1. Trigonometry:

i. sin θ = Perpendicular / Hypotenuse

ii. cos θ = Base / Hypotenuse

iii. tan θ = Perpendicular / Base

iv. cosec θ = 1/sin θ = Hypotenuse/Perpendicular

v. sec θ = 1/cos θ = Hypotenuse /Base

vi. cot θ = 1/tan θ = Base /Perpendicular

2. Trigonometrical Identities:

i. sin^2 θ + cos^2 θ = 1

ii. 1 + tan^2 θ = sec^2 θ

iii. 1 + cot^2 θ = cosec^2 θ

### General Tips

Example 1:

1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Solution:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, angle(ACB) = 30º and angle(ADB) = 45º

AB/AC = tan 30º =1/√3

=> AC = AB x √3 = 100√3 m

AB/AD = tan 45º = 1

=> AD = AB = 100 m

CD = (AC + AD) = (100√3 + 100) m

= 100(√3 + 1)

= (100 x 2.73) m

= 273 m

Example 2:

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

Solution:

Let AB be the wall and BC be the ladder

Then, angle(ACB) = 60º and AC = 4.6 m

AC/BC = cos 60º = 1/2

BC = 2 x AC

= (2 x 4.6) m

= 9.2 m

Example 3:

From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Solution:

Let AB be the tower

Then, angle(APB) = 30º and AB = 100 m

AB/AP = tan 30º =1/√3

AP = (AB x √3) m

= 100√3 m

= (100 x 1.73) m

= 173 m.

Example 4:

The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

Solution:

Let AB be the tree and AC be its shadow.

Let angle(ACB) = θ

Then, AC/AB = √3

cot θ= √3

θ = 30º.