Lesson 6.1- Law of
Sines
Provided by Vivian Skumpija and Amy Gimpel
When to use the Law of Sines
• To solve an oblique triangle
• Given:
– Two angles and any side (AAS or ASA)
– Two sides and an angle opposite one of them (SSA or ASS)
***Always remember to keep your calculator in
degree mode ***
AAS
• If C=102.3 °, B=28.7°, b=27.4
• A=180-102.3-28.7  49°
Since you have an angle measurement and its opposite side you
can set up the equation…
a
27.4

 43.06
sin 49 sin 28.7
c
27.4

 57.75
sin 102.3 sin 28.7
a= 43.06
c=57.75
NOTE: To solve for c, you
always go back to your
original equation to avoid
using rounded values.
ASS (Single Solution)
B
51.85 °
775.6
989.19
• If b= 795.1, c=775.6, B= 51.85°
795.1
775.6

 50.09
sin 51.85 sin c
A  180  51.85  50.09  78.06
795.1
a

 989.19
sin 51.85 sin 78.06
A
78.06 °
50.09 °
795.1
NOTE: To solve for sin c, you
must use the answer with
inverse sine (on calculator:
2nd Sin, 2nd ANS)
Since side b is greater than the other given side,
there is only one solution
C
ASS (No Solution)
• a=7, b=15, A=98°
7
15

 DOMAIN ERROR
sin 98 sin b
This triangle can not be solved for any answers,
which would make it NO SOLUTION.
ASS (Two Solutions)
• First: solve for the regular triangle
B
c=12.4
a=10
10
52 ©
A
77.7 ©
77.7 ©
C
C '  180  77.7  102.3
10
b

 5.5
sin 52 sin 25.7
b=9.8
10
12.4

 77.7 
sin 52 sin C
B  180  52  77.7  50.3
10
b

 9.8
sin 52 sin 50.3

Because the side
across from the
given angle is
smaller than the
other side, there is
another set of
solutions…
B’
25.7 ©
c’=12.4
A’
a’=10
52© 102.3 ©
b’=5.5
C’
Finding the Area
1
1
1
Area  bc sin A  ab sin C  ac sin B
2
2
2
• Since sides a and b are given and so is angle C…
b=52
102°
a=90
1
9052sin 102  2289sqmeters
2
Lesson 6.2- Law of
Cosines
When to use the Law of
Cosines
• To solve an oblique triangle
• Given:
– Three sides (SSS)
– Two sides and their included angle (SAS)
***Always remember to keep your calculator in
degree mode ***
• a=3, b=5, c=7
SSS
pick the largest
7 2  32  5 2  235 cos c
49  9  25  30 cos c
49  34  30 cos c
15  30 cos c
solve for the rest of the triangle
 30
7
3



21
.
79
1
sin 120 sin a
  cos c
2
1
1 
cos     C
7
5
2

 38.21


sin 120 sin b

C  120
SAS
• a=4.2, c=7.5, B=32°
Because you have angle B, you want to solve for side b…
b 2  4.2 2  7.52  24.27.5 cos 32
b 2  73.89  53.43
b 2  20.46
b 2  20.46
b  4.52
…Then you solve for the rest using the Law of Sines
4.52
4.2

 29.5
sin 32 sin a
Heron’s Area Formula
s( s  a)( s  b)( s  c)
whereS  a  b  c  / 2
If a=43m, b=53 m, c=72 m
43  53  72 / 2  84
Substitute for S
84(84  43)(84  53)(84  72)
84(41)(31)(12)  1131.89meters
2
THAT’S ALL
THERE IS TO IT