The document discusses using trigonometric ratios (sine, cosine, tangent) to find missing lengths and angles in right-angled triangles. It provides examples of using sine to find the opposite side when given the hypotenuse and angle, using cosine to find the adjacent side, and using tangent to find the opposite side when given the adjacent side and angle. The key trigonometric ratios SOH CAH TOA are emphasized to remember which ratio uses the opposite, adjacent, or hypotenuse sides of a right triangle. Multiple practice problems are worked through as examples.

1. Wednesday 8 August 2012
Trigonometry
Using all 3 trig ratios to find missing
lengths
Outcomes
Must: Use label sides of triangles correctly
Should: Be able to do calculations involving trig
functions
Could: Use trig ratios to find missing lengths in
triangles.

2. Wednesday 8 August 2012
Right-angled triangles
A right-angled triangle contains a right angle.
The longest side opposite
the right angle is called the
hypotenuse.

3. The opposite and adjacent sides
The two shorter sides of a right-angled triangle are named
with respect to one of the acute angles.
The side opposite the
marked angle is called
the opposite side.
The side between the x
marked angle and the
right angle is called
the adjacent side.

4. Label the sides

5. Calculate the following ratios
Use your calculator to find the following to 3 significant figures.
1) sin 79° = 0.982 2) cos 28° = 0.883
3) tan 65° = 2.14 4) cos 11° = 0.982
5) sin 34° = 0.559 6) tan 84° = 9.51
7) tan 49° = 1.15 8) sin 62° = 0.883
9) tan 6° = 0.105 10) cos 56° = 0.559

6. The three trigonometric ratios
Opposite
O
H
Sin θ =
Hypotenuse SOH
Y
P P
P O
O T
Adjacent
S
I
E
N Cos θ =
Hypotenuse CAH
T U
E S
E
θ Opposite
Tan θ =
Adjacent TOA
ADJACENT
Remember: S O H C A H T O A

7. The sine ratio
the length of the opposite side
The ratio of is the sine ratio.
the length of the hypotenuse
The value of the sine ratio depends on the size of the angles
in the triangle.
O H
P
Y
P
We say:
P O
T
O E opposite
S
N
U sin θ =
I S
E
hypotenuse
T
E θ

8. Using sine to find missing lengths
S O H C A H TO A Sin θ =
opposite
hypotenuse
opp
hyp Sin θ =
hyp
11 cm 65° Sin 65 x 11 = x
x
Sin 65 = 9.97 (2dp) = x
11
x cm
opposite
x x x
Sin 32 = Sin 72 = Sin 54 =
6 12 15
Sin 32 x 6 = x Sin 72 x 12 = x Sin 54 x 15 = x
3.18cm (2dp) = x 11.41cm (2dp) = x 12.14cm (2dp) = x
6 cm 32° 12 cm 72° 15 cm 54°
(1) (2) (3)
x cm x cm

9. Using sine to find missing lengths
S O H C A H TO A Sin θ =
opposite
hypotenuse
opp
hyp Sin θ =
hyp
x cm 47° 8
8 x =
Sin 47 = Sin 47
x
8 cm opposite x = 10.94 cm (2dp)
6
Sin 32 = x 12 3
Sin 72 = x Sin 54 = x
x = 6 12
Sin 32 x = x = 3
Sin 72 Sin 54
x = 11.32 cm(2dp)
x = 12.62 cm(2dp) x = 3.71 cm(2dp)
x cm 32° x cm 72° x cm 54°
hyp
(1) (2) (3)
6 cm opp 12 cm 3 cm

10. Using cosine to find missing lengths
adjacent
cos θ =
S O H C A H TO A hypotenuse
adj
cos θ = cos θ =
adj
hyp hyp
hyp hyp
cos 53 = x 12
10 cm 10 x cm cos 66 = x
53° cos 53 x 10 = x x = 12
66° cos 66
x cm 6.02 (2dp) = x 12 cm
adj x = 29.50 cm(2dp)
adj
adj
cos θ = hyp cos θ
adj
= hyp
hyp (2)
(1) cos 31 = x 22
9 cm 9 x cm cos 49 = x
31° cos 31 x 9 = x 49° x = 22
cos 49
x cm 7.71 (2dp) = x 22 cm
x = 33.53 cm(2dp)
adj

