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The Sine Law
In the previous trigonometry units we have only been able to work with right-angle
triangles. Although right angle triangles are really useful triangles, there are lots of
other types of triangles besides those with right angles. We need a way to deal with
those. If we're working with a non-right angle triangle, that means there's no
hypotenuse, so we can't use Pythagorean theorem. Instead, we have two
mathematical tools called the Sine Law and the Cosine Law. Let's take a look at the
Sine Law first.
An oblique triangle is a triangle that does not have a right angle. The Sine Law (or
Law of Sines) can be used to solve oblique triangles when the following information is
given:
 Two angles and any side (AAS or ASA)
 Two sides and an angle opposite one of the given sides (SSA)
The Sine Law
If A, B, C are the angle measures of a triangle and a, b, c are the lengths of the
corresponding sides opposite these angles, then
a
b
c


sin A  sin 
B  sin 
C
Or
sin 
A
sin 
B  sin 
C


a
b
c
Proof
First Construct triangle ABC to be a non right triangle.
Draw a perpendicular from C to meet AB at E
C
b
A
h
E
Now writing h in terms of sin(A) and sin(B).
a
B
h
sin A 
b
b sin A h
h
sin B 
a
a sin 
B h
Setting the h equations equal to each other.
b sin A a sin 
B
b
a

sin 
B  sin 
A
Now repeat by drawing a perpendicular from A to CB, and writing h in terms of
sin(C) and sin(B).
h
sin 
C 
b
b sin 
C h
h
sin B 
c
c sin 
B h
Setting the h equations equal to each other.
b sin 
C c sin 
B
b
c

sin 
B
sin 
C
Therefore putting these together
a
b
c


.
sin 
A sin 
B  sin 
C
Example
Solve ABC , if
A 44.3
, a 11.5m, and b=77m. Round the side length to the
nearest tenth of a metre and the angles to the tenth of a degree.
Solution:
This is Sine Law problem (ASS)
First draw a diagram:
C
a=11.5
b=7.7
44.3
A
B
c
Pick the two terms that have 3 of the 4 pieces of given information, and solve for the
remaining term.
sin 
B  sin 
A

b
a
sin 
B  sin 
44.3 


7.7
11.5
sin 
B 0.4676345824
B 27.9
Solve for the remaining angle
C 180
44.3
27.9

107.8
Use sine law to find c
c
a

sin 
C  sin 
A
c
11.5

sin 
107.8 
 sin 44.3
c 15.7
Summarizing:
A
44.3
B
27.9
C
107.8
a
b
c
11.5m
77m
15.7m
Example
If P= 20, Q= 31
, p=210m, solve
PQR .
Q
31
r
210m
q
R
20
P
Solution:
First let’s find R
R 180

20
31

129
Sum of angles
in triangle
Now apply sine law to find r
r
p

sin 
R  sin 
P
r
210

sin 
129 sin 
20
210sin 
129
r
sin 
20
477.2
Now apply the Sine Law to find q
q
p

sin 
Q sin P 
q
210

sin 
31
 sin 20
210sin 
31

q
sin 
20 

316.2
Summarizing:
P
20
Q
31
R
129
p
q
r
210m
316.2m
477.2 m
Example
An 8 metre radio antenna is located on top of an office building. At a distance d from
the bottom of the building, the angle of elevation to the top of the antenna is 25,
and the angle of elevation to the bottom of the antenna is 15. Determine the height
of the office building.
B
C

A
1525
h
D
Solution:
From the diagram, determine angles
 and B
B 902565
251510
Apply the Sine Law to determine AC
BC
AC

sin 
 sin 
B
8
AC

sin 
10
 sin 65
8sin 
65

AC 
sin 
10
41.75375
Now apply primary trigonometric ratio to determine h
h
sin 
15

41.75375
h 
41.75375
sin 
15

10.80667
Therefore the height of the building is about 10.8m
Example
A pilot, flying over a straight road, measures the angles of depression from km
markers on opposite sides of the airplane: the angle of depression to one km
marker is 85
, while the angle of depression to a km marker 4 km from the first
marker is 6
. Determine the altitude of the plane to the nearest 100 metres.
A
85
6
b
c
B
h
85
6
D
4 km
C
Solution:
We need to determine either c for ABD or side b for ADC . This way we have
a right angle triangle to work with to determine h.
Let’s choose c, with
ABD .
We will use the Sine Law with ABC .
First we need
A : A 180

856

89
Now,
a
c

Sin 
A  sin 
C
4
c

sin 
89 sin 
6

4
sin 
6

c
sin 
89

0.41818 km
418.18m
Now we can use right triangle
1000 m = 1 km
ABD to determine h
h
sin 
B 
c
h
sin 
85

418.18
h 418.18 
sin 
85

416.59
To the nearest 100 m, the height is 400 m.
Example:
An observatory is 30 metres tall, and is situated on the top of a hill. From the
top of the observatory, the angle of depression to a restaurant at the base of the
hill is 46. From the base of the observatory the angle of depression to the
restaurant is 35. Determine the distance, to the nearest metre, from the
restaurant to the base of the observatory.
46
30 m
35 
Solution:
This is a job for the sine law, with x representing the required distance.
Always try to first find as
many angles as you can
in the triangle.
The angle opposite the longest side is 3590 125
The angle between the top of the observatory and the restaurant is 90
4644
The angle at the restaurant is
180
44125 
 11
Applying the sine law
30
x

sin 
11
 sin 44
30 
sin 
44
x
sin 
11

109.218
The distance from the restaurant to the observatory is about 109 m.
Example
A hot air balloon is kept at a constant altitude by two ropes anchored at two points on
the ground. One rope is 150 feet long and makes an angle of 68° with the ground.
How long is the second rope, to the nearest ten feet, if it makes an angle of 75° with
the ground?
150 ft
x ft
75°
68°
Solution:
From the diagram, this is a job for the Sine Law
x
150

sin 
68
 sin 75
150 
sin 
68 

x
sin 
75
143.98
The second rope is about 144 feet long.