Heights and Distances

In the previous chapter, we have studied about trigonometric ratios. In this chapter, we will be looking at some ways in which trigonometry is used in the life around you.

Looking at a tower

Say we are looking at the top of a tower. From our eye level a line can be drawn to the top of the tower which is called the line of sight. This line of sight makes an angle with the horizontal, which is called the angle of elevation of the top of the tower from the eye of the observer i.e. when we raise our head to look at the object.

Now, look at the reverse case. Say, somebody is looking at us from the top of the tower In this case, the line of sight (drawn from the eye of the observer) is below the reference horizontal line.

The angle so formed by the line of sight with the horizontal is called the angle of depression.

Thus, the angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when the observer lowers their head to look at the point being viewed.

Now, let's look at the figure given below. A person is looking at the top of the hill from the ground level and another person is watching the ground from the top of the hill.

Can you identify the lines of sight, and the angles so formed in the figure.

What are the angles of elevation (or) angles of depression?

Angle of elevation : ∠CAB∠ABC Angle of depression : ∠ABH∠CAB Line of Sight : ABBCCA

Assuming that the above three conditions are known, can we determine the height of the mountain? YesNo

Given that we know the distance AC and the angle: the trigonometric ratio that can be used to find the height is tan(angle of elevation)sin(angle of elevation)cos(angle of elevation)

So, the height of the mountain is :

tanangleofelevation x DistanceofAfromthemountain + height of A

1. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Solution: First let us draw a simple diagram to represent the problem. Here AB represents the tower, CB is the distance of the point from the tower and ∠ACB is the angle of elevation.

We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angledacute-angled at B.

To solve the problem, we choose the trigonometric ratio tan (or cot), as the ratio involves AB and BC.

Now, tan 60° = ABBCBCAB

Putting values,

33 = AB15

AB = 15315

Hence, the height of the tower is 153 m.

2. An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (take 3 = 1.73)

Solution: The electrician is required to reach the point B on the pole AD.

So, BD = –

= 5 – 1.3 = m.

Here, BC represents the ladder. We need to find its length of the of the right triangle BDC.

Now, which trigonometic ratio should we consider?

It should be sin 60°cos 60°.

So, BDBCBCBD = sin 60°

/BC = 3213

Therefore, BC = 3.7x23 = m (upto two decimal places)

Thus, the length of the ladder should be 4.28 m

Now, DCBD = cot 60°tan 60° = 133

DC = 3.73 = m (upto two decimal places)

Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

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3. An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?

Solution : Here, AB is the chimney, CD the observer and ∠ ADE the angle of elevation. In this case, ADE is a triangle, -angled at E and we are required to find the height of the chimney.

We have AB = + BE = + and DE = CB = m

To determine AE, we choose a trigonometric ratio, which involves both AE and DE.

So, tan 45°sin 45° = AEDE

= AE28.5

Therefore, AE = m

So the height of the chimney (AB) = AE + BE = + m = m.

4. From a point P on the ground the angle of elevation of the top 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. (You may take 3 = 1.732)

Solution : AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e. DB and the distance of the building from the point P, i.e. PA.

Since, we know the height of the building AB, we will first consider the right triangle PAB.

We have tan 30°cot 30° = ABAP

133 = /AP

Therefore, AP = 103310

i.e. the distance of the building from P is 103 m = m. (upto two decimal places)

Next, let us suppose DB = x m. Then AD = 10+x10x m.

Now, in right triangle PAD, tan 45° = ADAPAPAD = 10+x10310310+x

Therefore, = 10+x103

i.e., x = 103−1103+1 = (upto two decimal places)

So, the length of the flagstaff is 7.32 m

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5. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Solution : AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is °.

Now, let AB be h m and BC be x m. According to the question, DB is m longer than BC.

So, DB = (40 + x) m

Now, we have two right triangles ABC and .

In triangle ABC, tan 60° = ABBCBCAB

313 = hxxh (1)

In triangle ABD, tan 30° = ABBDBDAB

133 = hx+40x+40h (2)

From (1), we have h = x33x

Putting this value in (2), we get x33 = x + 40, i.e. = x + 40

Thus, x =

So, h = 203203 [From (1)]

Therefore, the height of the tower is 203 m.

