Trigonometry
Exact Values
Angles greater than 90o
Useful Notation & Area of a triangle
Using Area of Triangle Formula
Sine Rule Problems
Cosine Rule Problems
Mixed Problems
23-Mar-16
Starter Questions
1. Factorise x2 - 36
2. A car depreciates at 20% each year.
How much is it worth after 4 years if it cost
£15 000 initially.
3. What sin30o as a fraction.
23-Mar-16
Exact Values
Learning Intention
1. To build on basic
trigonometry values.
Success Criteria
1. Recognise basic triangles and
exact values for sin, cos and
tan 30o, 45o, 60o .
2. Calculate exact values for
problems.
23-Mar-16
Exact Values
Some special values of Sin, Cos and Tan are useful
left as fractions, We call these exact values
60º
2
2
60º
60º
2
30º
2
3
1
This triangle will provide exact values for
sin, cos and tan 30º and 60º
60º
Exact Values
x
Sin xº
Cos xº
Tan xº
0º
0
30º
45º
60º
½
3
3
½
2
1
3
90º
2
3

Exact Values
Consider the square with sides 1 unit
1
1
45º
2
45º
1
1
We are now in a position to calculate
exact values for sin, cos and tan of 45o
Exact Values
x
Sin xº
Cos xº
Tan xº
0º
0
30º
45º
60º
½
1
2
3
3
2
1
2
½
1
3
1
3
90º
2

