# Trigonometry- Height and Distance

In this section, we will study the role of trigonometry in measuring height and distance.

It is evident, that trigonometry is one of the most studied subjects all over the world.

Trigonometry is being used in many fields like astronomy, geography, surveying, navigation, etc. There is no doubt that trigonometry is a vast subject but an interesting one.

##### Let us understand some terminologies that we are going to use.

Line of sight-  It is the line drawn from the eye of the observer to the object viewed by the observer.

Angle of elevation– When the object viewed is above the horizontal level, then the angle between the horizontal level and the line of sight is called the angle of elevation.

Angle of depression – When the object viewed is below the horizontal level, then the angle between the horizontal level and line of sight is called the angle of depression.

Too many definitions!! Let me make it simpler for you with an example.

Suppose you are standing still. Your eye level is at a fixed position, you are neither looking up nor down. Then, this level is called the horizontal level.

Now you started walking, you stopped at some distance from a building. The name at the top of the building grabs your attention. You look upward. Thus, the line drawn from your eye to the name at the top of the building is called the line of sight. The angle between the line of sight and the horizontal level is the angle of elevation.

Now again you started walking and you saw some coins on the ground(Lucky you!! ). You look down at the coins. Thus, the line drawn from your eye to the coins is called line of sight. The angle between the line of sight and horizontal level is called angle of depression.

If you notice, there are 3 vertices, 3 angles and 3 lines involved.(If you suppose the height of the viewer is negligible). What geometrical figure you get?? Triangle – This the point when trigonometry comes into play.

Now assuming that you have understood the concepts, you must know the following for solving height and distance questions.

#### Questions on height and distance

##### Question 1 : A point C on the ground, is some distance away from the foot of a tower AB. The angle of elevation from the point C to the top of the tower at point A is 30º. The distance between A and C is 20 m. What is the height of the tower?Solution: Given,
AC = 20 m
and Angle of elevation ∠ACB = 30°
We have to find the length of the tower= AB=?

In right ΔACB,

sin 30° = \frac { AB }{ AC }

⇒\frac { 1 }{ 2 } =\frac { AB }{ 20 }

AB = 10 m

Therefore, the height of the tower is 10 meter.

##### Question 2: At 129 meter away from the foot of a cliff on a level ground, the angle of elevation of the top of the cliff is 30° . The height of the cliff is??Solution:

Let, Height of the cliff = AB
Distance of foot B from point C = 129 m
angle of elevation ∠ACB = 30°

In ΔABC,
tan 30° =\frac { AB }{ BC }

\frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ 129 }

⇒AB =\frac { 129 }{ \sqrt { 3 } }

⇒AB = \frac { 129\sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } }

thus, AB = 43\sqrt { 3 }  m

Therefore, the height of the cliff is 43\sqrt { 3 }  meter.

##### Question 3: A man 1.6 m tall, is 45 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. Then the height of the tower is?Solution: The height of the man =AB= 1.6 m
also AB = EC = 1.6 m
Height of the tower = CD = h meter
Angle of elevation ∠DAE =30°
Distance of boy from the tower = BC =AE = 45 m
And DE = (h – 1.6) meter

In ΔDAE,

tan 30°=\frac { DE }{ AE }

\frac { 1 }{ \sqrt { 3 } }=\frac { h-1.6 }{ 45 }

\frac { 45 }{ \sqrt { 3 } } = h-1.6

⇒ 15\sqrt { 3 }= h-1.6

h = (15\sqrt { 3 } + 1.6)m
Therefore, height of the tower is (15\sqrt { 3 } + 1.6)m

##### Question 4 : From a point C 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 C is 45°. Find the height of the flagstaff and the distance of building from the point C.Solution: Here ,
AB is the height of the building = 10 m
and BD is the height of the flag staff
∠ACB= 30°
∠ACD= 45°
In ΔACB
tan 30° =\frac { AB }{ AC }

\frac { 1 }{ \sqrt { 3 } }=\frac { 10 }{ AC }

⇒AC = 10\sqrt { 3 } m

Thus, distance of point C from the building is 10\sqrt { 3 } m.

In ΔACD

tan 45° =\frac { AD }{ AC }

⇒1=\frac { AD }{ 10\sqrt { 3 } }

⇒AD = 10\sqrt { 3 } m

= 10\sqrt { 3 } – 10

= 10(\sqrt { 3 } – 1)m
Therefore, the height of the flag staff is 10(\sqrt { 3 } – 1)m

##### Question 5 : A person is standing on the bank of a river. He observes that angle of elevation of the top of the tree standing opposite of the bank is 60°. When he moves 30 m away from the bank of the river, he founds that the angle of elevation is 30°. Find the width of the river.Solution: Here,
Height of the tree = AB = h meter
and BC is the width of the river= x meter
distance travelled is CD = 30 m
∠ACB = 60°

In ΔACB

tan 60° =\frac { h }{ x }

\sqrt { 3 }=\frac { h }{ x }

⇒h = \sqrt { 3 }x

tan 30°=\frac { h }{ 30+x }

\frac { 1 }{ \sqrt { 3 } } =\frac { \sqrt { 3 } x }{ 30+x }

⇒30+x = 3x

⇒30 =2x

⇒x=15 m

Therefore, width of the river is 15 m.

