Transcript 6.2

Digital Lesson
Law of Cosines
2014
An oblique triangle is a triangle that has no right angles.
C
a
b
A
c
B
To solve an oblique triangle, you need to know the
measure of at least one side and the measures of any
other two parts of the triangle – two sides, two angles,
or one angle and one side.
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2
The following cases are considered when solving oblique triangles.
1. Two angles and any side (AAS or ASA)
A
A
c
c
B
C
2. Two sides and an angle opposite one of them (SSA)
c
C
3. Three sides (SSS)
b
a
c
a
4. Two sides and their included angle (SAS)
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c
B
a
3
The last two cases (SSS and SAS) can be solved using the
Law of Cosines.
(The first two cases can be solved using the Law of Sines.)
Law of Cosines
Standard Form
Alternative Form
a  b  c  2bc cos A
2
2
2
b

c

a
cos A 
2bc
b  a  c  2ac cos B
2
2
2
a

c

b
cos B 
2ac
c  a  b  2ab cos C
2
2
2
a

b

c
cos C 
2ab
2
2
2
2
2
2
2
2
2
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Example:
Find the three angles of the triangle.
2
2
2
a

b

c
cos C 
2ab
2
2
2
8

6

12

2(8)(6)
 64  36  144
96
 44
96
C  117.3
Law of Sines:
C
117.3
8
6
36.3
A
12
 6
sin 117.3 sin B
26.4
B
12
Find the angle
opposite the longest
side first.
B  26.4
A  180 117.3  26.4  36.3
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5
C
Example:
Solve the triangle.
9.9
Law of Cosines:
b 2  a 2  c 2  2ac cos B
A
67.8
6.2
75
37.2
9.5
B
 (6.2)2  (9.5)2  2(6.2)(9.5) cos 75
 38.44  90.25  (117.8)(0.25882)
 98.20
b  9.9
Law of Sines: 9.9
 6.2
sin 75 sin A
A  37.2
C  180  75  37.2  67.8
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Example:
29.69
Find the area of the triangle.
10
8
5
1
A  10  8  sin 29.69
2
Area  19.8 square units
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You Try:
The pitcher’s mound on a women’s softball field is 43 feet
from home plate and the distance between bases is 60 feet, as
shown. (The pitcher’s mound is not halfway between home
plate and second base)
How far is the pitcher’s
mound from first base?
42.43 ft.
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9
A ship travels 60 miles due east, then adjusts its course northward
as shown in the figure. After traveling 80 miles in that direction, the
ship is 139 miles from its point of departure. Describe the bearing
from point B to point C.
N
76.150
802  602  1392
cos B 
2  80  60 
B  166.150
166.150  900  76.15
N 76.15 E
A ship leaves port at noon and heads due west at 20 knots, or 20
nautical miles (nm) per hour. At 2 pm the ship changes course to
N 540 W. Find the ship’s bearing and distance from the port of
departure at 3 pm.
c 2  a 2 b 2 2ab  cosC
c 2  402 202 2(40)(20)  cos144 0
N
A
20nm
b
c  57.4 nm
N
c
540
1440
C
78.180
20nm
40nm 2
a
N 78.18 W
0
57.4 nm
B
57.4
c
C0
sinsin
144
20
b

sin B
20  sin144 0  57.4  sinB
57.4
B  11.820
57.4
Homework
Pg. 405 1-9 odd, 19-31 odd
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