Laws of Sines and Cosines

1
Laws of Sines and Cosines

• Sections 6.1 and 6.2

2
Objectives

• Apply the law of sines to determine the lengths
  of side and measures of angle of a triangle.
• Solve word problems requiring the law of sines.
• Apply the law of cosines to determine the lengths
  of side and measures of angle of a triangle.
• Solve word problems requiring the law of cosines.
  
• Solve a word problem requiring Heron's formula.

3
The formulas listed below will allow us to more
easily deal with triangles that are not right
triangles.

• Law of sines
• Law of cosines
• Herons formula

4
Formulas

• Law of sines
• Law of cosines
• Herons formula

or
or
a, b, and c are the lengths of the sides of the
triangle P is the perimeter of the triangle
5
Law of Sines
6
Law of Cosines
7
Use the Law of Sines to find the value of the
side x.
We are told to use the law of sines to find x.
In order to use the law of sines, we need to have
the lengths of two sides and the measures of the
angle opposite those sides. In this case we have
one side and the side we are looking for. We
have the measure of the angle opposite the side
we are looking for, but are missing the measure
of the angle opposite the side we have.
continued on next slide
8
Use the Law of Sines to find the value of the
side x.
Since we have the measures of two of the three
angles, we can use the fact that the sum of the
measures of the angles of a triangle add up to
180 degrees. This will give us
Now that we have the measure of the angle
opposite the side AB, we can apply the law of
sines to find the value of x.
continued on next slide
9
Use the Law of Sines to find the value of the
side x.
10
Use the Law of Cosines to find the value of the
side x.
x
In order to use the law of cosines, we need the
lengths of two sides and the measure of the angle
between them. We have that here. We can let
side a be x and angle a be the 39 degree angle.
Sides b and c are the lengths 21 and 42.
continued on next slide
11
Use the Law of Cosines to find the value of the
side x.
x
Now we plug into the law of cosine formula to
find x.
Since length is positive, x is approximately
28.88104097
12
Two ships leave a harbor at the same time,
traveling on courses that have an angle of 140
degrees between them.  If the first ship travels
at 26 miles per hour and the second ship travels
at 34 miles per hour, how far apart are the two
ships after 3 hours?
For this problem, the first thing that we should
do is draw a picture. Once we have the picture,
we may be able to see which formula we can use to
solve the problem.
continued on next slide
13
Two ships leave a harbor at the same time,
traveling on courses that have an angle of 140
degrees between them.  If the first ship travels
at 26 miles per hour and the second ship travels
at 34 miles per hour, how far apart are the two
ships after 3 hours?
harbor
harbor
140
34mph3hr 102 miles
x
ship 2
ship 1
Looking at the labeled picture above, we can see
that the have the lengths of two sides and the
measure of the angle between them. We are
looking for the length of the third side of the
triangle. In order to find this, we will need
the law of cosines. x will be side a. Sides b
and c will be 78 and 102. Angle a will be 140.
continued on next slide
14
Two ships leave a harbor at the same time,
traveling on courses that have an angle of 140
degrees between them.  If the first ship travels
at 26 miles per hour and the second ship travels
at 34 miles per hour, how far apart are the two
ships after 3 hours?
harbor
harbor
140
34mph3hr 102 miles
x
ship 2
ship 1
Since distance is positive, the ships are
approximately 169.3437309 miles apart after 3
hours.
15
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
We could draw a picture for this problem or we
can just use the picture on the next slide.
continued on next slide
16
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
Based on the information in the problem, we know
that the side AB of the triangle is 48 miles.
continued on next slide
17
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
48
Our next step is to determine if we should use
the law of sines or the law of cosines to find
the distance from the satellite to station A
(side AC). In the triangle ABC, we know the
measure of one angle and the length of one side.
Using a little geometric knowledge, we can find
the measures of the other two angles of the
triangle. This means that we would know the
measures of all three angles and the measure of
one side. With this information, we should use
the law of sines to find the length of side AC.
continued on next slide
18
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
48
The length of side AC (the distance between the
satellite and the station A) can be represented