11. Using tangent to find missing lengths
opposite
S O H C A H TO A tan θ =
adjacent
opp
opp
tan θ 8cm xcm
= adj
adj 31
71°
10 cm
adj tan 71 = x
10 opp
tan θ = adj
opp x cm tan 71 x 10 = x
8
29.04cm (2dp)= x tan 31 = x
x = 8
tan 31
7c
m (2) 43 x = 13.31 cm(2dp)
(1) c m
m
31
xc
4.21cm (2dp) = x
48 x = 38.72 cm(2dp)
xc
m

12. Using all 3 trig ratios to find missing lengths
S O H C A H TO A hyp
10 cm
opp opp
53°
7cm 7cm x cm
47 adj
33°
adj
xcm hyp x cm adj cos θ =
hyp
opp opp cos 53 = x
sin θ = tan θ = adj 10
hyp
7 7 cos 53 x 10 = x
sin 47 = x tan 33 = x
6.02cm (2dp) = x
x = 7 x = 7
sin 47 tan 33
x = 9.57 cm(2dp) x = 10.78 cm(2dp)

13. Finding Angles using Trig
S O H C A H TO A 6cm
opp
θ
8 cm opp hyp 10cm
opp
5 cm
tan θ =
opp sin θ =
adj hyp
θ adj
6
tan θ =
8 sin θ =
10
5
6
θ = sin-1
θ = 8 tan-1
5
10
57.99 (2dp) θ = 36.87 (2dp)

15. Wednesday 8 August 2012
Trigonometry 2
Objective: Use trig to find missing lengths
and angles in right angled triangles for
worded questions. Grade A
Outcomes
Must: Use trig ratios to find missing lengths and
angles in triangles.
Should: Use trig to answer worded problems.
Could: Use trig to answer more difficult worded
problems.

16. Finding side lengths
A 5 m long ladder is resting against a wall. It makes an
angle of 70° with the ground.
What is the distance between the
base of the ladder and the wall?
We are given the hypotenuse and we want
to find the length of the side adjacent to
5m the angle, so we use:
adjacent
cos θ =
hypotenuse
70° x
x cos 70° =
5
x = 5 × cos 70°
= 1.71 m (to 2d.p.)

18. Wednesday 8 August 2012
Area of a triangle
(using ½ ab Sin C)
Outcomes
Must: Understand when you
Objective
can use this formula for
Find the area of a triangle area.
triangle using
Area = ½ ab sin C
Should: Be able to find the
area of a triangle using
½ ab sin C.
Could: Answer exam questions

19. Triangle Area : When to use ½ ab Sin C ?
• When you have not been given a perpendicular height
(straight height). 4cm 40
6cm
• When you have been given two lengths and an
angle between them.
A
A = ½ ab Sin C
5cm A = ½ x 4 x 6 Sin 35
35
A = ……….cm2
B C
4cm
1 6cm
Area of triangle ABC = ab sin C
2 23
9cm
A = ½ x 4 x 5 Sin 35 A = ½ ab Sin C
A = ……….cm2 A = ½ x 6 x 9 Sin 35
A = ……….cm2

20. Find the area of the triangles
(1) (2)
6cm 4cm 6cm 4cm
50 35
(3)
7cm 7cm
5cm 45
A = ½ ab Sin C A = ½ ab Sin C 3cm
A= ½ x 7 x 4 Sin 50 A= ½ x 7 x 4 Sin 35
A= ……….cm2 A= ……….cm2
7cm
A = ½ ab Sin C
A= ½ x 5 x 3 Sin 45
A= ……….cm2

22. Wednesday 8 August 2012
Sine Rule
Outcomes
Objective
Use sine rule to find
angles and lengths in Must: Use sine rule to find
triangles. (not right lengths in triangles.
angles triangles)
Should: Use sine rule to find
angles in triangles.
Could: Answer mixed
questions

23. Sine Rule Examples: Find y using
sine rule
• Find the sides and angles of a triangle
whether it’s a right angle or not.
b 6cm
C y cm
a
25 40
A B
b a a b
=
Sin A Sin B
A B y
c 3
=
Sin 25 Sin 40
a b c
= = 3
sin A sin B sin C x Sin 40 = y
Sin 25
4.56 cm (2dp) = y

24. Sine Rule a b c
C = =
Sin A Sin B Sin C
b a
A c B (3) C
Exercise: Find lengths y using sine rule b
8.9cm
(2) C
(1) C
b
34 4.2cm 41 36 62
B
y cm y c
a
15 22
B
A y c
6cm c
Y = 5.9 cm
y 6 4.2 y
= =
Sin 15 Sin 34 Sin 22 Sin 41
6 4.2
y = x Sin 15 x Sin 41 = y
Sin 34 Sin 22
y = 2.8 cm (1dp) 7.4 cm (1dp) = y