Exact Values
Now try Exercise 1
Ch8 (page 94)
23-Mar-16
Starter Questions
1. Write down the Compound Interest Formula
and identify each term.
2. A house increases by 3% each year.
How much is it worth in 5 years if it cost
£40 000 initially.
3. What is the exact value of sin 45o.
23-Mar-16
Angles Greater than 90o
Learning Intention
1. Introduce definition of
sine, cosine and tangent
over 360o using triangles
with the unity circle.
23-Mar-16
Success Criteria
1. Find values of sine, cosine
and tangent over the range 0o
to 360o.
2. Recognise the symmetry and
equal values for sine, cosine
and tangent.
r
x
y
Angles Greater than 90o
We will now use a new definition to cater for ALL angles.
New Definitions
y-axis
y
O
23-Mar-16
P(x,y)
r
Ao
y
sin A 
r
x
cos A 
x
r
tan A 
y
x
x-axis
11
Trigonometry
Angles over 900
Example 1
The radius line is 2cm.
The point (1.2, 1.6).
(1.2, 1.6)
Find sin cos and tan for
the angle.
sin 53o 
1.6
 0.8
2
cos 53o 
1.2
 0.6
2
tan 53o 
23-Mar-16
1.6
 1.33
1.2
Check answer
with calculator
53o
Trigonometry
Example 1
Angles over 900
The radius line is 2cm.
The point (-1.8, 0.8).
Find sin cos and tan for
the angle.
0.8
sin127o 
 0.4
2
cos127o 
tan 53o 
23-Mar-16
1.8
 0.9
2
0.8
 0.44
1.8
Check answer
with calculator
(-1.8, 0.8)
127o
What Goes In The Box ?
Write down the equivalent values of the following
in term of the first quadrant (between 0o and 90o):
1) Sin 135o
sin 45o
1) Sin 300o
- sin 60o
2) Cos 150o
-cos 45o
2) Cos 360o
cos 0o
3) Tan 135o
-tan 45o
3) Tan 330o
- tan 30o
4) Sin 225o
-sin 45o
4) Sin 380o
sin 20o
5) Cos 270o
-cos 90o
5) Cos 460o
- cos 80o
Trigonometry
Angles over 900
Now try Exercise 2
Ch8 (page 97)
23-Mar-16
Trigonometry
Angles over 900
Extension for unit 2 Trigonometry
GSM Software
23-Mar-16
Angles Greater than 90o
Two diagrams display same data in a different format
Sin +ve
180o - xo
(0,1)
A0
(-1,0)
180o
+
xo
Tan +ve
All +ve
(1,0)
360o - xo
(0,-1)
Cos +ve
Starter Questions
1. Find the area of the triangle.
3cm
2. Factorise x - 4x + 3
2
8cm
3. Find the exact value of cos 120 o.
23-Mar-16
Area of a Triangle
Learning Intention
Success Criteria
1. To show the standard way
of labelling a triangle.
1. Be able to label a triangle
properly.
2. Find the area of a triangle
using basic trigonometry
knowledge.
2. Find the area of a triangle
using basic trigonometry
knowledge.
23-Mar-16
Labelling Triangles
In Mathematics we have a convention for labelling triangles.
B
a
c
A
23-Mar-16
C
b
Small letters a, b, c refer to distances
Capital letters A, B, C refer to angles
Labelling Triangles
Have a go at labelling the following triangle.
E
d
f
D
23-Mar-16
F
e
Area of a Triangle
Example 1 : Find the area of the triangle ABC.
B
10cm
A
50o
23-Mar-16
(i)
Draw in a line from B to AC
(ii)
Calculate height BD
BD
Sin50 =
10
o
7.66cm
D
12cm
BD = 10 Sin50o = 7.66
(iii)
C
Area
1
Area  base  height
2
 0.5 12  7.66  46cm2
Area of a Triangle
Example 2 : Find the area of the triangle PQR.
P
12cm
Q
40o
23-Mar-16
(i)
Draw in a line from P to QR
(ii)
Calculate height PS
PS
Sin40 =
10
o
7.71cm
S
20cm
PS = 12 Sin40o = 7.71
(iii)
R
Area
1
Area  base  height
2
 0.5  20  7.71  77.1cm2
Constructing Pie Charts
Now try Exercise 3
Ch8 (page 99)
23-Mar-16
Starter Questions
1. Multiply out and simplify 2( x  3)  (4  x)
2. Find the volume of a cylinder 15cm in height
and 10cm in diameter.
3. Write down the two values for sin
that give a value of 0.5
23-Mar-16
Area of ANY Triangle
Learning Intention
1. To explain how to use the
Area formula for ANY
triangle.
23-Mar-16
Success Criteria
1. Know the formula for the
area of any triangle.
2. Use formula to find area of
any triangle given two length
and angle in between.
General Formula for
Area of ANY Triangle
Co
Consider the triangle below:
Ao
Area = ½ x base x height
1
A  ch
2
1
A   c  b sin Ao
2
1
A  bc sin Ao
2
b
a
h
Bo
c
What does the sine of Ao equal
h
o
sin A 
b
Change the subject to h.
h = b sinAo
Substitute into the area formula
Key feature
Area of ANY Triangle
To find the area
you need to knowing
The area
of ANY
can be found
2 sides
andtriangle
the angle
byinthe
following
formula.
between
(SAS)
B
a
c
A
23-Mar-16
1
Area= bc sin A
2
C
b
Another version
1
Area= ac sin B
2
Another version
1
Area= ab sin C
2
Area of ANY Triangle
Example : Find the area of the triangle.
B
c
A
23-Mar-16
The version we use is
1
Area= ab sin C
2
20cm
30o
25cm
C
1
Area   20  25  sin 30o
2
Area  10  25  0.5  125cm 2
Area of ANY Triangle
Example : Find the area of the triangle.
The version we use is
E
60o
8cm
1
Area= df sin E
2
10cm
F
1
Area   8 10  sin 60o
2
D
Area  40  0.866  34.64cm2
23-Mar-16
Key feature
What Goes In The Box
?
Remember
(SAS)
Calculate the areas of the triangles below:
(1)
12.6cm
A =36.9cm2
23o
15cm
(2)
5.7m
71o
6.2m
A =16.7m2
Area of ANY Triangle
Now try Exercise 4
Ch8 (page 100)
23-Mar-16
Starter Questions
1. Multiply out the brackets and simplify
5(y - 5) - 7(5 - y)
2. Find the gradient and the y - intercept
3
for the line with equation y = 5x 4
3. Factorise x2 - 100
23-Mar-16
Sine Rule
Learning Intention
1. To show how to use the
sine rule to solve REAL
LIFE problems involving
finding the length of a
side of a triangle .
23-Mar-16
Success Criteria
1. Know how to use the sine
rule to solve REAL LIFE
problems involving lengths.
Sine Rule
Works for any Triangle
The Sine Rule can be used with ANY triangle
as long as we have been given enough information.
B
a
b
c
=
=
SinA SinB SinC
c
A
23-Mar-16
a
C
b
The Sine Rule
Consider a general triangle ABC.
Deriving the rule
C
b
a
B
P
CP
 CP  aSinB
a
CP
also SinA 
 CP  bSinA
b
 aSinB  bSinA
aSinB

