Other slides in this set

Slide 2

Preview of page 2

Here's a taster:

The Cosine Rule
B
Pythagoras' Theorem allows us to
a calculate unknown lengths in
c right-angled triangles using the
relationship a2 = b2 + c2
A b C
It would be very useful to be able to calculate unknown
sides for any value of the angle at A. Consider the square
on the side opposite A when angle A is not a right-angle.
1 2 3
2
a2 a
a2
A A
A Angle A
acute
Angle A
obtuse
a2 = b 2 + c 2 a2 > b 2 + c 2 a2 < b 2 + c 2…read more

Slide 3

Preview of page 3

Here's a taster:

The Cosine Rule
1
The Cosine Rule generalises Pythagoras' Theorem and
takes care of the 3 possible cases for Angle A.
A
Deriving the rule Consider a general triangle ABC. We
require a in terms of b, c and A.
B a2 = b 2 + c 2
BP2 = a2 ­ (b ­ x)2
Also: BP2 = c2 ­ x2 2
c a a2 ­ (b ­ x)2 = c2 ­ x2
a2 ­ (b2 ­ 2bx + x2) = c2 ­ x2 A
P a2 ­ b2 + 2bx ­ x2 = c2 ­ x2
A x b b-x C 2 2 2 a2 > b 2 + c 2
a = b + c ­ 2bx*
b
a2 = b2 + c2 ­ 2bcCosA 3
Draw BP perpendicular to AC
*Since Cos A = x/c x = cCosA
A
o 2 2 2
When A = 90 , CosA = 0 and reduces to a = b + c 1 Pythagoras
When A > 90o, CosA is negative, a2 > b2 + c2 Pythagoras + a bit
2
a2 < b 2 + c 2
When A < 90o, CosA is positive, a2 > b2 + c2 3 Pythagoras - a bit…read more

Slide 4

Preview of page 4

Here's a taster:

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 c a
Applying the same method as
A b C
earlier to the other sides
produce similar formulae for b2 = a2 + c2 ­ 2acCosB
b and c. namely:
c2 = a2 + b2 ­ 2abCosC…read more

Slide 5

Preview of page 5

Here's a taster:

The Cosine Rule a2 = b2 + c2 ­ 2bcCosA
To find an unknown side we need 2 sides and the
included angle.
1. Not to 2.
7.7 cm 65o
a 9.6 cm scale 5.4 cm
40o
8 cm m
2 2 2 o m2 = 5.42 + 7.72 ­ 2 x 5.4 x 7.7 x Cos 65o
a = 8 + 9.6 ­ 2 x 8 x 9.6 x Cos 40
2 2 o m = (5.42 + 7.72 ­ 2 x 5.4 x 7.7 x Cos 65o)
a = (8 + 9.6 ­ 2 x 8 x 9.6 x Cos 40 )
m = 7.3 cm (1 dp)
a = 6.2 cm (1 dp)
3. 15o 100 m p2 = 852 + 1002 ­ 2 x 85 x 100 x Cos 15o
85 m p = (852 + 1002 ­ 2 x 85 x 100 x Cos 15o)
p = 28.4 m (1 dp)
p…read more

Slide 6

Preview of page 6

Here's a taster:

The Cosine Rule a2 = b2 + c2 ­ 2bcCosA
Application Problem
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 onto a bearing of 035o and
sails to a lighthouse (L) 24 miles away. It then returns to harbour.
(a) Make a sketch of the journey
(b) Find the total distance travelled by the boat. (nearest mile)
HL2 = 402 + 242 ­ 2 x 40 x 24 x Cos 1250
HL = (402 + 242 ­ 2 x 40 x 24 x Cos 1250)
L
= 57 miles
Total distance = 57 + 64 = 121 miles.
24 miles
H
40 miles 125o
B…read more

Slide 7

Preview of page 7
Preview of page 7

Slide 8

Preview of page 8
Preview of page 8

Slide 9

Preview of page 9
Preview of page 9

Slide 10

Preview of page 10
Preview of page 10

Comments

daviesg


Lots of illustrated applications of the use of the Sine Rule

EllieDudley

Good job babes

niyi0706

davie more u need a shavey 

Similar Mathematics resources:

See all Mathematics resources »See all resources »