##### Question 6 : From the top of a 60 m tall building , the angle of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. The height of the tower is ? Solution:

Let, height of building = AB = 60 m

Height of tower = CD = h meter

and Angle of depressions

and

∠OAC =∠ACB= 60°

in ΔACB,

tan 60° =\frac { AB }{ BC }

\sqrt { 3 }=\frac { 60 }{ BC }

⇒BC =\frac { 60 }{ \sqrt { 3 } }

Now ,

tan 30°=\frac { AE }{ ED }

\frac { 1 }{ \sqrt { 3 } }=\frac { AE }{ (60/\sqrt { 3 } ) }

⇒AE =\frac { 60 }{ 3 }

⇒AE = 20 m

And AB = BE + AE

⇒60 = BE + 20

⇒BE = 40 m

and BE =CD = 40 m

Therefore, height of the tower is 40 m.

##### Question 7: From a point on a bridge across a river the angle of depression of the banks on the opposite sides of river are 30 and 45 respectively. If the bridge is at a height of 5 m from the bank. Find the width of the river.Solution: Height of bridge from river = AD = 5 m

Angle of depression from a point A on bridge to point B on bank of the river = ∠EAB =∠ABD = 45°

Angle of depression from a point A on bridge to point C

on bank of the river = ∠FAC =∠ACD = 30°

In ΔABD,

tan 45° =\frac { AD }{ BD }

⇒1=\frac { 5 }{ BD }

⇒BD = 5 m

tan 30° =\frac { AD }{ CD }

\frac { 1 }{ \sqrt { 3 } }=\frac { 5 }{ CD }

⇒CD =5\sqrt { 3 }

Thus, width of river = BD + CD

= 5 + 5\sqrt { 3 }

= 5(1+\sqrt { 3 })m
Therefore, width of the river is 5(1+\sqrt { 3 })m.

##### Question 8: A 2 m tall boy spots an airplane moving in a horizontal line with a height of 89 m from the ground. The angle of elevation of the airplane from the eyes of the boy is 60°. After some time the angle of elevation changes to 30°.Find the distance travelled by the airplane in this interval.Solution: Height of boy AB = 2 m

also AB = GH =CD

Height of airplane from the ground

EH = DF = 89 m

Distance traveled by airplane E to F

= EF

And EG = FC = (EH – GH)

=  89-2 = 87 m

In ΔEAG,

tan 60°=\frac { EG }{ AG }

\sqrt { 3 }=\frac { 87 }{ AG }

⇒AG =\frac { 87 }{ \sqrt { 3 } }  m

tan 30° =\frac { FC }{ AC }

\frac { 1 }{ \sqrt { 3 } }=\frac { 87 }{ AG+GC }

⇒AG+GC = 87\sqrt { 3 }

\frac { 87 }{ \sqrt { 3 } } + GC = 87

⇒GC =87\sqrt { 3 }\frac { 87 }{ \sqrt { 3 } }

⇒GC = 87(\sqrt { 3 }  – \frac { 1 }{ \sqrt { 3 } })

= 87 × \frac { 2 }{ \sqrt { 3 } }

GC = 58\sqrt { 3 } m

Therefore, the distance travelled by the airplane is 58\sqrt { 3 } m.

##### Question 9 : The angle of elevation of a helicopter from a point on the ground is 60°. After flying for 30 seconds the angle of elevation changes to 30°. If the helicopter is flying ar a height of 4500 m, then what is the speed of the helicopter in m/s?Solution: Helicopter travels from point A to D in 30 seconds

Height of helicopter from the ground

AB = CD = 4500 meter

∠AEB =60°

∠DEC =30°

In ΔAEB

tan 60°= \frac { AB }{ BE }

\sqrt { 3 }=\frac { 4500 }{ BE }

⇒BE = \frac { 4500 }{ \sqrt { 3 } }m

In ΔDEC

tan 30° =\frac { CD }{ CE }

\frac { 1 }{ \sqrt { 3 } }=\frac { 4500 }{ CE }

⇒CE =4500\sqrt { 3 }

And  BC = EC – EB

= 4500\sqrt { 3 }\frac { 4500 }{ \sqrt { 3 } }

= 4500(\sqrt { 3 }\frac { 1 }{ \sqrt { 3 } } )

BC = 3000\sqrt { 3 } m

=\frac { 3000\sqrt { 3 } }{ 30 }

= 100\sqrt { 3 } m/s
Therefore, the speed of the helicopter is 100\sqrt { 3 } m/s.

##### Question 10: The cliff of a mountain is 180 m high. The angle of depression of two ships on either side of cliff are 30° and 60°. What is the distance between the two ships?Solution: Height of cliff = AD = 180 m

∠EAB = ∠ABD = 60°

& ∠FAC =∠ACD =30°

In ΔABD,

tan 60° =\frac { AD }{ BD }

\sqrt { 3 }=\frac { 180 }{ BD }

⇒BD =\frac { 180 }{ \sqrt { 3 } }

In ΔACD

tan 30° =\frac { AD }{ CD }

\frac { 1 }{ \sqrt { 3 } } =\frac { 180 }{ CD }

⇒CD = 180\sqrt { 3 }

Distance between ships = BD + CD

= 180\sqrt { 3 } +180\sqrt { 3 }

=240\sqrt { 3 } m

Therefore, the distance between the ships is240\sqrt { 3 } m.