**ML Aggarwal ICSE Solutions Class 10 Math 20th Chapter ****Height and Distance**

Class 10 Chapter 20 Height & Distance**Chapter – 20 [Height and Distance]**

**(1) An electric pole is 10m high. If its shadow is 10√3 m in length, find the elevation of the sun.**

**Solution:**

Given that, An electric pole is 10m high and its shadow is 10√3m in length.

In fig. AB is the pole and OB is the shadow of pole.

=> AB = 10m and OB = 10√3m and θ is the angle of elevation of the sun.

we have, tanθ = AB/OB = 10/10√3 = 1/√3

θ = tan^{-1}(1/√3) = 30°

θ = 30° is the required elevation of the sun.

**(2) The angle of elevation of the top of tower from a point on the ground and at a distance of 150 from its foot is 30°. Find the height of the tower correct to one place of decimal.**

**Solution:**

Given that,

The angle of elevation of the top of tower from a point on the ground and at a distance of 150m from its foot is 30°.

In fig. BC is the tower of height x.

A is the point on ground so that ∠BAC = 30° and AC = 150m

we have, tanθ = BC/AC = x/150

tan30° = x/150

=> 1/√3 = x/150

x = 150/√3

x = 50√3m

(x = 86.6m is the height of tower.

**(3) A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5m away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall.**

**Solution:**

Given that,

A ladder is placed against a wall so that it reaches the top of the wall.

The foot of the ladder is 1.5m away from the wall & the ladder is inclined at an angle of 60° with the ground.

In fig. AB is the wall of height ‘x’

AC is the ladder and BC = 1.5m

and angle of inclination =>∠ACB = 60°

Then, tanθ = AB/BC

tan60° = x/1.5

√3 = x/1.5

=> x = √3 × 1.5 = 1.732 × 1.5

x = 2.6m is the required height of the wall.

**(4) What is the angle of elevation of the sun when the length of the shadow of a vertical pole is equal to its height.**

**Solution:**

Given that,

The length of the shadow of a vertical pole is equal to its height.

In fig. AB is the pole & BC is its shadow

and θ is the angle of elevation of the sun.

Let us consider, AB = BC = xm

We have,

tanθ = AB/BC = x/x = 1

θ = tan-1(1)

θ = 45° is the required angle of elevation of the sun.

**(5) From a point P on level ground, the angle of elevation of the top of a tower is 30°. If tower is 100m high, how far is P from the foot of the tower?\**

**Solution:**

Given that,

A point P on ground level, the angle of elevation of the top of a tower is 30°.

and the tower is 100m high.

In fig. AB is the tower => AB = 100m

P is the point on ground at a distance of xm from the ground.

=> PB = xm

and θ = 30°

We have,

tanθ = AB/PB = 100/x

tan30° = 100/x

1/√3 = 100/x

=> x = √3 × 100 = 1.732 × 100

x = 173.2m is the required distance between point P and foot of the tower.

**(6) From the top of a cliff 92m high, the angle of depression of a buoy is 20°. Calculate the nearest meter, the distance of the buoy from the foot of the cliff.**

**Solution:**

In fig. AB is the cliff => AB = 92m C is the point making depression angle of 26°

∴∠ACB = 20°

Let consider distance BC = xm

Then, In △ABC cotθ = BC/AB

cot20° = x/92

=> x = 92 cot20°

x = 92 × 2.7475

x = 252.77m is the required distance of the buoy from the foot of the cliff.

**(8) An electric pole is 10m high A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire?**

**Solution:**

In fig. let us consider, AB be the pole & AC be the wire

and AB = 10m, AC = xm

We have, In △ABC,

sinθ = AB/AC = 10/x

sin45° = 10/x

=> x = 10/√2 = √2 × 10 = 1.414 × 10 = 14.14m.

x = 14.14m is the required length of the wire.

**(9) A vertical tower is 20m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is the standing from the foot of the tower?**

**Solution:**

Given that,

A vertical tower is 20m high.

A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53.

In fig. AB is the tower => AB = 20m

BC = xm and cosθ = 0.53

cosθ = 0.53

θ = cos^{-1}(0.53)

θ = 58°

We have,

tanθ = AB/BC

=> tan58° = 20/x

x = 20/1.6003

x = 12.49 = 12.5

x = 12.5m is the required answer.

**(11) An observer 1.5m tall is 20.5m away from a tower 22m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution:**

Given that,

An observer 1.5m tall is 20.5m away from a tower 22m high.

In fig. AB is the tower and CD is the observer.