6. The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multistoreyed building and the distance between the two buildings.

Solution : PC denotes the multistoryed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e. PC and the distance between the two buildings i.e. .

Look at the figure carefully. Observe that PB is a to the parallel lines PQ and BD.

Therefore, ∠ QPB and ∠ PBD are angles and thus are .

So ∠ PBD = 30°45°. Similarly, ∠ PAC = °.

In right triangle PBD, we have:

PDBDBDPD = tan 30° = 133

BD = 3PD

In right triangle PAC, we have:

PCACACPC = tan 45° =

PC =

Also, PC = PD + , _{span.reveal(when="blank-24")}_therefore, PD + DC =

Since, AC = BD and DC = AB = m, we get:

PD + 8 = BD = PD 3

This gives PD = 83−1 = 83+13−13+1 = 43+143−1

So, the height of the multi-storeyed building is 43+1+8 = 43+343−3 and the distance between the two buildings is also 43+343−3 m.

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Text based adventure game

Percy is on a quest. His goal is to find the flower of life so that he can take it back to help his friend who was stabbed by a minotaur.

He starts on the journey and walks through the vast country side. Suddenly he comes upon a big chasm. He is at the edge of the chasm and the other edge is above him at some distance. His keen sense of angles tells him that if he looks up to the top of the other end, the angle of elevation is 60°. But unfortunately Percy is bad at trigonometry.

This is where you come in. Athena has appointed you as the guardian angel of Percy. Percy can reach out to at any time and ask you for help.

Percy: I am stuck. I am at the end of the chasm. What should I do?

You: What is the angle of elevation?

Percy: I think it is around 60°.

You: On what mountain are you standing?

Percy: On mount Sicarus.

You: What is the mountain in front of you?

Percy: Mount Janus.

You: Good. I find from the record books that mount Sicarus is 700 meters in height and mount Janus is 737 meters in height. So the height of the chasm you need to scale is meters.

Percy: But how will the height help me. Should I set the magic ladder to 37 meters? You know we have to keep the right setting. If we set it to too long or too short the magic ladder will self destruct.

You: NoYes

You: That's the height you need to scale. But the length you need to cross is different. Let's see what facts we have. We know the angle is 60° and the opposite side is 37 meters. So let me apply my trigonometric knowledge. We can use the following trigonometric ratio: sin 60 cos 60 tan 60.

You: Solving for the sin 60 equation we get the length as 42.843.743.0.

Percy: Thanks.

He sets the setting on the magic ladder to 42.8 and crosses the chasm. The flower of life is on mount Janus. he picks it up and goes to his friend and saves him.

Exercise 9.1

1. A circus artist is climbing a 20 m long rope, which istightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, ifthe angle made by the rope with the ground level is 30° (see Fig. 9.11).

Solution:

Given:

Length of rope = m

The angle of rope with ground = °= ∠ACBBAD

We have to find the height of the pole.

AB = Height of the Pole

BC = Distance between the point on the ground and the pole.

AC = Length of the Rope (Hypotenuse)

In ΔABC,

sinC = ABACBCAC

sin30° = AB20AC20

12 = AB20 [Since sin30° = 12]

AB = (12) ×

AB = m

Height of pole AB = 10 m

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

Solution:

Height of the tree = AB +

Trigonometric ratio which involves AB, BC and ∠C is tan θ, where AB can be measured.

Trigonometric ratio which involves AB, AC and ∠C is sin θ, where AC can be measured.

Distance between the foot of the tree to the point where the top touches the ground = BC = m

In triangle ABC,

tan C = ABBCABAC

tan 30° = AB8

13 = AB8 [Since tan 30° = 13]

AB =83

sin C = ABACBCAC

sin 30° = 83AC

12 = 83 × 1AC

AC = 83 × 212

AC = 163

Height of tree = AB + AC

= 83 + 163

= 243

= 24×33 × 313. [On rationalizing ]

= 2433

= 83

So, the height of tree is meters.

