Transcript Right Triangle Trigonometry

Laws of Sines and Cosines
Sections 6.1 and 6.2
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.
The formulas listed below will allow us to more easily deal with
triangles that are not right triangles.
• Law of sines
• Law of cosines
• Heron’s formula
Formulas
• Law of sines
sin  sin  sin 


a
b
c
• Law of cosines
a 2  b 2  c 2  2bc cos   or b 2  a 2  c 2  2ac cos   or c 2  a 2  b 2  2ab cos  
• Heron’s formula
1
A
P (P  2a )(P  2b )(P  2c )
4
a, b, and c are the lengths of the sides of the triangle
P is the perimeter of the triangle
Law of Sines
sin  sin  sin 


a
b
c
Law of Cosines
a 2  b 2  c 2  2bc cos  
b 2  a 2  c 2  2ac cos  
c  a  b  2ab cos  
2
2
2
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
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:
angle ACB  180  52  70
angle ACB  58
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
Use the Law of Sines to find the
value of the side x.
sin 58 sin 52

26.7
x
x sin( 58)  26.7 sin( 52)
26.7 sin( 52)
sin( 58)
x  24.8097805
x 
Use the Law of Cosines to find
the value of the side x.
x
a 2  b 2  c 2  2bc cos  
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 α be the
39 degree angle. Sides b and c are the lengths 21 and 42.
continued on next slide
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.
x 2  212  422  2(21)(42) cos 39
x 2  441  1764  1764 cos 39
x 2  2205  1764 cos 39
x 2  834.114524
x   834.114524
x  28.88104091
Since length is positive, x is approximately 28.88104097
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
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
26mph*3hr = 78 miles
ship 1
140°
34mph*3hr = 102 miles
x
ship 2
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 α will be 140°.
a 2  b 2  c 2  2bc cos  
continued on next slide
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
26mph*3hr = 78 miles
140°
34mph*3hr = 102 miles
x
ship 1
ship 2
x 2  782  1022  2(78)(102) cos 140 
x 2  6084  10404  15912 cos 140 
x  16488  15912 cos 140 
2
x 2  28677.29918
x   28677.29918
x  169.3437309
Since distance is positive, the
ships are approximately
169.3437309 miles apart after
3 hours.
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
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
48
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?
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.
sin  sin  sin 


a
b
c
continued on next slide
48
sin  sin  sin 


a
b
c
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?
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 α.
The length of side AB (distance between the two stations) and be
represented by b. This would make the angle ACB the angle β. 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
48
sin  sin  sin 


a
b
c
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?
angle CAB = 180°-87°=93°.
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°-93°=3°.
Now we are ready to plug everything we know into the law of sines.
continued on next slide
48
sin  sin  sin 


a
b
c
sin 84 sin 3

AC
48
48 sin( 84)  AC sin(3)
48 sin( 84)
 AC
sin(3)
AC  912.1272334
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?
Thus we have answered the first
question. The satellite is
approximately 912.1272334 miles
away from tracking station A.
continued on next slide
48
sin  sin  sin 


a
b
c
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?
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
x
48
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?
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
x
sin( 87) 
912.1272334
912.1272334 sin( 87)  x
x  910.8771948
Thus the satellite is
approximately 910.8771948
miles above the ground.
continued on next slide
x
48
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?
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
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.
1
A
P (P  2a )(P  2b )(P  2c )
4
Heron’s Formula
continued on next slide
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.
Heron’s Formula
1
A
P (P  2a )(P  2b )(P  2c )
4
In Heron’s 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 680+320+802 = 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 Heron’s Formula.
1
A
1802(1802  2(680))(1802  2(320))(1802  2(802))
4
continued on next slide
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.
Heron’s Formula
1
1802(1802  2(680))(1802  2(320))(1802  2(802))
4
1

1802(1802  1360)(1802  640)(1802  1604)
4
1

1802( 442)(1162)(198)
4
1

183251852800
4
1
 ( 428079.2599)
4
continued on next slide
 107019.815
A
A
A
A
A
A
1
A
P (P  2a )(P  2b )(P  2c )
4
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.
Heron’s Formula
1
A
P (P  2a )(P  2b )(P  2c )
4
A  107019.815 square feet
We now the area in square feet that we have calculated to the area in acres.
A  107019.815 square feet 
A
1 acre
43560 square feet
107019.815 acre
43560
A  2.456836891 acres
continued on next slide
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.
Heron’s Formula
1
A
P (P  2a )(P  2b )(P  2c )
4
Now that we have the number of acres that the triangular parcel covers, we
can calculate the value by multiplying by 2100.
value  2.456836891 2100
value  $5159.36
Note that the value is rounded to dollars and cents.