Transcript Cosine Rule - kcpe-kcse

Trigonometry
Sine Rule Finding a length
Sine Rule Finding an Angle
Cosine Rule Finding a Length
Cosine Rule Finding an Angle
Area of ANY Triangle
Mixed Problems
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 .
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
a
c
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
C
Match up corresponding sides and angles:
a
b
c


sin Ao sin B sin C
10
a

o
sin 34 o
sin 41
Rearrange and solve for a.
10sin 41o
a
sin 34o
a
10  0.656
 11.74m
0.559
Calculating Sides
Using The Sine Rule
Example 2 : Find the length of d in this triangle.
D
10m
o
133
37o
C
E
Match up corresponding sides and angles: d
d
10

o
sin133
sin 37 o
c
d
e


sin C o sin D sin E
Rearrange and solve for d.
10sin133o
d
sin 37o
d
10  0.731
= 12.14m
0.602
What goes in the Box ?
Find the unknown side in each of the triangles below:
12cm
(1)
a
32o
(2)
47o
b
72o
93o
16mm
A = 6.7cm
B = 21.8mm
Sine Rule
Learning Intention
1. To show how to use the
sine rule to solve problems
involving finding an angle
of a triangle .
Success Criteria
1. Know how to use the sine
rule to solve problems
involving angles.
Calculating Angles
Using The Sine Rule
B
Example 1 :
Find the angle
45m
38m
Ao
23o
A
Match up corresponding sides and angles:
45
38

sin Ao sin 23o
a
b
c


sin A sin B sin C
Rearrange and solve for sin Ao
45sin 23o
sin A 
= 0.463
38
o
1
Use sin-1 0.463 to find Ao
A  sin 0.463  27.6
o
o
C
Calculating Angles
Using The Sine Rule
Example 2 :
75m
X
Find the angle Xo
143o
Y
Match up corresponding sides and angles:
75
38

o
sin143o
sin X
Z
38m
x
y
z


sin X sin Y sin Z
Rearrange and solve for sin Xo
o
38sin143
sin X o 
75
1
Use sin-1 0.305 to find Xo
= 0.305
X  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
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 .
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
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 negative,  a2 > b2 + c2
2
Pythagoras + a bit
When A < 90o, CosA is positive,  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 (SAS).
2. An unknown angle when 3 sides are given (SSS).
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
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
Example 2 :
12.2 m
137o
17.5 m
Find the length of side M.
a = M b = 12.2 C = 17.5
a2 = b2 +
c2 -2bccosAo
Ao = 137o
M
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
Learning Intention
1. To show when to use the
cosine rule to solve REAL
LIFE problems involving
finding an angle of a
triangle .
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
C
b
Finding Angles
Using The Cosine Rule
Works for any Triangle
Consider the Cosine Rule again:
a2 = b2 +
c2 -2bccosAo
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 Ao .
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 yo 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 =
Ao = yo =
- 0.723
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
Area of ANY Triangle
Learning Intention
1. To explain how to use the
Area formula for ANY
triangle.
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.
Labelling Triangles
In Mathematics we have a convention for labelling triangles.
B
a
c
A
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
F
e
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
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
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
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
Mixed problems
Learning Intention
1. To use our knowledge
gained so far to solve
various trigonometry
problems.
Success Criteria
1. Be able to recognise the
correct trigonometric
formula to use to solve a
problem involving triangles.
Exam Type Questions
Angle TDA = 180 – 35 =
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.
145o
Angle DTA = 180 – 170 = 10o
T
36.5
B
t
d
a


sin T sin D sin A
TD
15

Sin 25o Sin10o
15Sin 25o
TD 
 36.5 m
Sin10
10o
35o
145o 25o
D
15 m
A
SOH CAH TOA
Sin 35o 
TB
36.5
 TB  36.5Sin 35o  20.9 m
Exam Type Questions
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
Exam Type Questions
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
T
o
o
o
Angle BCA =180 – 110 = 70 Angle ACT = 180 – 70 = 110 Angle ATC = 180 – 115 = 65
t
d
a


sin T sin D sin A
 TC 
65o
110o
TC
53.21

o
Sin 5
Sin 65o
C
70o
53.21 Sin 5
 5.1 m (1dp )
Sin 65o
5o
A
20o
25o
SOH CAH TOA
B
50 m
Cos 20o 
50
AC
 AC 
50
Cos 20o
 53.21 m (2dp )
Exam Type Questions
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