A triangle has two angles that measure and . The length of the side between these two angles is 10. What are the lengths of the other two sides?

### Guidance

Consider the non right triangle below. We can construct an altitude from any one of the vertices to divide the triangle into two right triangles as show below.

from the diagram we can write two trigonometric functions involving :

Since both are equal to , we can set them equal to each other to get:

and finally divide both sides by to create the proportion:

If we construct the altitude from a different vertex, say , we would get the proportion: . Now, the transitive property allows us to conclude that . We can put them all together as the Law of Sines: . In the examples that follow we will use the Law of Sines to solve triangles.

#### Example A

Solve the triangle.

**
Solution:
**
Since we are given two of the three angles in the triangle, we can find the third angle using the fact that the three angles must add up to
. So,
. Now we can substitute the known values into the Law of Sines proportion as shown:

Taking two ratios at a time we can solve the proportions to find and using cross multiplication.

To find :

To find :

This particular triangle is an example in which we are given two angles and the non-included side or AAS (also SAA).

#### Example B

Solve the triangle.

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Solution:
**
In this example we are given two angles and a side as well but the side is between the angles. We refer to this arrangement as ASA. In practice, in doesn’t really matter whether we are given AAS or ASA. We will follow the same procedure as Example A. First, find the third angle:
.

Second, write out the appropriate proportions to solve for the unknown sides, and .

To find :

To find :

Notice that and . This illustrates a property of isosceles triangles that states that the base angles (the angles opposite the congruent sides) are also congruent.

#### Example C

Three fishing ships in a fleet are out on the ocean. The Chester is 32 km from the Angela. An officer on the Chester measures the angle between the Angela and the Beverly to be . An officer on the Beverly measures the angle between the Angela and the Chester to be . How far apart, to the nearest kilometer are the Chester and the Beverly?

**
Solution:
**
First, draw a picture. Keep in mind that when we say that an officer on one of the ships is measuring an angle, the angle she is measuring is at the vertex where her ship is located.

Now that we have a picture, we can determine the angle at the Angela and then use the Law of Sines to find the distance between the Beverly and the Chester.

The angle at the Angela is .

Now find ,

The Beverly and the Chester are about 27 km apart.

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Concept Problem Revisit
**
The measure of the triangle's third angle is

### Guided Practice

Solve the triangles.

1.

2.

3. A surveying team is measuring the distance between point on one side of a river and point on the far side of the river. One surveyor is positioned at point and the second surveyor is positioned at point , 65 m up the riverbank from point . The surveyor at point measures the angle between points and to be . The surveyor at point measures the angle between points and to be . Find the distance between points and .

#### Answers

1.

2.

3.

### Vocabulary

- Law of Sines
- For any triangle, the ratio of the sine of an angle over its opposite side is equal to the sine of any other angle in the triangle over its opposite side.

### Practice

Solve the triangles. Round your answers to the nearest tenth.

Using the given information, solve .

- .

- .

- .

- .

Use the Law of Sines to solve the following world problems.

- A surveyor is trying to find the distance across a ravine. He measures the angle between a spot on the far side of the ravine, , and a spot 200 ft away on his side of the ravine, , to be . He then walks to the angle between and his previous location to be . How wide is the ravine?
- A triangular plot of land has angles and . The side opposite the angle is 35 m long. How much fencing, to the nearest half meter, is required to enclose the entire plot of land?