by a. This would make the angle ABC the angle a.
The length of side AB (distance between the two
stations) and be represented by b. This would
make the angle ACB the angle ß. We need to find
the measure of angle ACB.
The 87 angle and the angle CAB are supplementary
angles. This means that the sum of measures is
180. Thus we can find out the measure of angle
CAB.
continued on next slide
19
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
48
angle CAB 180-8793.
Now the three angles of the triangle add up to
180. We know that two are the angles are 93
and 84, thus we can find the third angle.
angle ACB 180-84-933.
Now we are ready to plug everything we know into
the law of sines.
continued on next slide
20
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
48
Thus we have answered the first question. The
satellite is approximately 912.1272334 miles away
from tracking station A.
continued on next slide
21
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
48
Now we need to answer the second question. How
far the satellite is above the ground is shown by
the red line drawn on the picture that is at a
right angle with the ground.
We could use the law of sines to find the length
of the red line, but there is an easier way. The
red line is part of a right triangle (made up of
the red and green lines). In the first part of
the problem, we found the length of the
hypotenuse of the right triangle. We know that
one of the angles is 87. The red line is
opposite the 87 angle.
continued on next slide
22
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
x
48
This means that we can use our basic
trigonometric functions for right triangles using
the opposite side, the hypotenuse and angle 87.
The sine function uses the opposite side and
hypotenuse. Thus we have
Thus the satellite is approximately 910.8771948
miles above the ground.
continued on next slide
23
The path of a satellite orbiting the earth causes
it to pass directly over two tracking stations A
and B, which are 48 miles apart.  When the
satellite is on one side of the two stations, the
angles of elevation at A and B are measured to be
87 degrees and 84 degrees.  How far is the
satellite from station A?  How far is the
satellite above the ground?
x
48
Just a note The second part of the problem used
the information calculated in the first part.
Generally, I try to avoid this in case I make a
mistake. If there is an error in part 1, the any
further answers using part 1 would also be
incorrect. In this case I chose to use part 1
since that was the simplest way to answer the
question in part 2. A more complicated method
to find the height of the satellite above the
ground can be found in the balloon example in the
Right Triangle Trigonometry power point with
solutions starting on slide 28
24
A triangular parcel of land has sides of length
680 feet, 320 feet, and 802 feet.  What is the
area of the parcel of land?  If land is valued at
2100 per acre (1 acre is 43560 square feet),
what is the value of the parcel of land.
Herons Formula
continued on next slide
25
A triangular parcel of land has sides of length
680 feet, 320 feet, and 802 feet.  What is the
area of the parcel of land?  If land is valued at
2100 per acre (1 acre is 43560 square feet),
what is the value of the parcel of land.
Herons Formula
In Herons Formula the a, b, and c are the
lengths of the sides of the triangle. The P is
the perimeter (sum of the lengths of the sides)
of the triangle. For our triangle P is
680320802 1802. For this question, we need
to first find the area of the triangular parcel.
Once we have the area in square feet, we can
convert it to acres and then calculate the value.
We will start by plugging into Herons Formula.
continued on next slide
26
A triangular parcel of land has sides of length
680 feet, 320 feet, and 802 feet.  What is the
area of the parcel of land?  If land is valued at
2100 per acre (1 acre is 43560 square feet),
what is the value of the parcel of land.
Herons Formula
continued on next slide
27
A triangular parcel of land has sides of length
680 feet, 320 feet, and 802 feet.  What is the
area of the parcel of land?  If land is valued at
2100 per acre (1 acre is 43560 square feet),
what is the value of the parcel of land.
Herons Formula
We now the area in square feet that we have
calculated to the area in acres.
continued on next slide
28
A triangular parcel of land has sides of length
680 feet, 320 feet, and 802 feet.  What is the
area of the parcel of land?  If land is valued at
2100 per acre (1 acre is 43560 square feet),
what is the value of the parcel of land.
Herons Formula
Now that we have the number of acres that the
triangular parcel covers, we can calculate the
value by multiplying by 2100.
Note that the value is rounded to dollars and
cents.