You know how to use trigonometric ratios to find missing sides in right triangles, but what about non-right triangles? For the triangle below, can you find \begin{align*}AB\end{align*}?

#### Guidance

Look at the triangle below. Based on the angles, can you tell which side is the shortest? Which side is the longest?

The smallest angle is \begin{align*}\angle A\end{align*}. It opens up to create the shortest side, \begin{align*}\overline{BC}\end{align*}. The largest angle is \begin{align*}\angle C\end{align*}. It opens up to create the longest side, \begin{align*}\overline{AB}\end{align*}. Clearly, angles and opposite sides within triangles are connected. In fact, the Law of Sines states that the ratio between the sine of an angle and the side opposite the angle is **constant** for each of the three angle/side pairs within a triangle.

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

In the past, you used the sine ratio to find missing sides and angles in *right* triangles. With the Law of Sines, you can find missing sides and angles in *any type* of triangle. In the examples you will learn how to prove the Law of Sines.

**Example A**

Consider \begin{align*}\Delta ABC\end{align*} below. Draw an altitude from vertex \begin{align*}B\end{align*} that intersects \begin{align*}\overline{AC}\end{align*} to divide \begin{align*}\Delta ABC\end{align*} into two right triangles. Find two equations for the length of the altitude.

**Solution:** Below, the altitude has been drawn and labeled \begin{align*}h\end{align*}.

Consider the right triangle with hypotenuse of length \begin{align*}c\end{align*}. \begin{align*}h\end{align*} is opposite \begin{align*}\angle A\end{align*} in this triangle. With an opposite side and hypotenuse you can use the sine ratio.

\begin{align*}\sin A &= \frac{h}{c}\\ h &= c \sin A\end{align*}

Now consider the right triangle with hypotenuse of length \begin{align*}a\end{align*}. \begin{align*}h\end{align*} is opposite \begin{align*}\angle C\end{align*} in this triangle. With an opposite side and hypotenuse you can use the sine ratio.

\begin{align*}\sin C &= \frac{h}{a}\\ h &= a \sin C\end{align*}

You have now found two equations for \begin{align*}h\end{align*}.

**Example B**

Use your work from Example A to prove the Law of Sines.

**Solution:** From Example A, \begin{align*}h=c \sin A\end{align*} and \begin{align*}h=a \sin C\end{align*}. This means that \begin{align*}c \sin A=a \sin C\end{align*}. Divide both sides by \begin{align*}ac\end{align*} and you will see the Law of Sines.

\begin{align*}c \sin A &= a \sin C\\ \frac{c \sin A}{ac} &= \frac{a \sin C}{ac}\\ \frac{\sin A}{a} &= \frac{\sin C}{c}\end{align*}

\begin{align*}A\end{align*} and \begin{align*}C\end{align*} were two random angles in the original triangle. This proves that in general, within any triangle the ratio of the sine of an angle to its opposite side is constant.

**Example C**

Use the Law of Sines to find \begin{align*}BC\end{align*} and \begin{align*}AC\end{align*}.

**Solution:** Look for an angle and side pair whose measurements are both given. \begin{align*}m \angle C=64^\circ\end{align*}, and its opposite side is \begin{align*}AB=12\end{align*}. Next, set up an equation using the two known measurements as one of the ratios in the Law of Sines. Make sure to match angles with opposite sides.

Solve for \begin{align*}BC\end{align*}:

\begin{align*}\frac{\sin 64^\circ}{12} &= \frac{\sin 38^\circ}{BC}\\ BC &= \frac{12 \sin 38^\circ}{\sin 64^\circ}\\ BC & \approx 8.22\end{align*}

Solve for \begin{align*}AC\end{align*}:

\begin{align*}\frac{\sin 64^\circ}{12} &= \frac{\sin 78^\circ}{AC}\\ AC &= \frac{12 \sin 78^\circ}{\sin 64^\circ}\\ AC & \approx 13.06\end{align*}

**Concept Problem Revisited**

To find \begin{align*}AB\end{align*}, you can use the Law of Sines. Match angles with opposite sides.

\begin{align*}\frac{\sin 27^\circ}{9} &= \frac{\sin 72^\circ}{AB}\\ AB &= \frac{9 \sin 72^\circ}{\sin 27^\circ}\\ AB & \approx 18.85\end{align*}

#### Vocabulary

The ** sine (sin)** of an angle within a right triangle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

The ** trigonometric ratios** are sine, cosine, and tangent.