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:

We have to find the height of the tower.

Let us consider the height of the tower as AB, the distance between the foot of the tower to the point on the ground as BC.

In ΔABC, trigonometric ratio involving AB, BC and ∠C is tan θ.

tan C = ABBCABAC

tan 30° = AB30BC30

13 = AB30

AB = 303

= 30×33×3

= 3033

= 1033

Height of tower AB = 103 m

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

We take the height of the flying kite as AB, the length of the string as AC, and the inclination of the string with the ground at ∠C.

Trigonometric ratio involving AB, AC and ∠C is sinθ.

In ΔABC,

sinC = ABACBCAC

sin 60° = 60AC

32−32 = 60AC

AC = (60 × 23)

= 120×33×3

= 12033

= 403203

Length of the string AC = 403m.

6. 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 increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.Find the height of the tower.

Solution:

Let the height of the building is BC, the height of the transmission tower which is fixed at the top of the building be AB.

D is the point on the ground from where the angles of elevation of the bottom B and the top A of the transmission tower AB are 4560° and 6090° respectively.

The distance of the point of observation D from the base of the building C is CD.

Combined height of the building and tower = AC = AB + BCAD

Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θ.

In ΔBCD,

tan 45° = BCCDCDBC

= 20CD

CD =

In ΔACD,

tan 60° = ACCDCDAC

313 = AC20

AC = 3

Height of the tower, AB = AC - BC

AB = 203 - m

= 20 (3 - 1) m

8. A statue, 1.6 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.

Solution:

Let the height of the pedestal be BC, the height of the statue, which stands on the top of the pedestal, be AB. D is the point on the ground from where the angles of elevation of the bottom B and the top A of the statue AB are 45° and 60° respectively.

The distance of the point of observation D from the base of the pedestal is CD. Combined height of the pedestal and statue AC = AB + BC

Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θ.

In ΔBCD,

tan 45° = BCCDCDBC

= BCCD

Thus, BC =

In ΔACD,

tan 60° = ACCDCDAC

tan 60° = AB+BCCD

3 = 1.6+BCBC [Since BC = CD]

3 BC = + BC

3 BC - = 1.6

BC (3 - 1) =

BC = 1.6×3+13−13+1

= 1.63+13−1

= 1.63+12

= 0.83+10.63+1

Height of pedestal BC =0.83+1 m.

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution:

Let the height of the tower be AB and the height of the building be CD.

The angle of elevation of the top of building D from the foot of tower B is 3060° and the angle of elevation of the top of tower A from the foot of building C is 6030°.

Distance between the foot of the tower and the building is BC.

Trigonometric ratio involving sides AB, CD, BC and angles ∠B and ∠C is tan θ.

In ΔABC,

tan 60° = ABBCBCAB

3 = 50BC30BC

BC = 503 ....(i)

In ΔBCD,

tan 30° = CDBCBCCD

13−13 = CDBC

13 = CD503 [from (i) BC = 503]

CD = 13 × 503

CD = 503503

Height of the building CD = 503 m.

10. 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 the poles and the distances of the point from the poles.

Solution:

Let us consider the two poles of equal heights as AB and DC and the distance between the poles as BC.

From a point O, between the poles on the road, the angle of elevation of the top of the poles AB and CD are 60° and 30° respectively.

Trigonometric ratio involving angles, distance between poles and heights of poles is tan θ.

Let the height of the poles be x

Therefore AB = DC = x

In ΔAOB,

tan 60° = ABBOBOAB

3 = xBO

BO = x3 ....(i)

In ΔOCD,

tan 30° = DCOCOCDC

13 = xBC−OB

13 = x80−x3 [from (i)]

- x3 = 3x

x3 + 3x =

x (13 + 3) =

x1+33 =

x (43) =

x = 8034

x = 3

Height of the poles x = 203 m.

Distance of the point O from the pole AB

BO = x3

= 33

=

Distance of the point O from the pole CD

OC = BC - BOOC

= 80 -

=

The height of the poles is 20√3 m and the distance of the point from the poles is 20 m and 60 m.

