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```Law of Sines and Cosines
The entire trigonometry that has been done thus far has dealt with the indirect measurement of
sides and angles of right triangles. The calculations were carried out using Pythagorean Theorem
and/or Trigonometric Ratios.
hyp 2  s1 2  s 2 2
sin  
opp
hyp
cos  
hyp
tan  
opp
This process can now be expanded to include oblique triangles. An oblique triangle is one that
has no right angle. The three angles can be acute (all angles less than 900) or one angle can be
obtuse (greater than 900) and the remaining two angles acute. Regardless of what scenario exists,
the measure of unknown angles and the lengths of unknown sides can be determined by using
either the Law of Sines or the Law of Cosines. Let us look first at the Law of Cosines.
The Law of Cosines can be used to find the length of an unknown side if we know the length of
two sides of the triangle and the measure of the angle they form. This angle is called the included
angle. Knowing two sides and the included angle is referred to as SAS. Another way to refer to
this is a “V”. The lines of the letter “V” represent the sides and the vertex of the “V” is the
included angle. Students can use their fingers to determine if these measurements are known.
Example 1:
Given ABC with b = 9cm, c = 12cm, and  A = 620, calculate the length of
side ‘a’.
Solution 1:
When the triangle has been sketched, it is obvious that SAS does exist.
To determine the length of side ‘a’ the following formula is used:
a  b 2  c 2  2bcCosA
a
92  122  2912(Cos62 0 )
a  123.5941424
a  11.1cm.
Notice that the lower case letters are used to name the sides of the triangle. These lower case
letters, used for the sides, are those of the corresponding angles. Once the values have been
substituted into the formula, the calculations can be done on the calculator by typing in exactly
what is written under the radical sign.
Example 2:
Given ABC with a = 15m, c = 28m and  B = 1120, calculate the length of
side ‘b’.
Solution 2:
When the triangle has been sketched, it is obvious that SAS does exist.
To determine the length of side ‘b’ the following formula is used:
b  a 2  c 2  2acCosB
b
152  282  21528Cos112 0 
b  1323.669538
b  36.4cm
Notice that the letter of the side that is being calculated only appears once in the formula and that
is for its corresponding angle. The remainder of the formula consists of the letters of the other
two sides – the sum of their squares minus twice their product times the cosine of the
included angle. If this highlighted statement is repeated when the formula is used in class, the
students will adapt it when using the formula themselves.
The Law of Cosines is also used to determine the measure of an unknown angle when the lengths
of all the sides of the triangle are known. This is referred to as SSS. Students should be reminded
of the fact that the longest side of a right triangle, the hypotenuse, is located opposite the right
angle which is the largest angle of the triangle. The same will follow for an oblique triangle –
longest side opposite largest angle and shortest side opposite smallest angle. This formula for
determining the measure of an angle is the result of solving a 2  b 2  c 2  2bcCosA in terms of
CosA.
Example 3: Given ABC with a = 18 cm, b = 21 cm, and c = 27 cm, calculate the
measure of the smallest angle (nearest tenth)
Solution 3:
The shortest side is side ‘a’ which makes  A the
smallest angle. This is the angle we must find.
b2  c2  a2
CosA 
2bc
2
2
(21)  27 2  18
CosA 
22127 
846
CosA 
1134
CosA  .7640
 
cos 1 CosA  cos 1 .7640 
A  41.8 0
Notice that the answer was determined by working with the formula as a fraction – result
of the numerator over result of the denominator. The other option is to use brackets
(numerator) / (denominator). However, students must be very careful when using the
brackets since failure to open and close all brackets will result in an error. To err on the
side of caution, the first option will not result in an error.
Example 4: Given ABC with a = 118 cm, b = 55 cm, and c = 65 cm, calculate the
measure of  A (nearest tenth).
Solution 4:
The triangle was sketched as being
an obtuse triangle because side ‘a’
was much longer than the other two
sides.
b2  c2  a2
CosA 
2bc
2
2
2
55  65  118
CosA 
25565
 6674
CosA 
7150
CosA  .9334
 
cos 1 CosA  cos 1  .9334
A  159 0
CosA is a negative value. This indicates that the angle is greater than 900 since the cosine
function is negative in the second quadrant. 90 0  A  180 0
Now that we have seen the Law of Cosines in action, it is time to explore the Law of Sines. In
order to use the Law of Sines to determine the length of a side, we must know the measure of an
angle and its corresponding side plus the measure of the corresponding angle of the unknown
side. To use the Law of Sines to determine the measure of an angle, we must know the measure
of an angle and its corresponding side plus the length of the corresponding side of the unknown
angle. In other words, we must have ASA or SSA. The Law of Sines uses ratios of the length of
a side to its corresponding angle.
a
b
c


sin A sin B sin C
OR
sin A sin B sin C


a
b
c
To calculate the measure of an angle or the length of a side, one ratio will be the ‘known’
pair and the other will consist of the other known side or angle.
Example 5:
Given ABC with A  42 0 , a = 52 cm, and C  73 0 cm, calculate the
length of side ‘c’.
Solution 5:
a
c

sin A sin C
52
c

0
sin 42
sin 73 0
52
c

.6691 .9563
.6691.9563 52  .6691.9563 c
.6691
.9563
.956352  .6691c
.956352  c
.6691
74.3cm.  c
OR
a
c

sin A sin C
52
c

0
sin 42
sin 73 0

52  sin 73 0
c
sin 42 0
c  74.3cm.