b
SinA
a
b


SinA SinB
SinB 
c
Draw CP perpendicular to BA
A
This can be extended to
a
b
c


SinA SinB SinC
or equivalently
SinA SinB SinC


a
b
c
Calculating Sides
Using The Sine Rule
Example 1 : Find the length of a in this triangle.
B
a
10m
A
34o
41o
Match up corresponding sides and angles:
10
a

o
sin 34 o
sin 41
Now cross multiply.
a sin 34o  10sin 41o
10sin 41o
a
sin 34o
a
Solve for a.
10  0.656
 11.74m
0.559
C
Calculating Sides
Using The Sine Rule
Example 2 : Find the length of d in this triangle.
D
10m
o
133
C
37o
Match up corresponding sides and angles: d
d
10

o
sin133
sin 37 o
Now cross multiply.
d sin 37o  10sin133o
Solve for d.
10sin133o
d
sin 37o
10  0.731
= 12.14m
0.602
d
E
What goes in the Box ?
Find the unknown side in each of the triangles below:
12cm
(1)
a
32o
72o
47o
b
93o
a = 6.7cm
23-Mar-16
(2)
16mm
b = 21.8mm
Sine Rule
Now try Ex 6&7
Ch8 (page 103)
23-Mar-16
Starter Questions
1. Factorise 9x - 36
2. Find the gradient and the y - intercept
1
3
for the line with equation y = - x +
5
4
3. Write down the two values of cos
1
that give you a value of
2
23-Mar-16
Sine Rule
Learning Intention
1. To show how to use the
sine rule to solve problems
involving finding an angle
of a triangle .
23-Mar-16
Success Criteria
1. Know how to use the sine
rule to solve problems
involving angles.
Calculating Angles
Using The Sine Rule
Example 1 :
Find the angle
Ao
45m
38m
23o
Ao
Match up corresponding sides and angles:
45
38

sin Ao sin 23o
Now cross multiply:
38sin Ao  45sin 23o
Solve for sin Ao
45sin 23o
sin A 
= 0.463
38
Use sin-1 0.463 to find Ao
o
1
A  sin 0.463  27.6
o
o
Calculating Angles
Using The Sine Rule
75m
Example 2 :
Find the angle Bo
Bo
143o
38m
Match up corresponding sides and angles:
75
38

