A

BC

900 adjacent

Cos = BC/ACTan = AB/ACSin = AB/AC

Trigonometry

Angle of

elevation

Angle of

depression

The peak point of a maszid

spire found to be 300 of

Angle of elevation with

60 mts distance from

foot of the maszid.

300

60 mts

h mts

Tan 300 = 1

31

3

h

60=

h = 203

150 mts clock tower

is found to be 450 from

the other bank of river.

Find the width of the

river?

450

150 m

d mts

Tan 450 = 1

150hd

= 1d

=

d = 150 m

A spire on the clock tower

Is found at 300 angle from

a distance. On receding towards

the tower by 20 mts again it is

found at 450 angle. Find the

Height of the Clock tower.

300 450

20 m d

300 450

20 m d

Tan 300 = h

(d + 20)13

=

h =3

(d + 20)

Tan 450 = 1 = hd

h = d

1

2

From 1 2&

h =3

(h + 20)

3 h = h + 20

(3 – 1) h = 20

h =20

(3 – 1)=10(3 – 1)

A flag staff found at 450 of

Angle From the ground floor

of a building.On climbing

5 mts first floor of the building

again the same Flag staff

found at 600 angle. Find the

Height of the flag staff and

Distance between building

and flag staff.

600

450

5 m

H m

5 m

d mts

(H–5) m

600

450

5 m

H m

5 m

d mts

Tan 600 = 3 = Hd

d =H3

1

Tan 450 = 1 = H–5

d d = H–5 2

From 1 2&H3

d = H–5 =

3H–53 = H

3H – H = 53

H = 53

3–1

A hundred meters tower found

To be 300 and 450 of angle of

Elevation at two different

locations opposite to each other.

Find the distance between

these two locations.

d1 d2

300 450

d1 d2

300 450

Tan 300 =13

100

d1=

Tan 450 = 1100

d2=

d1 = 100 3 1

d2 = 100 2

From 1 2&

D = d1 + d2

D =100 3 + 100

D =100 (3 + 1)

Two different locations

apart from each other and

to the one side of the 30 mts

light post have found at 450

and 300 of angle of depression.

Find distance between

the locations.

300

300450

450

30 m

d md1m

300450

30 m

d m

Tan 450 = 1 =

d1m

hd1

d1 = h = 30 m

Tan 300 =13

30

(d + 30)=

d + 30 = 30 3

d = 30 3–30 = 21.96

An aeroplane at an altitude 2500 mts

observes the angles of depression of

opposite points on the two banks of

river to be 410 20' and 520 10'. Find

the width of the river.

Tan 0' 6' 12' 18' 1' 2' 3' 4' 5'

410 0.8693 8724 8754 8785 5 10 16 21 26

520 1.2799 2846 2892 2938 8 16 24 31 39

2500 mts

D mts

410 20'

520 10'

520 10' 410 20'

w mtsd1 mts

Tan 410 20’ = 0.8785 + 0.0010

hD

= = 0.8795 2500

D

D = = 2842.522500

0.8795

Tan 520 10’ = 1.2846 + 0.0031hd1

= = 1.2877 2500

d1

d1 = = 1941.452500

1.2877

1

2

From 1 2&

w = D – d1

w = 2842.52 – 1941.45w = 901.07