# 5.8: Angle-Side-Angle Triangles

**At Grade**Created by: CK-12

**Practice**Angle-Side-Angle Triangles

You're eating lunch in the cafeteria one afternoon while working on your math homework. Lately you seem to notice the triangular shapes in everything. At home, at school, with your friends. It seems like triangles are everywhere. And you find yourself trying to apply what you are learning in math class to all of the triangles around you. And today is no exception. As you start to take a bite of your chip, you suddenly recognize that familiar shape - the triangle.

You estimate the length of one of the sides of the chip to be 3 cm. You also can tell that the angle adjacent to the 3 cm side is \begin{align*}50^\circ\end{align*}

Read on, and at the end of this Concept, you'll be able to solve this problem.

### Watch This

Law of Sines - Solving ASA Triangle

### Guidance

The Law of Sines states: \begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}

One case where we use the Law of Sines is when we know two angles in a triangle and the **included** side (ASA). For instance, in \begin{align*}\triangle{TRI}\end{align*}

\begin{align*}\angle{T},\angle{R}\end{align*}

\begin{align*}\angle{T},\angle{I}\end{align*}

\begin{align*}\angle{R},\angle{I}\end{align*}

In this case, the Law of Sines allows us to find either of the non-included sides.

#### Example A

In the triangle above, \begin{align*}\triangle{TRI},\angle{T} = 83^\circ, \angle{R} = 24^\circ\end{align*}

**Solution:** Since we know two angles and the included side, we can find either of the non-included sides using the Law of Sines. Since we already know two of the angles in the triangle, we can find the third angle using the fact that the sum of all of the angles in a triangle must equal \begin{align*}180^\circ\end{align*}

\begin{align*}\angle{I} & = 180 - (83 + 24) \\
\angle{I} & = 180 - 107 \\
\angle{I} & = 73^\circ\end{align*}

Now that we know \begin{align*}\angle{I} = 73^\circ\end{align*}

\begin{align*}\frac{\sin 73}{18.5} & = \frac{\sin 83}{t} \\
t (\sin 73) & = 18.5 (\sin 83) \\
t & = \frac{18.5(\sin 83)}{\sin 73} \\
t & \approx 19.2\end{align*}

Notice how we wait until the last step to input the values into the calculator. This is so our answer is as accurate as possible.

#### Example B

In order to avoid a large and dangerous snowstorm on a flight from Chicago to Buffalo, pilot John starts out \begin{align*}27^\circ\end{align*}

- What is the total distance of the modified flight path?
- How much further did he travel than if he had stayed on course?

**Solution, Part 1:** In order to find the total distance of the modified flight path, we need to know side \begin{align*}x\end{align*}

\begin{align*}Missing Angle & = 180 - (27 + 88) = 65^\circ && \text{The sum of angles in a triangle is}\ 180 \\
\frac{\sin 65}{412} & = \frac{\sin 27}{x} && \text{Law of Sines} \\
x (\sin 65) & = 412 (\sin 27) && \text{Cross multiply} \\
x & = \frac{412(\sin 27)}{\sin 65} && \text{Divide by sin}\ 65 \\
x & \approx 206.4\ miles\end{align*}

The total distance of the modified flight path is \begin{align*}412 + 206.4 = 618.4\ miles\end{align*}

**Solution, Part 2:** To find how much farther John had to travel, we need to know the distance of the original flight path, \begin{align*}y\end{align*}

\begin{align*}\frac{\sin 65}{412} & = \frac{\sin 88}{y} && \text{Law of Sines} \\
y (\sin 65) & = 412 (\sin 88) && \text{Cross multiply} \\
y & = \frac{412 (\sin 88)}{\sin 65} && \text{Divide by}\ \sin 65 \\
y & \approx 454.3\ miles\end{align*}

John had to travel \begin{align*}618.4 - 454.3 = 164.1\ miles\end{align*}

#### Example C

In the triangle shown here:

The sides given are \begin{align*}\angle{A} = 14^\circ\end{align*}

**Solution:** Since we know two angles and the included side, we can find either of the non-included sides using the Law of Sines. Since we already know two of the angles in the triangle, we can find the third angle using the fact that the sum of all of the angles in a triangle must equal \begin{align*}180^\circ\end{align*}

\begin{align*}\angle{C} & = 180 - (18 + 14) \\
\angle{C} & = 180 - 32 \\
\angle{C} & = 148^\circ\end{align*}

Use the Law of Sines to find the length of side "a":

\begin{align*}\frac{\sin 14}{a} & = \frac{\sin 148}{11} \\
11 (\sin 14) & = a (\sin 148) \\
a & = \frac{11(\sin 14)}{\sin 148} \\
a & \approx 5.02\end{align*}

### Vocabulary

**Angle Side Angle Triangle:** An ** angle side angle triangle** is a triangle where two angles and the length of the side in between them are known quantities.

### Guided Practice

1. Find side "a" in the triangle below using the following information: \begin{align*}b = 16, A = 11.7^\circ, C = 23.8^\circ\end{align*}

2. Find side "a" in the triangle below using the following information: \begin{align*}k = 6.3, J = 16.2^\circ, L = 40.3^\circ\end{align*}

3. Even though ASA and AAS triangles represent two different cases of the Law of Sines, what do they both have in common?

**Solutions:**

1. \begin{align*}\frac{\sin 11.7^\circ}{a} = \frac{\sin 144.5^\circ}{16}, a = 5.6\end{align*}

2. \begin{align*}\frac{\sin 40.3^\circ}{l} = \frac{\sin 123.5^\circ}{6.3}, l = 4.9\end{align*}

3. Student answers will vary but they should notice that in both cases you know or can find an angle and the side across from it.

### Concept Problem Solution

You can use the Law of Sines to find the length of either of the other 2 sides. However, first it is good to note that since the sum of the interior angles of a triangle must equal \begin{align*}180^\circ\end{align*}

Now to set up the ratios:

\begin{align*}\frac{\sin 60}{a} & = \frac{\sin 70}{3} \\
3 (\sin 60) & = a (\sin 70) \\
a & = \frac{3(\sin 60)}{\sin 70} \\
a & \approx 2.7647\end{align*}

The length of one of the other 2 sides is approximately 2.7647 centimeters.

To find the length of the last side:

\begin{align*}\frac{\sin 50}{b} & = \frac{\sin 70}{3} \\
3 (\sin 50) & = b (\sin 70) \\
b & = \frac{3(\sin 50)}{\sin 70} \\
b & \approx 2.46\end{align*}

The length of the other one of the 2 unknown sides is approximately 2.46 centimeters.

### Practice

In \begin{align*}\triangle ABC\end{align*}

- Find \begin{align*}m\angle C\end{align*}
m∠C . - Find the length of a.
- Find the length of b.

In \begin{align*}\triangle DEF\end{align*}

- Find \begin{align*}m\angle F\end{align*}.
- Find the length of d.
- Find the length of e.

In \begin{align*}\triangle BIG\end{align*}, \begin{align*}m\angle B=56^\circ\end{align*}, \begin{align*}m\angle I=71^\circ\end{align*}, and g=23.

- Find \begin{align*}m\angle G\end{align*}.
- Find the length of b.
- Find the length of i.

In \begin{align*}\triangle APL\end{align*}, \begin{align*}m\angle A=79^\circ\end{align*}, \begin{align*}m\angle P=40^\circ\end{align*}, and l=15.

- Find \begin{align*}m\angle L\end{align*}.
- Find the length of a.
- Find the length of p.

In \begin{align*}\triangle SAU\end{align*}, \begin{align*}m\angle S=5^\circ\end{align*}, \begin{align*}m\angle A=99^\circ\end{align*}, and u=21.

- Find \begin{align*}m\angle U\end{align*}.
- Find the length of s.
- Find the length of a.

### Image Attributions

Here you'll learn to find an unknown side of a triangle using the Law of Sines when two angles and the length of the side between them are known.