o
sin143o
sin B
Now cross multiply:
75sin B o  38sin143o
Solve for sin Bo
o
38sin143
sin Bo 
75
Use sin-1 0.305 to find Bo
1
= 0.305
B  sin 0.305  17.8
o
o
What Goes In The Box ?
Calculate the unknown angle in the following:
(1)
100o
8.9m
Ao
(2)
12.9cm Bo
14.5m
Ao = 37.2o
14o
14.7cm
Bo = 16o
Sine Rule
Now try Ex 8 & 9
Ch8 (page 106)
23-Mar-16
Starter Questions
1. Find the gradient of the line that passes
through the points ( 1,1) and (9,9).
2. Find the gradient and the y - intercept
for the line with equation y = 1 - x
3. Factorise x - 64
2
23-Mar-16
Cosine Rule
Learning Intention
1. To show when to use the
cosine rule to solve
problems involving finding
the length of a side of a
triangle .
23-Mar-16
Success Criteria
1. Know when to use the cosine
rule to solve problems.
2. Solve problems that involve
finding the length of a side.
Cosine Rule
Works for any Triangle
The Cosine Rule can be used with ANY triangle
as long as we have been given enough information.
a =b +c - 2bc cos A
2
2
2
B
a
c
A
23-Mar-16
C
b
The Cosine Rule
The Cosine Rule generalises Pythagoras’ Theorem and
takes care of the 3 possible cases for Angle A.
Deriving the rule
B
Consider a general triangle ABC. We
require a in terms of b, c and A.
BP2
– (b –
x)2
Also: BP2 = c2 – x2
a
c
=
a2
1
A
a2 = b2 + c2
2
 a2 – (b – x)2 = c2 – x2
 a2 – (b2 – 2bx + x2) = c2 – x2
A
x
P
b
b
b-x
Draw BP perpendicular to AC
C
 a2 – b2 + 2bx – x2 = c2 – x2
 a2 = b2 + c2 – 2bx*
 a2 = b2 + c2 – 2bcCosA
*Since Cos A = x/c  x = cCosA
When A = 90o, CosA = 0 and reduces to a2 = b2 + c2
1
Pythagoras
When A > 90o, CosA is positive,  a2 > b2 + c2
2
Pythagoras + a bit
When A < 90o, CosA is negative,  a2 > b2 + c2
3
Pythagoras - a bit
A
a2 > b2 + c2
3
A
a2 < b2 + c2
The Cosine Rule
The Cosine rule can be used to find:
1. An unknown side when two sides of the triangle and the
included angle are given.
2. An unknown angle when 3 sides are given.
B
Finding an unknown side.
a2 = b2 + c2 – 2bcCosA
Applying the same method as
earlier to the other sides
produce similar formulae for
b and c. namely:
a
c
A
b
b2 = a2 + c2 – 2acCosB
c2 = a2 + b2 – 2abCosC
C
Cosine Rule
Works for any Triangle
How to determine when to use the Cosine Rule.
Two questions
1. Do you know ALL the lengths.
OR
SAS
2. Do you know 2 sides and the angle in between.
If YES to any of the questions then Cosine Rule
Otherwise use the Sine Rule
23-Mar-16
Using The Cosine Rule
Works for any Triangle
Example 1 : Find the unknown side in the triangle below:
L
5m
43o
Identify sides a,b,c and angle Ao
12m
a= L
b= 5
a2 = b2 +
c = 12
Ao = 43o
c2 -2bccosAo
Write down the Cosine Rule.
2
a2 = 52 + 122 - 2 x 5 x 12 cos 43o Substitute values to find a .
a2 = 25 + 144 - (120 x 0.731 )
a2 = 81.28
a = L = 9.02m
Square root to find “a”.
Using The Cosine Rule
Works for any Triangle
12.2 m
Example 2 :
Find the length of side M.
a = M b = 12.2 C = 17.5
a2 = b2 +
c2 -2bccosAo
137
17.5 m
o
M
Ao = 137o Identify the sides and angle.
Write down Cosine Rule
a2 = 12.22 + 17.52 – ( 2 x 12.2 x 17.5 x cos 137o )
a2 = 148.84 + 306.25 – ( 427 x – 0.731 )
Notice the two negative signs.
a2 = 455.09 + 312.137
a2 = 767.227
a = M = 27.7m
What Goes In The Box ?
Find the length of the unknown side in the triangles:
43cm
(1)
78o
31cm
L
L = 47.5cm
(2)
M
5.2m
M = 5.05m
38o
8m
Cosine Rule
Now try Ex 11.1
Ch11 (page 142)
23-Mar-16
Starter Questions
1. If lines have the same gradient
What is special about them.
2. Factorise x + 4x - 12
2
3. Find the missing angles.
23-Mar-16
Cosine Rule
Learning Intention
1. To show when to use the
cosine rule to solve REAL
LIFE problems involving
finding an angle of a
triangle .
23-Mar-16
Success Criteria
1. Know when to use the cosine
rule to solve REAL LIFE
problems.
2. Solve REAL LIFE problems
that involve finding an angle
of a triangle.
Cosine Rule
Works for any Triangle
The Cosine Rule can be used with ANY triangle
as long as we have been given enough information.
a =b +c - 2bc cos A
2
2
2
B
a
c
A
23-Mar-16
C
b
Finding Angles
Using The Cosine Rule
Works for any Triangle
Consider the Cosine Rule again:
a2 = b2 +
c2 -2bc cosAo
We are going to change the subject of the formula to cos Ao
b2 + c2 – 2bc cos Ao = a2
Turn the formula around:
-2bc cos Ao = a2 – b2 – c2
Take b2 and c2 across.
2
2
2
a