∴ AB = 22m, CD = 1.5m, BD = 20.5

In fig. CE || DB then,

AE = 22 – 1.5 = 20.5m

and CE = DB = 20.5m

We have, tanθ = AE/CE = 20.5/20.5 = 1

=> tan45° = 1

=>θ = 45° is the required angle of elevation of the top of the tower from the eye of the observer.

**(12) In the fig, the angle of elevation from a point P of the top of a tower QR, 50m high is 60° and that of the tower PT from a point Q is 30°. Find the height of the tower PT correct to the nearest metre.**

**Solution:**

In fig. QR is the tower => QR = 50m

height of tower PT = hm

∠RPQ = 60° and ∠TQP = 30°

We have, tan60° = RQ/PQ

√3 = 50/PQ

=> PQ = 50/√3m

and tan30° = PT/PQ

PT = PQ × tan30° = 50/√3 = 50/3 = 16.67m

Thus, the height of the tower PT is found to be 17m.

**(13) From a point P on the ground, the angle of elevation of the top of a 10m tall building and a helicopter hovering over the top of the building are 30****° and 60° respectively. Find the height of the helicopter above the ground.**

**Solution:**

Given that,

From a point P on the ground, the angle of elevation of the top of a 10m tall building and a helicopter hovering over the top of building are 30° and 60° respectively.

In fig. AB is the building and H is the helicopter hovering over it.

let P be the point on the ground so that ∠BPA = 30° and ∠HPA = 60°

AB = 10m, PA = xm, BH = hm and PA = ym.

In △ABP, tanθ = AB/PA

tan30° = 10/x

=> x = 10/ 1/√3 = 10√3m

= x = 10√3m

In △APH,

tan60° = AH/PA

tan60° = (10 + h)/x

=>√3 = (10 + h)/10√3

10 ×√3 ×√3 = 10 + h

30 = 10 + h

= h = 20m

Thus, height of the helicopter from the ground is found to be AH = 10 + 20 = 30m.

**(15) A man observes the angle of elevation of the top of the tower to be 45****°. He walks towards it in a horizontal line through its base. On covering 20m the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figure.**

**Solution:**

Let us consider, AB is the height of tower => AB = hm

and P, Q are the two observing points so that

∠APB = 45°, ∠AQB = 60°, PQ = 20m

In fig, tan60° = AB/QB

√3 = h/QB

QB = h/√3

In △QBA,

tan45° = AB/PB

1 = h/(PQ + QB)

=> h = PQ + QB

h = 20 + h/√3

=> h (√3-1/√3) = 20

h = 20√3/√3 – 1 ×√3 + 1/√3 + 1

= 20(3+√3)/2 = 10(3 + 1.73) = 47.3m

Thus, the height of the tower is found to be 47.3m.

**(16) The shadow of a vertical tower on a level ground increases by 10m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.**

**Solution:**

In fig. let AB is the tower.

let BD = xm and AB = hm

tan45° = AB/BD = h/x

1 = h/x

=> h = x

In △ABC, tan30° = h/(x + 10)

1/√3 = h/(x + 10)

x + 10 = √3h

h + 10 = √3h

=> (√3 – 1)h = 10

=> h = 10/(√3 – 1) × (√3 + 1)/( √3 + 1)

= 10(√3+1)/2

= 5(√3+1) = 5 (1.73 + 1)

h = 5 × 2.73

h = 13.65m is the required height of the tower.

**(18) A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60m the angle of elevation changes to 60°. Find the height of the building correct to the nearest meter.**

**Solution:**

Given that,

A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base.

And on covering the distance of 60m the angle of elevation changes to 60°.

In fig, AB is the building CD = 60m.

**In △ABC,**

tan60° = AB/BC

√3 = AB/BC

=> BC = AB/√3

BC = AB/√3 …. (i)

**In △ABD,**

tan30° = AB/BD

1/√3 = AB/(BC + 60) => BC + 60 = √3AB

=> BC = √3AB – 60 ……(ii)

from (i) < (ii) => AB/√3 = √3AB – 60

AB = 3AB – 60√3

60√3 = 2AB

AB = 30√3

= 30 × 1.732 = 51.96m

AB = 51.96m

Thus, the height of the building is found to be 51.96m.

**(19) At a point on level ground, the angle of elevation of a vertical lower is found to be such that its tangent is 5/12. On walking 192m towards the tower, the tangent of the angle is found to be ¾. Find the height of the tower.**

**Solution:**

In fig,

Let us consider,

TR be the tower and tanθ = 5/12

tan ∝ = ¾, PQ = 192m

Let us take, TR = x and QR = y

Then, tan∝ = TR/QR = x/4

¾ = x/4

4 = 4/3x

In △TPR, tanθ = TR/PR

=> x/(y + 192) = 5/12

=> x = (y + 192) × 5/12

x = (4/3x + 192) × 5/12

= x = 5/9x + 80 => x – 5/9x = 80

4/9x = 80

4x = 720

x = 180m is the required height of the tower.

