Angle of Elevation

Ram

Ravi

)Eye level

Ram looks up above Ravi

‘’ Is the angle of elevation

Angle of depression

(Eye level

Ravi looks down at Ram

Ravi

Ram

Is the angle of ‘depression’

Angle of elevatio

n

Angle of depression

Eye level)))

))

If the object is above the horizontal level of the eyes we raise our head

upwards to view the object. Our eyes move through an angle to view the

object. The angle between the horizontal line and the line of sight

is called the angle of elevation

θ

If the object is below the horizontal level of the eyes we

move down our eyes through an angle to view the object. The

angle between the horizontal line and the line of sight is called the

angle of depression.Ф

Steps in Solving problems To solve problems the first step is to

draw the correct labeled diagram based on the question.

Mark the sides and angles given in the question.

Take the unknown quantities to be

found out as ‘x’ , ‘y’ , ‘p’ etc , and mark in the diagram.

Identify the correct trigonometric ratio which can be formed with the given quantities and the unknown quantity to be found out.

How will the man find the height of tip of the flag post from the ground?

1500m 45°(

Solution Let the height of

the hill be ‘h’ m Tan 45° = h/1500 Or, 1 = h/1500 h = 1500 m

1500m 45°(

øb

h2

tan =h1/a

tan ø=h2/b

a,b, ø & can be measured to find h1 & h2

a

h1

Hill, far away from the measuring point Take two points A,B (say) the distance between them is

measurable(500m) The angle of elevation at A is (45°) the angle of elevation at B is (60°)

Hill height= h

BA 500m)60 )45

C

D

height= h

BA500m

)60o )45 o

Tan45°=h/AD

1=h/AD h=AD

(or) h=AB + BD = 500+BD---(1)

Also tan 60°=h/BD

3=h/BD

BD=h/ 3 -------(2)

From (1) & (2) , find ‘h’

h = 500+BD--- (1) BD = h/ 3 -------(2) Substitute (2) in (1) h = 500 + h/ 3 (or) 3h – h = 500 3 (or) h (3 – 1 ) = 500 3 (or) h = 500 3 (3 – 1 ) = 500 3 (3 + 1 ) (3 – 1 ) (3+1 ) = (1500-500

x1.732)/2Simplify & find the answer.

AB is light house C and D are

positions of boats A man in a boat

observes that angle of elevation changes from 60 to 45 in 2 minutes

B

A

CD45°

60°

150m

To Find Speed

AD \ AB = cot 45 ° = 1

AD \ 150 = 1AD = 150 m

Let AC = X meters CD = (150 -x)

AC \ AB = cot 60° = 1 \ 3 x \ 150 = 1 \ 3

A

45° 60°

B

45°CD A

60°

B

X = ( 150 \ \/3 )m = ( 50 3 ) m

CD = (150 - 50 3) m = 63.4 m

Boat covers 63.4 m in 2 minutes.

Speed of boat = ( 63.4 ÷120 ) m\s = 0.53 m\s

To Find Height

An observer is 1.6 m tall and is 45 meters away .

The angle of elevation from his eye to top of tower is 30

Determine height of tower.

30°

45m

1.6

m

Solution

Height of the tower = h+1.6m

h=45 tan30° h=45 X 1/3 h=45 X 3/3 =15X1.73 =25.95

Height=27.55 m

1.6

30°45m

h45m

From the top of a tower 50m high the angles of depression of the top and bottom of a pole are observed to be 45º and 60 º respectively. Find the height of the pole , and the tower stand in the same line.

tower

pole

50m

60°

45 °

Solution A

tower

pole

50m

60°

45 ° AB is the tower

B

C

D

CD is the pole

60°

45 °

EB=CD & EC =BD Let Height of the pole CD =h m

Take AEC & ABD to solve for “h”

50m

B

C

D60°

45 °

A

E

h

There is a small island in the middle of a 100m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively30 º and 45º,find the height of the tree.

100 mP Q30° 45°

Tree height =h m (say)h/PR =tan30° =1/33 h =PR3 h =100-RQ………..

(1)h/RQ =tan45° =1h =RQ………..(2)

There is a small island in the middle of a 100m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively30 º and 45º,find the height of the tree.

tree

P Q30 45

R

From (1) & (2)

3 h =100-h

h(3 -1) =100

H =100/(3 -1)

= 100 (3 +1) /(3 -1) /(3 +1)

100x2.732/2

27.32/2

13.66 m

Height of the tree =13.7 m

A

θθ

фф

h’

h

W P

From a window ( h meters high above the ground) of a house in a street ,the angle of elevation and depression of the top and the foot of another house on opposite side of the street are θ and ф respectively. show that the height of the opposite house ish(1+tan θ cot ф ).

street

house

Let W be the window and AB be the house on the opposite side. then WP is the width of the street. In ΔBPW, tan ф =PB/WP

h/WP = tan ф or WP = h cot ф In Δ AWP, tan θ = AP/WP or h’/WP = tan θ or

h’ = WP tan θ h’ = h cot ф tan θ Height of the house = h + h’ =h+ h tan θ cot ф =h(1+ tan θ cot ф )