1. TRIGONOMETRY
2.4

2. PYTHAGORAS h2 = a2 + b2 Can only occur in a right angled triangle
hypotenuse
Pythagoras Theorem states: h a
h2 = a2 + b2 b
right angle
e.g. x
9.4 cm
7.65 m y
11.3 m
8.6 cm
x2 =
y = 9.42
9.42 = y
square root undoes squaring
smaller sides should always be smaller than the hypotenuse
x2 =
- 8.62
y2 = 9.42 – 8.62
- 8.62
x = √
y2 = 14.4
x = m (2 d.p.)
y = √14.4
y = 3.79 cm (2 d.p.)

3. TRIGONOMETRY (SIN, COS & TAN)
- Label the triangle as follows, according to the angle being used.
to remember the trig ratios use
SOH CAH TOA
Hypotenuse (H)
Opposite (O)
and the triangles A
Always make sure your calculator is set to degrees!!
Adjacent (A) O A O S H C H T A
means divide
1. Calculating Sides
means multiply
e.g. O
7.65 m x H h T A O O
29°
50°
h = tan50 x 6.5
6.5 cm A O
x = sin29 x 7.65
h = 7.75 cm (2 d.p.)
x = 3.71 m (2 d.p.) S H

4. e.g.
d = 455 ÷ sin32 O d H
455 m
d = m (2 d.p.) O S H
32°
2. Calculating Angles
-Same method as when calculating sides, except we use inverse trig ratios.
e.g. B
4.07 m
23.4 mm H
sin-1 undoes sin
2.15 m
16.1 mm H A O A
cosB = 2.15 ÷ 4.07
sinA = 16.1 ÷ 23.4 A O
B = cos-1(2.15 ÷ 4.07)
A = sin-1(16.1 ÷ 23.4) C H S H
B = 58.1° (1 d.p.)
A = 43.5° (1 d.p.)
Don’t forget brackets, and fractions can also be used

5. TRIGONOMETRY APPLICATIONS
e.g. A ladder 4.7 m long is
leaning against a wall. The
angle between the wall and
ladder is 27°.
Draw a diagram and find the
height the ladder extends up
the wall. A
Wall (x) C H
27° A
Ladder (4.7 m)
x = cos27 x 4.7 H
x = 4.19 m (2 d.p.) A
e.g. A vertical mast is held by a
48 m long wire. The wire is
attached to a point 32 m up
the mast.
Draw a diagram and find the
angle the wire makes with C H H A
48 m
cosA = 32 ÷ 48
32 m
A = cos-1(32 ÷ 48) A
A = 48.2° (1 d.p.)

6. NON-RIGHT ANGLED TRIANGLES
1. Naming Non-right Angled Triangles
- Capital letters are used to represent angles
- Lower case letters are used to represent sides
e.g. Label the following triangle c A
The side opposite the angle is given the same letter as the angle but in lower case. B b C a

7. 2. Sine Rule
a = b = c .
SinA SinB SinC
a) Calculating Sides
Only 2 parts of the rule are needed to calculate the answer
e.g. Calculate the length of side p
p =
Sin Sin46
52° A
× Sin52
p = × Sin52
Sin46
× Sin52 B
46°
6 m b
p = m (2 d.p.) p a
To calculate you must have the angle opposite the unknown side.
Re-label the triangle to help substitute info into the formula

8. For the statement: 1 = 3 is the reciprocal true? 2 6 Yes as 2 = 6 1 3
b) Calculating Angles
For the statement: 1 = 3 is the reciprocal true?
Yes as 2 = 6
Therefore to calculate angles, the Sine Rule is reciprocated so the unknown angle is on top and therefore easier to calculate.
You must calculate Sin51 before dividing by 6 (cannot use fractions)
SinA = SinB = SinC
a b c
a = b = c .
SinA SinB SinC
e.g. Calculate angle θ
Sinθ = Sin51 θ A
Sinθ = Sin51 × 7 6
× 7
× 7 B
51°
6 m b
θ = sin-1( Sin51 × 7) 6
7 m a
θ = 65.0° (1 d.p.)
To calculate you must have the side opposite the unknown angle
Re-label the triangle to help substitute info into the formula