**(20) In the fig. not drawn to scale TF is a tower The elevation of T from A is x° where tanx = 2/5 and AF = 200m. The elevation of T from B, where AB = 80m, is calculate**

**(i) The height of the tower TF**

**(ii) The angle y, correct to the nearest degree.**

**Solution:**

In fig. TF is a tower & the elevation of T from A is x° where tan x = 2/5 and AF = 200m.

The elevation of T from B, where AB = 80m is y°.

let us consider, the tower be TF

TF = xm

tanx = 2/5, AF = 200m, AB = 80m (Photo)

**(i) △ATF, tan x° = TF|AF**

=> 2/5 = x/200

=> x = (2 × 200)/5 = 400/5 = 80m

x = 80m is the required height of tower.

**(ii) In △TBF, tan y = TF|BF**

tan y = 80| (200 – 80) = 80/200

tan y = 2/3 = 0.6667

y = tan^{-1}(0.6667) = 33° 41^{I} = 34°

y = 34° is the required angle of elevation.

**(22) The horizontal distance between two towers is 140m. The angle of elevation 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 60m, find the height of the first tower.
Solution:**

**Given that,**

The horizontal distance between two towers is 140m.

The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°.

And the height of the second tower is 60m.

In fig, TR is the tower ⇒ TR = xm

Height of second tower ⇒ PQ = 60m

Distance between two tower ⇒QR = 140m

Here, PL||QR ⇒ PQ = LR = 60m

PL = QR = 140m

In △TPL, tanθ = TL|PL

tan30° = (x – 60)|140

1/√3 = (x – 60)/140 => (x – 60) = 140/√3

x – 60 = 140/√3 ×√3/√3 = 140√3/3

x = 140√3/3 + 60 = (140 × 1.732)/3 + 60 = 80.83 + 60

x = 140.83 is the required height of first tower.

**(23) As observer from the top of a 80m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse in horizontal line with its base are 30° & 40° respectively. Find the distance between two ships. Give your answer correct to nearest metre.
Solution:**

In fig.

let us consider, AB be the lighthouse and C, D are the two ships.

In △ADB,

tan 30° = AB/BD

1/√3 = 80/BD => BD = 80√3

In △ACB, tan40° = AB/BC = 80/BC

0.84 = 80/BC => BC = 80/0.84 = 95.25

Now, BD = 80 √3 = 80 × 1.73 = 138.4

and DC = BD – BC = 138.4 – 95.25 = 43.15

DC = 43.15m is the required distance between two ships.

**(25) From the two points A and B on the same side of a building the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10m, find the distance between A and B correct to two decimal places.
Solution:**

In fig.

A and B are the point on the same side of a building.

The angles of elevation of top of the building are 30°& 60° respectively.

And the height of the building is 10m.

In △DBC,

tan 60 = 10|BC

√3 = 10|BC

BC = 10√3

In △DBC, tan30° = 10/(BC + AB)

1/√3 = 10/(10/√3 + AB)

(10/√3 + AB) = 10√3

AB = 10√3 0 10/√3 = (30-10)/√3 = 20/√3

AB = 20/1.732 = 11.54m

AB = 11.54m

This is the required distance between A and B.

**(26) The angles of depression of two ships A and B as observed from the top of a lighthouse 60m high are 60°& 45° respectively. If the two ships are on the opposite sides of the lighthouse, find the distance between two ships. Give your answer correct to the nearest whole number.**

**Solution:**

Given that,

The angles of depression of two ships A and B as observed from the top of a light house 60m high are 60°& 45° respectively.

Also, the two ships are on the opposite side of the lighthouse.

In fig. let CD is the lighthouse => CD = 60m

Let us consider, AD = xm, BD = ym

Now, In △ACD,

tan 60° = CD/AD

√3 = 60/x => X = 60/√3 ×√3/√3 = 60√3/3

x = 20√3 = 20 × 1.732 = 34.64m x = 34.64m

and In △BCD,

tan45° = CD/BD

1 = 60/y => y = 60m

Then, distance between two ships = x + y = 34.64 + 60 = 94.64m

Thus, the distance between two ships is found to be 95m.

