30o

Hence, height of the tower is m.

Solution 14

Let A be the aeroplane and B be the observer on the ground. The vertical height will be AC = 1 km = 1000 m. After 10 seconds, let the aeroplane be at point D.

Let the speed of the aeroplane be x m/sec.

CE = 10x

Hence, speed of the aeroplane is 415.67 km/hr.

Solution 15

Let AB be the hill of height 'h' km and C and D be the two consecutive stones such that CD = 1 km, ACB = 30o and ADB = 45o.

Hence, the two stones are at a distance of 1.366 km and 2.366 km from the foot of the hill.

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 22 - Heights and Distances Page/Excercise 22(C)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Given, CA = CB = 15 cm, ACB = 131o

Drop a perpendicular CP from centre C to the chord AB.

Then CP bisects ACB as well as chord AB.

Solution 6

Let AB be the vertical tower and C and D be two points such that CD = 192 m. Let ACB = and ADB = .

Hence, the height of the tower is 180 m.

Solution 7

Let AB be the tower of height x metre, surmounted by a vertical flagstaff AD. Let C be a point on the plane such that and AD = h.

Solution 8

Let AD be the height of the man, AD = 2 m.

Solution 9

Let AB be the tower of height h metres.

Let C and D be two points on the level ground such that BC = b metres, BD = a metres, .

Solution 10

Solution 11

Let AB be the tower of height 20 m.

Let be the angle of elevation of the top of the tower from point C.

Solution 12

Let AB be the tree and AC be the width of the river. Let D be a point such that CD = 50 m. Given that

Solution 13

Solution 14

Let AB be the tower and CD be the pole.

Then

Solution 15

Let A be a point 36 m above the surface of the lake and B be the position of the bird. Let B' be the image of the bird in the water.

Solution 16

Let AB be a building and M and N are the two positions of the man which makes angles of elevation of top of building as 30o and 60o respectively.

MN = 60 m

Let AB = h and NB = x m

Solution 17

Let AB represent the lighthouse.

Let the two ships be at points D and C having angle of depression 30° and 40° respectively.

Let x be the distance between the two ships.

The distance between the two ships is 43 m.

Solution 18

Solution 19

Let A be the position of the airplane and let BC be the river. Let D be the point in BC just below the airplane.

B and C be two boats on the opposite banks of the river with angles of depression 60° and 45° from A.

Solution 20

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TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 22 - Heights and Distances  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.

Text Book Solutions

ICSE X - Mathematics

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