** Trigonometry** is the study of triangles.

\begin{align*}\theta\end{align*}, or ** “theta”,** is a Greek letter. In geometry, it is often used as a variable to represent an unknown angle measure.

The ** Law of Sines** states that the ratio between the sine of an angle and the side opposite the angle is the

*same*for each of the three angle/side pairs within a triangle.

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

#### Guided Practice

1. The triangle below is drawn to scale. Use the Law of Sines to solve for the measure of angle \begin{align*}B\end{align*}. Does your answer seem correct?

2. Recall that SSA was not a criterion for triangle congruence. This was because two non-congruent triangles could have the same “side-side-angle” pattern. What does this have to do with the seemingly wrong answer to #2?

3. The triangle below is drawn to scale. Use the Law of Sines to solve for the measure of angle C.

**Answers:**

1. From looking at the triangle, \begin{align*}\angle B\end{align*} looks to be an obtuse angle. Match angles with opposite sides.

\begin{align*}\frac{\sin 36^\circ}{14} &= \frac{\sin B}{22}\\ \sin B &= \frac{22 \sin 36^\circ}{14}\\ m \angle B &= \sin^{-1} \left(\frac{22 \sin 36^\circ}{14}\right)\\ m \angle B & \approx 67.5^\circ\end{align*}

This answer doesn't seem correct because the angle appears to be obtuse (greater than \begin{align*}90^\circ\end{align*}).

2. There are two triangles that fit the criteria given in #2 (angle measures have been slightly rounded).

The two possible measures of \begin{align*}\angle B\end{align*} are supplementary \begin{align*}(112^\circ+68^\circ=180^\circ)\end{align*}. Also note that \begin{align*}\sin 112^\circ=0.927=\sin 68^\circ\end{align*}. The inverse sine function on your calculator will only produce angles between \begin{align*}0^\circ\end{align*} and \begin{align*}90^\circ\end{align*}. In a sense, the calculator was imagining the triangle on the right while you were imagining the triangle on the left. **This is a problem when using the Law of Sines to solve for missing angles**. If the information you are given fits SSA, it is possible that there are two answers. If your picture is drawn to scale, you can determine whether your answer should be the acute angle or the obtuse angle by looking at the picture.

3. Angle C appears to be obtuse. Match angles with opposite sides.

\begin{align*}\frac{\sin 28^\circ}{17} &= \frac{\sin C}{33}\\ \sin C &= \frac{33 \sin 28^\circ}{17}\\ m \angle C &= \sin^{-1} \left(\frac{33 \sin 28^\circ}{17}\right)\\ m \angle C & \approx 65.7^\circ\end{align*}

This is the acute version of the answer. You know the angle should be obtuse, so find the angle supplementary to \begin{align*}65.7^\circ\end{align*}:

\begin{align*}180^\circ-65.7^\circ=114.3^\circ\end{align*}

*Note that \begin{align*}\sin 65.7^\circ=\sin 114.3\end{align*}*

**The final answer is: \begin{align*}m \angle C=114.3^\circ\end{align*}.**

#### Practice

For each triangle, find the measure of all missing sides and angles.

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9. What does SSA have to do with the Law of Sines? What type of problems do you have to think extra carefully about to make sure you have the correct answer?

10. The triangle below is drawn to scale. Determine the measure of each of the missing angles.

11. The triangle below is drawn to scale. Determine the measure of each of the missing angles.

12. The triangle below is drawn to scale. Determine the measure of each of the missing angles.

13. Use the picture below to derive the Law of Sines.

14. What type of problems can you solve with the Law of Sines?

15. Explain why you *cannot* use the Law of Sines to solve for \begin{align*}x\end{align*} in the triangle below.