A triangle has two sides of lengths 2 (a) and 5 (b). The non-included angle (B) of the triangle measures . What are possible measures for the two other angles and the remaining side?

### Guidance

Recall that the sine ratios for an angle and its supplement will always be equal. In other words, . In Geometry you learned that two triangles could not be proven congruent using SSA and you investigated cases in which there could be two triangles. In Example A, we will explore how the Law of Sines can be used to find two possible triangles when given two side lengths of a triangle and a non-included angle.

#### Example A

Given with , , and , solve for the other angle and side measures.

**
Solution:
**
First, let’s make a diagram to show the relationship between the given sides and angles. Then we can set up a proportion to solve for angle
:

From here we can find , since the three angles must add up to . We can also find the third side using another Law of Sines ratio:

Putting these measures in the triangle, we get:

But, we know that so when we solved for we only got one of the two possible angles. The other angle will be . Next we need to determine the measure of angle for and the length of the third side in this second possible triangle. The sum of the three angles must still be , so . Now set up a proportion to solve for the third side just as before:

The second triangle would look like this:

In this instance there were two possible triangles.

#### Example B

Given with , and , solve for the other angle and side measures.

**
Solution:
**
Again we will start with a diagram and use the law of sines proportion to find a second angle measure in the triangle.

Now find the third angle, and solve for the third side:

Because we used the inverse sine function to determine the measure of angle , the angle could be the supplement of or so we need to check for a second triangle. If we let and then attempt to find the third angle, we will find that the sum of the two angles we have is greater than and thus no triangle can be formed.

This example shows that two triangles are not always possible. Note that if the given angle is obtuse, there will only be one possible triangle for this reason.

### More Guidance

In both examples we simply tested to see if there would be a second triangle. There are, however, guidelines to follow to determine when a second triangle exists and when it does not. The “check and see” method always works and therefore it is not necessary to memorize the following table. It is interesting, however, to see to pictures and make the connection between the inequalities and what if any triangle can be formed.

First, consider when is obtuse:

If
, then
**
one
**
triangle can be formed.

If
, then
**
no
**
triangle can be formed.

Now, consider the possible scenarios when is acute.

If
, the
**
one
**
triangle can be formed.

For the following cases, where , keep in mind that we would be using the proportion:

and that

If
,
**
no
**
triangle can be formed because
.

If
,
**
one right
**
triangle can be formed because
.

If
(and
),
**
two
**
triangles can be formed because
.

#### Example C

Given with , and , use the rules to determine how many, if any, triangles can be formed and then solve the possible triangle(s).

**
Solution:
**
In this case,
is acute and
, so we need to look at the value of
. Since
, there will be two triangles. To solve for these triangles, use the Law of Sines extended proportion instead of making a diagram. Plugging in what we know, we have:

Take the first and last ratios to solve a proportion to find the measure of angle .

So, the or and or respectively.

Solve for the measure of side in each triangle:

Putting it all together, we have:

Triangle 1:

Triangle 2:

**
Concept Problem Revisit
**

Now find the third angle, and solve for the third side:

### Guided Practice

1. Use the given side lengths and angle measure to determine whether zero, one or two triangles exists.

a. .

b. .

c. .

2. Solve the following triangles.

a.

b.

3. Given , and , find .

#### Answers

1. a. Since is obtuse and , no triangle can be formed.

b. Since is acute, and , two triangles can be formed.

c. Since is acute and , there is one possible triangle.

2. a. There will be two triangles in this case because is acute, and .

Using the extended proportion: , we get:

b. Since is acute and , there is one possible triangle.

Using the extended proportion: , we get:

3. In this instance is acute, and so two triangles can be formed. So, once we find the two possible measures of angle , we will find the two possible measures of angle . First find :

Now that we have , use the triangle sum to find .

### Practice

For problems 1-5, use the rules to determine if there will be one, two or no possible triangle with the given measurements.

Solve the following triangles, if possible. If there is a second possible triangle, solve it as well.