**(27) An aero plane at an altitude of 250m observes the angle of depression of two boats on the opposite banks of a river of to be 45°& 60° respectively. Find the width of the river. Write the answer correct to nearest whole number.
Solution:**

Given that,

An aero plane at an altitude of 250m observes the angle of depression of two boats on the opposite banks of river to be 45°& 60° respectively.

In fig. AD is the height of aero plane.

AD = 250m

B and C are the two boats.

Let us take, BD = xm and

DC = ym

In△ ADB, cot45° = x/250

1 = x/250

x = 250m

and, cot60° = y/250

1/√3 = y/250

y = 250/√3

The width of river is BC = BD + DC = x + y

= 250 + 250/√3

= 250 (1 + 1/√3)

= 250(√3 + 1/√3)

= 250(1.732 + 1/1.732)

= 250 (2.732/1.732)

= 250 × 1.577

BC = 394.25m

Thus, the width of the river is found to be 394m.

**(29) A man 1.8m high stands at a distance of 3.6m from a lamp post and casts a shadow of 5.4m on the ground. Find the height of the lamp post.**

**Solution:**

Given that,

A man 1.8m high stands at a distance of 3.6m from a lamp post and casts a shadow of 5.4m on the ground.

In fig. AB is the lamp post & CD is the height of man from the foot of the lamp.

And FD is the shadow of man.

Here, CE||BD

let us consider, AB = x and CD = 1.8m

EB = CD = 1.8m

AE = x – 1.8, FD = 5.4m

In△ACE, tanθ = AE|CE

tanθ = (x – 1.8)/3.6 …. (i)

In △CFD, tanθ = CD/FD

tanθ = 1.8/5.4 = 1/3 ….. (ii)

From (i) & (ii)

x – 1.8/3.6 = 1/3

3(x – 1.8) = 3.6

3x – 5.4 = 3.6

3x = 9

x = 3m is the required height of the lamp post.

**(31) A pole of height 5m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°. Find the height of the tower. (Take √3 = 1.732)**

**Solution:**

Given that,

A pole of height 5m is fixed on the top of a tower.

The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°.

In fig. Let QR be the tower & PQ is the pole on it

∠PAR = 60°&∠× QA = 45° or ∠QAR = 45°

PQ = 5m

In △QAR,

tanθ = QR/AR

tan45° = h/AR

= h/AR

AR = h

In △PAR,

tan60° = PR/AR

√3 = (5^{th})|h

√3h = 5 + h

h(√3 – 1) = 5

h = 5|0.732

h = 6.83 is the required height of the tower.

**(32) A vertical pole and a vertical tower are on the same level ground. From the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the height of the pole is 20m.
Solution:**

In fig.

Let us consider, TR be the tower.

PL is the pole => PL = 20m

Here, PQ||LR

∠TPQ = 60° and ∠QPR = 30°

∠PRL = ∠QPR = 30°

let us take, LR = x and TR = h

Then, TQ = TR – QR = (h – 20)m

In △PRL, tanθ = PL|LR

tan30° = 20|x

1|√3 = 20|x

x = 20√3m

In △PQT, tan60° = TQ/PQ

√3 = (h – 20)|x

20√3 ×√3 = h – 20

60 = h – 20

h = 80m is the required height of the tower.

**(33) From the top of a building 20m high, the angle of elevation of the top of a monument is 45° and the angle of depression of its foot is 15°. Find the height of the monument.
Solution:**

Given that,

The building is 20m high, the angle of elevation of the top of a monument is 45° and the angle of depression of its foot is 15°.

In fig. let us consider, AB is the building and AB = 20m

let us take CD is the monument and CD = xm

Let ‘y’ be the distance between building & monument.

**In △ BCD,**

tanθ = CD/BD

tan45° = x/y

1 = x/y

x = y

In △ ABD,

tan15° = AB/BD

0.2679 = 20/x

x = 20/0.2679

x = 74.65m

Thus, the height of the monument is found to be 74.65m.

**(35) An aircraft is flying at a constant height with a speed of 360km/h. from a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed as to be 30°. Determine the height at which the aircraft is flying.**

**Solution:**

Given that,

Speed of aircraft = 360km/h

Then, distance covered in 20 sec = 360 × 20/60 × 60 = 2km

Let ‘E’ be the fixed point on the ground.

Let us take, AB = CD = h km

In △ARB,

tanθ = AB/EB

tan45° = h/EB

1 = h/EB

h = EB

In fig.

ED = EB + BD

ED = h + 2

**In △CED,**

tan30° = CD/ED

1/√3 = h/(h + 2)

√3 h = (h + 2)

1.732h – h = 2

2 = 0.732h

h = 2/0.732

h = 2732m is the required height of flying aircraft.