9. Sine Rule Applications
e.g. A conveyor belt 22 m in length drops sand onto a cone-shaped heap. The sides of the cone measure 7 m and the cone’s sides make an angle of 32° with the ground. Calculate the angle that the belt makes with the ground (θ), and the diameter of the cone’s base (x).
SinA = SinB = SinC
a b c
Conveyor belt : 22 m
116° A b
Sinθ = Sin148 b
7 m
7 m a
32° B
32°
148° B A θ
Sinθ = Sin148 × 7 22
× 7
× 7 x a
a = b = c .
SinA SinB SinC
θ = sin-1( Sin148 × 7) 22
x =
Sin Sin32
θ = 9.7° (1 d.p.)
× Sin116
x = × Sin116
Sin32
× Sin116
x = m (2 d.p.)

10. 3. Cosine Rule
-Used to calculate the third side when two sides and the angle between them (included angle) are known.
a2 = b2 + c2 – 2bcCosA
a) Calculating Sides
e.g. Calculate the length of side x
x2 = – 2×13×11×Cos37
13 m b
x2 = 61.59 A
37°
x = √61.59 x
x = 7.85 m (2 d.p.) a
11 m c
Remember to take square root of whole, not rounded answer
Re-label the triangle to help substitute info into the formula

11. b) Calculating Angles
- Need to rearrange the formula for calculating sides
CosA = b2 + c2 – a2
2bc
Watch you follow the BEDMAS laws!
e.g. Calculate the size of the largest angle P
CosR = – 242
2×13×17
24 m a Q
CosR =
13 m b A
R = cos-1(-0.267)
17 m c
R = 105.5° (1 d.p.) R
Re-label the triangle to help substitute info into the formula
Remember to use whole number when taking inverse

12. Cosine Rule Applications
e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °.
Calculate the distance from the tee to the hole (x) and the angle (θ) at which the golfer hit the ball away from the correct direction. b x a
Hole
Tee
a2 = b2 + c2 – 2bcCosA A θ
x2 = – 2×130×245×Cos60
x2 = 45075
130 m A
245 m
60°
x = √45075 b a c c
Ball
x = m (2 d.p.)
CosA = b2 + c2 – a2
2bc
Cosθ = – 1302
2×212.31×245
Remember to use whole number from previous question!
Cosθ = 0.848
θ = cos-1(0.848)
θ = 32.0° (1 d.p.)

13. 3D FIGURES - Pythagoras and Trigonometry can be used in 3D shapes
e.g. Calculate the length of sides x and w and the angles CHE and GCH A B
x2 =
x = √ G F
x = √61
x = 7.8 m (1 d.p.) w O
Make sure you use whole answer for x in calculation
w2 = D C
7 m
w = √ A x
5 m
w = √110 O
w = m (1 d.p.) H E
6 m A
tanCHE = 5 ÷ 6
tanCHE = 7 ÷ 7.8 O O
CHE = tan-1(5 ÷ 6)
CHE = tan-1(7 ÷ 7.8) T A T A
CHE = 39.8° (1 d.p.)
CHE = 41.9° (1 d.p.)

14. 4. Area of a triangle
- can be found using trig when two sides and the angle between the sides (included angle) are known
Area = ½abSinC
e.g. Calculate the following area
Area = ½×8×9×Sin39
9 m b C
Area = 22.7 m2 (1 d.p.)
39°
52°
8 m a
89°
Re-label the triangle to help substitute info into the formula
Calculate size of missing angle using geometry (angles in triangle add to 180°)

15. Area Applications
e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °.
Calculate the area contained in between the tee, hole and ball.
Hole
Tee
Area = ½abSinC C
Area = ½×130×245×Sin60
130 m
245 m
60° a b
Area = m2 (1 d.p.)
Ball