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# laws of sines and cosines - PowerPoint PPT Presentation

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.

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### Laws of Sines and Cosines

Sections 6.1 and 6.2

• 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 with triangles that are not right triangles.

• Law of sines

• Law of cosines

• Heron’s formula

or

or

a, b, and c are the lengths of the sides of the triangle

P is the perimeter of the triangle

Law of Sines with triangles that are not right triangles.

Law of Cosines with triangles that are not right triangles.

Use the Law of Sines to find the value of the side with triangles that are not right triangles.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 with triangles that are not right triangles.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

Use the Law of Sines to find the value of the side with triangles that are not right triangles.x.

Use the Law of Cosines to find the value of the side with triangles that are not right triangles.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 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 with triangles that are not right triangles.x.

x

Now we plug into the law of cosine formula to find x.

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

harbor 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

140°

140°

34mph*3hr = 102 miles

34mph*3hr = 102 miles

26mph*3hr = 78 miles

26mph*3hr = 78 miles

x

x

ship 2

ship 2

ship 1

ship 1

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°

34mph*3hr = 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 α will be 140°.

continued on next slide

harbor 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

140°

140°

34mph*3hr = 102 miles

34mph*3hr = 102 miles

26mph*3hr = 78 miles

26mph*3hr = 78 miles

x

x

ship 2

ship 2

ship 1

ship 1

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°

34mph*3hr = 102 miles

x

ship 2

ship 1

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

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

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 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

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°-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

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

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

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

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

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

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

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.

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

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

We now the area in square feet that we have calculated to the area in 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

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.