You have previously learned that the area of a triangle is \begin{align*}A=\frac{1}{2}bh\end{align*} where \begin{align*}b\end{align*} is a base of the triangle and \begin{align*}h\end{align*} is the corresponding height. Why is it helpful to have another formula for calculating the area of a triangle?

#### Watch This

http://www.youtube.com/watch?v=WEOVjRMh_vI James Sousa: Determine the Area of a Triangle Using the Sine Function

#### Guidance

One way to find the area of a triangle is by calculating \begin{align*}\frac{1}{2}bh\end{align*}. This formula works when you know or can determine the base and the height of a triangle. What if you wanted to find the area of the following triangle:

The two given sides are *not* the base and the height. In the examples you will derive a formula for calculating the area of a triangle given this type of information.

**Example A**

Below, the altitude of \begin{align*}\Delta ABC\end{align*} from \begin{align*}A\end{align*} has been drawn. Write an equation that relates \begin{align*}h, 34^\circ\end{align*} and 7.

**Solution:** Notice that two right triangles have been formed. This means you can use the trigonometric ratios to relate the sides.** **\begin{align*}h\end{align*} is opposite the \begin{align*}34^\circ\end{align*} angle and 7 is the length of the hypotenuse of the right triangle. This is the sine relationship.

\begin{align*}\sin 34^\circ=\frac{h}{7}\end{align*}

**Example B**

Solve the equation from Example A for \begin{align*}h\end{align*}. Then, use \begin{align*}A=\frac{1}{2}bh\end{align*} to find the area of the triangle using your value for \begin{align*}h\end{align*}. Can you generalize the formula?

**Solution:** If \begin{align*}\sin 34^\circ=\frac{h}{7}\end{align*} then \begin{align*}h=7 \sin 34^\circ\end{align*}. For the height that has been drawn, \begin{align*}\overline{BC}\end{align*} is the base.

\begin{align*}A &= \frac{1}{2}bh\\ A &= \frac{1}{2} (12)(7 \sin 34^\circ)\\ A &= \frac{1}{2} (12)(7) (\sin 34^\circ)\\ A & \approx 23.45 \ un^2\end{align*}

Notice the calculation: \begin{align*}A =\frac{1}{2} (12) (7) (\sin 34^\circ)\end{align*}. You found half the product of the two sides and the sine of the included angle. Consider the general triangle below:

If you know \begin{align*}\angle B\end{align*}, the area will be \begin{align*}Area = \frac{1}{2} (a)(c) \sin B\end{align*}. If you know \begin{align*}\angle C\end{align*}, the area will be \begin{align*}Area=\frac{1}{2} (a)(b) \sin C\end{align*}.

**Example C**

Previously, you have only used the sine ratio on acute angles. Can you find the area of a triangle when given an obtuse angle? Consider the triangle below. Find an equation to determine the value of \begin{align*}h\end{align*}. Then, find the area of the triangle.

**Solution:** First find the measure of the exterior angle of the triangle at vertex \begin{align*}B\end{align*}. Remember that the interior and exterior angles at vertex \begin{align*}B\end{align*} must sum to \begin{align*}180^\circ\end{align*}.

Now, consider the right triangle with \begin{align*}\overline{AB}\end{align*} as the hypotenuse. \begin{align*}h\end{align*} is the side opposite the \begin{align*}55^\circ\end{align*} angle and 7 is the hypotenuse. Once again, this is the sine relationship.

\begin{align*}\sin 55^\circ &= \frac{h}{7}\\ \rightarrow h &= 7 \sin 55^\circ\end{align*}

For the height that has been drawn, \begin{align*}\overline{BC}\end{align*} is the base.

\begin{align*}A &= \frac{1}{2} bh\\ A &= \frac{1}{2} (12)(7 \sin 55^\circ)\\ A &= \frac{1}{2} (12) (7)(\sin 55^\circ)\\ A & \approx 34.4 \ un^2\end{align*}

In general, if your given angle is obtuse, you can use the angle supplementary to your given angle in your area calculation.