b

c
cos Ao 
2bc
b c a
cos A 
2bc
2
o
2
2
Divide by – 2 bc.
Divide top and bottom by -1
You now have a formula for
finding an angle if you know all
three sides of the triangle.
Finding Angles
Using The Cosine Rule
Works for any Triangle
Example 1 : Calculate the
unknown angle xo .
2
2
2
b

c

a
cos Ao 
2bc
a = 11 b = 9
Ao = ?
92  162  112
cos A 
2  9 16
9cm
11cm
Ao
16cm
Write down the formula for cos Ao
c = 16 Label and identify Ao and a , b and c.
o
Substitute values into the formula.
Cos Ao = 0.75
Calculate cos Ao .
Ao = 41.4o
Use cos-1 0.75 to find Ao
Finding Angles
Using The Cosine Rule
Works for any Triangle
Example 2: Find the unknown
yo
15cm
13cm
Angle in the triangle:
26cm
2
2
2
b

c

a
cos Ao 
2bc
Ao = yo
a = 26
b = 15
Write down the formula.
c = 13
2
2
2
15

13

26
cos Ao 
2 15 13
cosAo =
- 0.723
Ao = yo = 136.3o
Identify the sides and angle.
Find the value of cosAo
The negative tells you
the angle is obtuse.
What Goes In The Box ?
Calculate the unknown angles in the triangles below:
(1)
5m
Ao
10m
Ao =111.8o
7m
(2)
12.7cm
Bo
8.3cm
7.9cm
Bo = 37.3o
Cosine Rule
Now try Ex 11.2
Ch11 (page 143)
23-Mar-16
Starter Questions
1. A washing machine is reduced by 10%
in a sale. It's sale price is £360.
What was the original price.
2. Factorise x - 7x +12
2
3. Find the missing angles.
23-Mar-16
61o
Mixed problems
Learning Intention
1. To use our knowledge
gained so far to solve
various trigonometry
problems.
23-Mar-16
Success Criteria
1. Be able to recognise the
correct trigonometric
formula to use to solve a
problem involving triangles.
The angle of elevation of the
top of a building measured
from point A is 25o. At point
D which is 15m closer to the
building, the angle of
elevation is 35o Calculate the
height of the building.
T
10o
36.5
35o
B
Angle TDA = 180 – 35 = 145o
Angle DTA = 180 – 170 = 10o
TD
15

Sin 25o Sin10o
15Sin 25o
TD 
 36.5 m
Sin10
145o
25o
D
15 m
Sin 35o 
A
TB
36.5
 TB  36.5Sin 35o  20.9 m
The angle of elevation of the top of a column measured from point A, is 20o.
The angle of elevation of the top of the statue is 25o. Find the height of the
statue when the measurements are taken 50 m from its base
Angle BCA = 180 – 110 = 70o
Cos 20o 
50
Angle ACT = 180 – 70 = 110o
TC
53.21

Sin 5o Sin 65o
AC
50
53.21 Sin 5
 5.1 m (1dp )
Cos 20o  TC 
Sin 65o
 53.21 m (2dp )
 AC 
Angle ATC =
T
180 – 115 = 65o
65o
110o
C
70o
5o
A
20o
25o
50 m
B
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a
marker buoy (B). At B the boat turns left and sails for 24 miles to a
lighthouse (L). It then returns to harbour, a distance of 57 miles.
(a) Make a sketch of the journey.
(b) Find the bearing of the lighthouse from the harbour. (nearest degree)
572  402  242
CosA 
2x 57x 40
A  20.4o
L
 Bearing  90  20.4  070o
57 miles
H
24 miles
A
40 miles
B
An AWACS aircraft takes off from RAF
Waddington (W) on a navigation
exercise. It flies 530 miles North to
a point (P) as shown, It then turns
left and flies to a point (Q), 670
miles away. Finally it flies back to
base, a distance of 520 miles.
Find the bearing of Q from point P.
b2 c 2 a2
CosA 
2bc
5302  6702  5202
CosP 
2x 530x 670
P  48.7o
 Bearing  180  48.7  229o
Not to Scale
P
670 miles
530 miles
Q
520 miles
W
Mixed Problems
Now try Ex 14
Ch8 (page 117)
23-Mar-16