Solving the equation in terms of the variable ‘c’.

Once the equation has been solved for ‘c’, the calculations can be done on the calculator.
Not all students will be able to perform this task efficiently so the first method, although more
cumbersome, may be the format that many students will follow.
Example 6:
Given ABC with B  39 0 , b = 5cm, and c  7.6cm. , calculate the
measure of C (nearest tenth).
Solution 6:
b
c

sin B sin C
5
7 .6

0
sin C
sin 39
7.6 sin 39 0
sin C 
5
sin C  .9565


sin 1 sin C   sin 1 .9565
C  73.10
The answer given on the calculator for arcsin .9565 is 73.10.
In the above example, the only measurement that was required was the measure of C .
Therefore, two solutions for the angle had no bearing on the rest of the triangle. However, if we
were solving the triangle, two solutions would have a direct impact on the triangle. To solve a
triangle means to determine the measurements for all the sides and angles. Remember that the
longest side of a triangle is opposite the largest angle.
Ambiguous Case for the Law of Sines
There is another possible answer to this question and that is the co-terminal angle of 106.90. The
sine function is positive in the first and second quadrant, but calculators are designed to display
the first angle as a result. To determine if the second angle is a possible solution, add 390 and
106.90. The sum is less than 1800, so both results are possible and acceptable solutions. If the
sum of the two angles is greater than 1800, then the larger angle is not an acceptable value. This
is referred to as the ambiguous case for the Law of Sines.
Example 1:
Given ABC with B  34 0 , b = 15cm, and c  20cm. , solve the triangle.
Solution 1:
b
c

sin B sin C
15
20

0
sin C
sin 34
20 sin 34 0
sin C 
15
sin C  .7455


sin 1 sin C   sin 1 .7455
C  48.2 0 or 131.8 0
If C = 48.20, then A = 97.80. This means that  A is the largest angle of the triangle and side
‘a’ is the longest side. However, if C = 131.80, then  A = 14.20. In this case  A is the
smallest angle of the triangle and then side ‘a’ is the shortest side. Therefore, the length of side
‘a’ must be calculated for each case.
b
a

sin B sin A
15
a

0
sin 34
sin 97.8 0
15 sin 97.8 0
a
sin 34 0
a  26.58cm.


OR
b
a

sin B sin A
15
a

0
sin 34
sin 14.2 0
15 sin 14.2 0
a
sin 34 0
a  6.58cm.


The two possible values for C creates two solutions for the measurements of the triangle
when it is solved. The two solutions are shown below.
A  97.8 0
A  14.2 0
B  34 0
B  34 0
C  48.2 0
a  26.58cm.
OR
C  131.8 0
a  6.58cm.
b  15cm.
b  15cm.
c  20cm.
c  20cm.
Both solutions must be shown as the solution.
Sometimes both the Law of Sines and the Law of Cosines can be used to determine the
measure of an angle or the length of a side. Students will use the one that they are more
proficient using.
Example 2: Given ABC with A  46 0 , b  32cm. , and c  53cm. , find the measure of
C (nearest tenth).
Solution 2:
Before the measure of C can be found, the length of side ‘a’
must be determined.
a  b 2  c 2  2bcCosA
a
322  532  23253 cos46 0 
a  1476.718807
a  38.43cm.
Now that the length of side ‘a’ has been determined, the measure of C can be found. The
lengths of the three sides are known so the Law of Cosines may be used or the Law of Sines may
be used. The choice is simply an individual preference.
a2  b2  c2
2ab
2
2
38.43 2  32   53
CosC 
238.4332 
 308.1351
CosC 
2459.52
CosC  .1252
CosC 


cos 1 CosC   cos 1  .1252 
C  97.2 0
Exercises:
1. For each of the following triangles, determine the length of the indicated side or the
measure of the indicated angle.(nearest tenth).
a) Find ‘b’.
b)
Find B
c) Find ‘a’.
d) Find A
Solutions:
a)
a
b

sin A sin B
17
b

0
sin 20.7
sin 118.3 0
17  sin 118.30
b
sin 20.7 0
b  42.35 cm.


b)
a2  c2  b2
2ac
2
7   82  52
CosB 
27 8
88
CosB 
112
CosB  .7857
CosB 
B  38.2 0
c) Find ‘a’.
a  b 2  c 2  2bcCosA
a
6.7 2  8.42  26.7 8.4cos75.2 0 
a  86.6970
a  9.31 cm.
d) Find A
a
b

sin A sin B
9.2
6.4

sin A sin 410
9.2 sin 410
sin A 
6.4
sin A  .9430


sin 1 sin A  sin 1 .9430
A  70.6 0
The Law of Sines and the Law of Cosines are also used to solve real-world problems that
can be represented by an oblique triangle.
Example 1:
An eight metre telephone pole has a very bad lean and creates an angle greater than 90
with the ground. A guide wire, 14 m long, is attached to the pole for support so the pole
will not fall down. The guide wire is anchored in the ground at a point 10m from the base
of the pole. Calculate the angle that the pole makes with the ground.
Solution 1:
a2  c2  b2
2ac
2
2
2