**Concept Problem Revisited**

You have previously learned that the area of a triangle is \begin{align*}A=\frac{1}{2}bh\end{align*} where \begin{align*}b\end{align*} is a base of the triangle and \begin{align*}h\end{align*} is the corresponding height. Why is it helpful to have another formula for calculating the area of a triangle?

Often you only know the sides and angles of a triangle and not the height. \begin{align*}A=\frac{1}{2} (a)(b) \sin C\end{align*} allows you to quickly find the area of acute or obtuse triangles when given two sides and an included angle.

#### 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.

#### Guided Practice

Find the area of each triangle.

1.

2.

The **sine ratio** is actually a **function** that takes *any* angle measure as an input (not just angles between \begin{align*}0^\circ\end{align*} and \begin{align*}90^\circ\end{align*}). One property of the sine function is that the sine of supplementary angles will always be equal. Therefore, when finding the area of a triangle given an obtuse angle, you can use the obtuse angle in your calculation instead of the acute supplementary angle and your answer will be the same. *Note: You will study the sine function in much more detail in future courses!*

3. Find the area of the triangle from #2 using the obtuse angle in your calculations. What do you notice?

**Answers:**

1. The \begin{align*}65^\circ\end{align*} angle is the included angle to the two given sides of length 7 and 10. Use the new area formula.

\begin{align*}A &= \frac{1}{2} (7)(10) \sin 65^\circ\\ A & \approx 31.72 \ un^2\end{align*}

2. Notice that the given angle is obtuse. Draw the altitude from vertex \begin{align*}C\end{align*} so that it intersects the extension of \begin{align*}\overline{AB}\end{align*} at point \begin{align*}D\end{align*}. The exterior angle at \begin{align*}\angle A\end{align*} is \begin{align*}30^\circ\end{align*}.

Use the new area formula with the \begin{align*}30^\circ\end{align*} angle.

\begin{align*}A &= \frac{1}{2} (9)(15) \sin 30^\circ\\ A &= 33.75 \ un^2\end{align*}

3. Use the area formula as in #2. Instead of using \begin{align*}30^\circ\end{align*}, use \begin{align*}150^\circ\end{align*}.

\begin{align*}A &= \frac{1}{2} (9)(15) \sin 150^\circ\\ A &= 33.75 \ un^2\end{align*}

The result is the same. \begin{align*}\sin 30^\circ=0.5\end{align*} and \begin{align*}\sin 150^\circ=0.5\end{align*}.

**This means that regardless of whether the given angle is acute or obtuse, you can always find the area of a triangle by finding the half the product of two sides and the sine of their included angle:**

**For all triangles: \begin{align*}A=\frac{1}{2} (a)(b) \sin C\end{align*}**

#### Practice

Find the area of each triangle.

1.

2.

3.

4.

5. Explain why the following triangle has the same area as the triangle in #4.

Find the area of each triangle.

6.

7.

8.

9. Use your calculator to find \begin{align*}\sin 90^\circ\end{align*}.

10. Find the area of the triangle below in two ways. First, use the formula \begin{align*}A=\frac{1}{2} bh\end{align*}. Then, use the new formula from this concept. What do you notice?

11. Find the area of the parallelogram:

12. Use your work from #11 to help you to describe a general method for calculating the area of a parallelogram given its sides and angles.

13. The area of the triangle below is \begin{align*}78 \ un^2\end{align*}. Find the measure of \begin{align*}\theta\end{align*} rounded to the nearest degree.

14. Use the triangle below to explain where the area formula \begin{align*}A=\frac{1}{2} ab \sin C\end{align*} comes from.

15. Why does the area formula \begin{align*}A=\frac{1}{2} ab \sin C\end{align*} work even if \begin{align*}\angle C\end{align*} is obtuse?