25. Sine Rule
C
b
C
2.3cm 43
b a x
3.5cm B
A B c
c
Sin B Sin C
Use this when finding a length =
b c
a b c Sin 43
= = Sin x
sin A sin B sin C =
2.3 3.5
Sin 43
or Sin x = x 2.3
3.5
Use this when finding an angle
sin A sin B sin C Sin x = 0.44817…..
= = x = 0.44817…..sin-1
a b c
x = 26.6 (1dp)

26. Exercise: Find angle y using sin rule
Sine Rule
C (2) C
(1)
63 y
C 2.9 cm 7.3 cm
b
a
b a y 53
A B
A c B 4.3cm c 8.4cm c
Sin A
= Sin B = Sin C Sin A
=
Sin C Sin B
=
Sin Y
a b c a c b c
Sin y Sin 63 Sin 53 Sin y
= =
2.9 4.3 7.3 8.4
Sin 63 Sin 53
Sin y = x 2.9 x 8.4 = Sin y
4.3 7.3
Sin y = 0.600911….. 0.9080… Sin y
=
y = 0.600911….. -1
sin 0.9080… sin-1 = y
y = 36.9 (1dp) 65.2 (1dp) = y

30. Sine Rule Practice
Higher GCSE for AQA (Oxford) Book
Page 387 Exercise 4r Q1 – Q19

31. Wednesday 8 August 2012
Cosine Rule
Outcomes
Objective
Use cosine rule to find
angles and lengths in Must: Use cosine rule to find
triangles. (not right lengths in triangles.
angles triangles)
Should: Use cosine rule to
find angles in triangles.
Could: Answer mixed
questions

32. The cosine rule Use when
finding a length
a2 = b2 + c2 – 2bc cos A of a side.
A
or
c b Use when
cos A = b2 + c2 – a2 finding an
B C 2bc Angle.
a
When you are given two lengths and an angle
Example 1 between them use;
Find a B
a2 = b2 + c2 – 2bc cos A
a 4 cm a2 = 72 + 42 - 2 x 7 x 4 cos 48
a2 = 27.52868…..
C 48° A a = 5.25 (2dp)
7 cm

33. When you are given two lengths and an angle
Example 1 between them use;
Find a B
a2 = b2 + c2 – 2bc cos A
a 4 cm a2 = 72 + 42 - 2 x 7 x 4 cos 48
a2 = 27.52868…..
C 48° A a = 5.25 (2dp)
7 cm
Exercise Find the length marked x
(1) (2)
2cm
x 9cm
52
98
x
5cm 5cm
a2 = 52 + 92 - 2 x 5 x 9 cos 98
a2 = 52 + 22 - 2 x 5 x 2 cos 52
a2 = 118.52557…..
a2 = 16.68677…..
a = 10.89 (2dp)
a = 4.08 (2dp)

34. The cosine rule Use when
finding a length
a2 = b2 + c2 – 2bc cos A of a side.
A (given 2 lengths
or and an angle)
c b
cos A = b2 + c2 – a2 Use when
B C 2bc finding an
a Angle.
(given 3 lengths)
Example 1 You are given 3 sides and asked for an angle.
Find angle A
B
cos A = b + c - a
2 2 2
a 2bc
8 cm c
6 cm cos A = 4 + 6 - 8
2 2 2
2x4x6
C 4 cm A cos A = - 0.25
b
A = - 0.25 cos-1 A = 104. 5 (1dp)

35. Example 1 You are given 3 sides and asked for an angle.
Find angle A
cos A = b + c - a
2 2 2
B
2bc
a
8 cm c cos A = 4 + 6 - 8
2 2 2
6 cm
2x4x6
A= 104. 5 (1dp)
C A cos A = - 0.25
b
4 cm
A = - 0.25 cos-1
Exercise Find the length marked x
(2) a
11cm b
(1) b 9cm
2cm
x
A x a A
4cm c 5cm
c 5cm cos A = 9 + 5 - 11
2 2 2
2x9x5
cos A = 2 + 5 - 4
2 2 2
2x2x5 cos A = - 0.1666….
cos A = 0.65 A = - 0.166..cos-1 A = 99. 6 (1dp)
A = 0.65 cos-1 A = 49.5 (1dp)