Objective: Be able to use the Law of Sines to
find unknowns in triangles!
Homework:
Lesson 12.3/1-10, 12-14, 19, 20
Quiz Friday – 12.1 – 12.3
Quick Review:
What does Soh-Cah-Toa stand for?
adj
opp
cos 
sin 
hyp
hyp
opp
tan 
adj
What kind of triangles do we use this for?
right triangles
What if it’s not a right triangle?
GASP!! What do we do then??
B


sin A sin B
sin C


a
b
c

A
Note:
 capital letters always
angles
stand for __________!
 lower-case letters
always stand for
sides
________!
c
a
b
C
Use the Law of Sines ONLY when:
 you DON’T have a right triangle AND
 you know an angle and its opposite side
For any triangle (right, acute or obtuse), you may
use the following formula to solve for missing
sides or angles:
Can also be written in this form:
a
b
c


sin A sin B sin C
If you have 3 dimensions of a triangle and you need to find
the other 3 dimensions - they cannot be just ANY 3
dimensions though, or you won’t have enough info to solve
the Law of Sines equation.
Use the Law of Sines if you are given:


AAS - 2 angles and 1 adjacent side
ASA - 2 angles and their included side
Ex. 1: Use the Law of Sines to find each missing angle or
side. Round any decimal answers to the nearest tenth.
63°
29
A
sin A sin B  sin C 


a
b
c
a
79˚
38˚
42
C
sin 63 sin C 

42
29
29 sin 63  42 sin C
29 sin 63
 sin C 
42
.6152 ...  sin C 
38.0  C
sin 63 sin 79

42
a
a sin 63  42 sin 79
42 sin 79
a
sin 63
a  46.3
Ex. 2: Use the Law of Sines to find each missing angle or
side. Round any decimal answers to the nearest tenth.
T


sin A sin B
sin C


a
b
c

s
40°
sin 51 sin 40

4.8
r
51˚ r
89°
4.8
sin 51 sin 89

4.8
s
r sin 51  4.8 sin 40
4.8 sin 40
r
sin 51
s sin 51  4.8 sin 89
4.8 sin 89
s
sin 51
r  4.0
s  6.2
Ex. 3: Draw ΔABC and mark it with the given information.
Solve the triangle. Round any decimal answers to the
nearest tenth.
B
a.
a  7, mA  37, mB  76
sin A sin B  sin C 


a
b
c
sin 37 sin 76

7
b
C  67
c
A
76˚
37˚
7
67˚
b
sin 37 sin 67 

7
c
b sin 37  7 sin 76 c sin 37  7 sin 67
7 sin 67
7 sin 76
c
b
sin 37
sin 37
b  11.3
c  10.7
C
b.
a  12, mA  70, c  3.1
B
sin A sin B  sin C 


a
b
c

sin 70
sin C

12
3.1
A
.2427 ...  sin C 
96˚
70˚
12
14˚ C
b

3.1sin 70  12 sin C
3.1sin 70
 sin C 
12
14.0  C
3.1
sin 70 sin 96

12
b
B  96
b sin 70  12 sin 96
12 sin 96
b
sin 70
b  12 .7
Ex. 3: Draw ΔABC and mark it with the given information.
Solve the triangle. Round any decimal answers to the
nearest tenth.
You are given a triangle, ABC, with angle A = 70°,
angle B = 80° and side a = 12 cm. Find the
measures of angle C and sides b and c.
* In this section, angles are named with capital
letters and the side opposite an angle is named
with the same lower case letter .*
Ex. 3: con’t
B
The angles in a ∆ total 180°, so
angle C = 30°.
80°
Set up the Law of Sines to find
side b:
a = 12
c
A 70°
b
C
12
b

sin 70 sin 80
12 sin 80  b sin 70
12 sin 80
b
 12.6cm
sin 70
Ex. 3: con’t
Set up the Law of Sines to
find side c:
B
80°
c
A 70°
a = 12
b = 12.6
30°
12
c

sin 70 sin 30
C 12 sin 30  c  sin 70
12 sin 30
c
 6.4cm
sin 70
Ex. 3: Solution
Angle C = 30°
B
Side b = 12.6 cm
80°
Side c = 6.4 cm
a = 12
Note:
A 70°
b = 12.6
30°
C
We used the given values of
A and a in both calculations.
Your answer is more
accurate if you do not used
rounded values in
calculations.
Ex. 4: Draw ΔABC and mark it with the given information.
Solve the triangle. Round any decimal answers to the
nearest tenth.
You are given a triangle, ABC, with angle C = 115°,
angle B = 30° and side a = 30 cm. Find the
measures of angle A and sides b and c.
B
30°
c
a = 30
115°
C
b
A
Ex. 4: con’t
To solve for the missing sides or
angles, we must have an angle and
opposite side to set up the first
equation.
B
30°
c
We MUST find angle A first
because the only side given is side
a.
a = 30
115°
C
b
A
The angles in a ∆ total 180°, so
angle A = 35°.
Ex. 4: con’t
Set up the Law of Sines to find side b:
30
b

sin 35 sin 30
B
30°
30 sin 30  b sin 35
30 sin 30
b
 26.2cm
sin 35
c
a = 30
115° 35°
C
b
A
Ex. 4: con’t
Set up the Law of Sines to find side c:
B
30
c

sin 35 sin 115
30°
c
a = 30
115° 35°
C b = 26.2
A
30 sin 115  c  sin 35
30 sin 115
c
 47.4cm
sin 35
Ex. 4: Solution
Angle A = 35°
Side b = 26.2 cm
B
Side c = 47.4 cm
30°
c = 47.4
a = 30
115° 35°
C b = 26.2
A
Note: Use the Law of Sines
whenever you are given 2
angles and one side!
a
b
c


sin A sin B sin C


sin A sin B
sin C


a
b
c
Use the Law of Sines to find
the missing dimensions of a
triangle when given any
combination of these
dimensions.

AAS
 ASA

A 46-foot telephone pole tilted at an angle of from
the vertical casts a shadow on the ground. Find the
length of the shadow to the nearest foot when the
angle of elevation to the sun is
Draw a diagram Draw
Then find the
Since you know the measures of two angles of the
triangle,
and the length of a side
opposite one of the angles
you
can use the Law of Sines to find the length of the shadow.
Law of Sines
Cross products
Divide each side by sin
Use a calculator.
Answer: The length of the shadow is about
75.9 feet.
A 5-foot fishing pole is anchored to the edge of a
dock. If the distance from the foot of the pole to the
point where the fishing line meets the water is 45 feet,
about how much fishing line that is cast out is above
the surface of the water?
Answer: About 42 feet of the fishing line that is cast out
is above the surface of the water.
A road slopes 15 above the horizontal, and a vertical telephone pole
stands beside the road. The angle of elevation of the Sun is 65 , and
the pole casts a 15 foot shadow downhill along the road. Find the height
of the pole.
Let x  the height of the pole.
A
65º
BAC  180  90  65  25
ACB  65  15  50
x
B
sin 25 sin 50

15º
15
x
C
15sin 50
x
 27.2
sin 25
The height of the pole is about 27.2 feet.
o
o
o
o
o
o
o
We can find the area of a triangle if we are given any two
sides of a triangle and the measure of the included angle.
(SAS)
1
Area = (side)(side)(sine of included angle)
2
Using two sides and an Angle.
1
Area  bc  SinA
2
1
Area  ab  SinC
2
1
Area  ac  SinB
2