Transcript Trigonometry 2 (Cosine Rule)

The Cosine Law
The Cosine Rule generalizes 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 = a2 – (b – x)2
Also: BP2 = c2 – x2
a
c
 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
The Cosine Law
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
C
b2 = a2 + c2 – 2acCosB
c2 = a2 + b2 – 2abCosC
The Cosine Law
a2 = b2 + c2 – 2bcCosA
To find an unknown side we need 2 sides and the
included angle.
1.
Not to
scale
9.6 cm
a
2.
7.7 cm
40o
a2 = 82 + 9.62 – 2 x 8 x 9.6 x Cos 40o
a=
+
9.62
– 2 x 8 x 9.6 x Cos
a = 6.2 cm (1 dp)
3.
15o
5.4 cm
m
8 cm
(82
65o
100 m
40o)
m2 = 5.42 + 7.72 – 2 x 5.4 x 7.7 x Cos 65o
m = (5.42 + 7.72 – 2 x 5.4 x 7.7 x Cos 65o)
m = 7.3 cm (1 dp)
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
p = 28.4 m (1 dp)
The Cosine Law
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 055o 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.
H
24 miles
40 miles
125o
B
The Cosine Law
To find unknown angles the 3 formula for sides need to be
re-arranged in terms of CosA, B or C.
a2 = b2 + c2 – 2bcCosA
a  b  c  2bcCosA
2
2
2
 2bcCosA  b  c  a
2
2
b2 = a2 + c2 – 2acCosB
2
c2 = a2 + b2 – 2abCosC
b2 c 2 a2
CosA 
2bc
B
Similarly
a2 c 2  b2
CosB 
2ac
CosC 
a  b c
2ab
2
a
c
2
2
A
b
C
b2 c 2 a2
CosA 
2bc
The Cosine Law
To find an unknown angle we need 3 given sides.
1.
9.6 cm
6.2
Not to
scale
2.
7.7 cm
P
5.4 cm
A
7.3 cm
8 cm
82  9.62  6.22
CosA 
2x 8x 9.6
5.42  7.72  7.32
CosP 
2x 5.4 x 7.7
P  65o
A  40o
3.
R
100 m
1002  852  28.42
CosR 
2x 100x 85
85 m
28.4 m
R  15o
The Cosine Law
Application Problems
b2 c 2 a2
CosA 
2bc
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