8  10   14 
Cos 
2810 
 32
Cos 
160
Cos 
Cos  0.2
cos 1 Cos   cos 1  0.2 
  101.5 0
The pole makes an angle of 101.50 with the ground.
Example 2:
A spider crawling down a wall spots its prey, a moth, on the ground at an angle of 160 with the
wall. After crawling downward 16 cm, the moth still hasn’t moved, but now the angle with the
wall is 280. How far is the moth from the wall?
Solution 2:
Before the distance CD can be calculated, some measurements
must be determined.
ACD  180 0  (90 0  28 0 )
ACB  74 0  62 0
ACD  62 0
ACB  12 0

BCD  180 0  90 0  16 0
BCD  74

0
The measurements of ABC will be used to calculate the length of ‘d’ ( AC ).
ABC
b
c

sin B sin C
b
16

0
sin 16
sin 12 0

16 sin 16 0
b
sin 12 0
b  21.21cm.


CD
AC
x
sin 28 0 
21.21
x
.4694 
21.21
21.21.4694 
The moth is 9.96cm. from the wall.
x
21.21
21.21
x  9.96cm
Exercises:
1. Josh, Mary and Evan are playing frisbee in the school field. Their current positions
form a triangle with the angle at Josh equal to 44o and the angle at Mary equal to 21o.
If Mary is 15 metres from Evan, how far apart are Josh and Mary?
2. While exploring the woods at the end of Bengal Road in Mira, two of Glace Bay’s
policemen, spotted a fire in the distance. From where they were standing, they
estimated an angle of elevation of 150 to the top of the tower. Moving 10 m closer to
the tower, they now estimate the angle of elevation to be 180. How high is the tower?
Solutions:
1.


The measure of E is 180 0  44 0  210  1150
e
j

sin E sin J
e
15

0
sin 115
sin 44 0

15 sin 115 0
e
sin 44 0
e  19.57 m

The distance between Josh and Mary is 19.57m

2.
Calculate the height of the tower BD.
BCA  180 0  18 0
BCA  162 0

ABC  180 0  162 0  15 0
ABC  3
0

ABC
a
b

sin A sin B
a
10

0
sin 15
sin 30
10 sin 15 0
a
sin 3 0
a  49.45m.

BCD
BD
sin C 
BC
x
sin 18 0 
49.45
x
.3090 
49.45

49.45.3090 
x
49.45
49.45
x  15.28m
The height of the tower is 15.28 m.
Area of an Oblique Triangle
The method that we will use to determine the area of an oblique triangle requires knowing the
measurement of two sides of the triangle and the included angle (SAS).
The formula used to calculate the area of ABC if b and c are the known sides and A
is the included angle is:
1
Area  bc sin A
2
Example 1: Determine the area of the following triangle:
Solution 1:
1
ac sin B
2
1
Area  14.113.6 sin 410
2
Area  62.90cm 2
Area 
 
Example 2: Determine the area of the following triangle:
The lengths of the three sides are known
but the measure of an included angle
must be determined. Choose one angle
and use the Law of Cosines to find its
measure.
Solution 2:
b2  c2  a2
cos A 
2bc
8.42  8.12  4.82
cos A 
28.4 8.1
cos A  .8313
cos 1 cos A  cos 1 .8313
1
bc sin A
2
1
Area  8.4 8.1sin 33.8 0
2
Area  18.93cm 2
Area 


A  33.8 0
If another angle were chosen, there may be a slight variation in the area of the triangle.
This is the result of rounding values for the function(s) and any measurements that were
calculated.
Exercises:
1. Your backyard is a great spot to build a skating rink this winter. The ground must be
covered with sand that costs \$16 per square metre. If the sides of the triangular plot of
land measure 8.24m., 7.67m. and 8.13m., what will be the cost of the sand needed to
cover this area?
2. Josh, Mary and Evan are playing frisbee in the school field. Their current positions
form a triangle with the angle at Josh equal to 44o and the angle at Mary equal to 21o.
If Mary is 15 metres from Evan, what is the area of the triangular plot formed by
Josh, Mary and Evan?
Solutions:
1.
b2  c2  a2
2bc
2
2
2

8.13  7.67   8.24 
cos A 
28.137.67 
cos A  .4572
cos A 
cos 1 cos A  cos 1 .4572 
A  62.8 0
1
bc sin A
2
1
Area  8.137.67 sin 62.8 0
2
Area  27.73m 2
Area 


The cost of the sand is (28m 2 )(\$16 / m 2 )  \$448.00
Area  28m 2
2. In a previous exercise, we solved this problem to determine the distance between
Mary and Josh. This distance was 19.57m.
1
ej sin M
2
1
Area  19.57 15 sin 210
2
Area  52.6m 2
Area 
 
```
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