Class 10 SELINA Solutions Maths Chapter 22 - Heights and Distances

Ex. 22(A)

Ex. 22(B)

Ex. 22(C)

Heights and Distances Exercise Ex. 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 60^o

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 68^o

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 30^o

The angle of elevation of B is 38^o

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 = 32^o24'.

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

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 = 48^o and BCP = 57^o

$therefore fraction numerator BP over denominator PC space end fraction equals tan space 57 to the power of straight o rightwards double arrow BP equals 40 cross times 1.539 equals 61.56 space straight m Also comma AP over PC equals tan space 48 to the power of straight o rightwards double arrow AP equals 40 cross times 1.110 equals 44.4 space straight m$

Hence, distance between the climbers = AB = BP - AP = 17.16 m

Solution 12

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

Given, AR = 9 m.

Also, PAR = 28^o and QAR = 13^o

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 = 20^o

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

Heights and Distances Exercise Ex. 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 = 30^o and ACB = 60^o

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 = 30^o and ACB = 45^o

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 = 36^o and ACB = 48^o

(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 = 60^o and CPD = 30^o

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 = 45^o and ACB = 60^o

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 = 30^o and ACB = 60^o

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 = 30^o and ACB = 60^o

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 = 45^o and ACB = 60^o

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 = 30^o and ACB = 60^o

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 = 45^o.

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 = 45^o and ACB = 30^o

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

$rightwards double arrow straight x equals 100 open parentheses square root of 3 minus fraction numerator 1 over denominator square root of 3 end fraction close parentheses equals 100 cross times 1.1547 equals 115.47 space straight m divided by sec equals 115.47 cross times 18 over 5 space km divided by hr space equals space 415.69 space km divided by hr$