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Solution:

Considering ΔABC,

tan 60° = ABBCBCAB

sqrt(3) = ABBC

AB = BC3 ...(i)

Considering ΔABD,

tan 30° = ABBDBDAB

tan 30° = ABCD+BC

13 = BC320+BC [from (i)]

+ BC = BC3 × 3

20 + BC = × BC

3BC - BC =

BC = 20

BC =

Substituting BC = 10 m in Equation (1), we get AB = 3 m

Height of the tower AB = 103 m

Width of the canal BC = m

12. 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 its foot is 45°. Determine the height of the tower.

Solution:

Let the height of the tower be CE and the height of the building be AB. The angle of elevation from the top E of the tower to the top A of the building is 60° and the angle of depression from the bottom C of the tower to the top A of the building is 45°.

Draw AD || BC.

Then, ∠DAC = ∠ACB = 45° (alternate interior angles)

In ΔABC,

tan 45° = ABBCBCAB

= 7BC

BC =

ABCD is a rectanglesquare,

Therefore, BC = AD = and AB = CD =

In ΔADE,

tan 60° = EDADADED

3 = ED7AD7

ED = 73

Height of tower = CE = ED + CD

= 73 +

= 7 (3 + )

Height of the tower = 7 (3 + 1) m.

13. As observed from the top of a 75 m high lighthouse from the sea-level, 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.

Solution:

Let the height of the lighthouse from the sea level be AB and the ships are C and D.

The angles of depression of the ships C and D from the top A of the lighthouse are 30° and ° respectively.

Distance between the ships = CD = BD − BCAC

In ΔABC,

tan 45° = ABBCBCAB

= 75BC

BC =

In ΔABD,

tan 30° = ABBDBDAB

13 = 75BD−75BD

BD = 753

Distance between two ships CD = BD - BC

CD = 753 -

= 75 (3 - )

Distance between two ships CD is 75 (3 - 1) m.

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13).Find the distance travelled by the balloon during the interval.

Solution:

Trigonometric ratio involving AB, BC, OD, OA and angles is tanθ. [Refer AB, BC, OA and OD from the figure.]

Distance travelled by the balloon OB = AB - OA

From the figure, OD = BC, and it can be calculated as

88.2 m - 1.2 m = m --- (1)

In ΔAOD,

tan 60° = ODOAOAOD

3 = 87OA87OD

OA = 8738713

= 87×33×3

= 87×33

= 3 m

In ΔABC,

tan 30° = BCABABBC

13 = 87AB87AC

AB = 3

Distance travelled by the balloon, OB = AB - OA

OB = 87283 - 29873

= 3

Distance travelled by the balloon = 58 3 m.

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

Let the height of the tower be AD and the starting point of the car be at point B and after 6 seconds let the car be at point C. The angles of the depression of the car from the top A of the tower at point B and C are 30° and 60° respectively.

Distance travelled by the car from the starting point towards the tower in 6 seconds = BC

Distance travelled by the car after 6 seconds towards the tower = CDAB

We know that, speed = distancetime

The speed of the car is calculated using the distance BC and time = 64 seconds.

Using Speed and Distance CD, the time to reach foot can be calculated.

In ΔABD,

tan 30° = ADBDBDAD

131−3 = ADBD

BD = 3 ....(1)

In ΔACD,

tan 60° = ADCDCDAD

313 = ADCD

AD = 3 ....(2)

From equation (1) and (2)

BD = 3 × 3

BC + CD = CD [∵ BD = BC + CD]

BC = CD ....(3)

Distance travelled by the car from the starting point towards the tower in 6 seconds = BC

Speed of the car to cover distance BC in 6 seconds = DistanceTime

= BC6AC6

= 2CD6 [from (3)]

= CD3BD3

Speed of the car = CD3 m/s

Distance travelled by the car from point C, towards the tower = CDAC

Time to cover distance CD at the speed of CD3 m/s

Time = Distancespeed

= CDCD3

= CDAB × 3CD

=

The time taken by the car to reach the foot of the tower from point C is 3 seconds.