<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Alternate Formula for the Area of a Triangle

## Area equals half the product of two sides and the sine of the included angle.

Estimated9 minsto complete
%
Progress
Practice Alternate Formula for the Area of a Triangle

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated9 minsto complete
%
Sine to find the Area of a Triangle

You have previously learned that the area of a triangle is A=12bh\begin{align*}A=\frac{1}{2}bh\end{align*} where b\begin{align*}b\end{align*} is a base of the triangle and h\begin{align*}h\end{align*} is the corresponding height. Why is it helpful to have another formula for calculating the area of a triangle?

### Finding the Area of a Triangle by using Sine

One way to find the area of a triangle is by calculating 12bh\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.

#### Writing Equations

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

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

sin34=h7\begin{align*}\sin 34^\circ=\frac{h}{7}\end{align*}

#### Solving for Unknown Values

Solve the equation for h\begin{align*}h\end{align*}. Then, use A=12bh\begin{align*}A=\frac{1}{2}bh\end{align*} to find the area of the triangle using your value for h\begin{align*}h\end{align*}. Can you generalize the formula?

If sin34=h7\begin{align*}\sin 34^\circ=\frac{h}{7}\end{align*} then h=7sin34\begin{align*}h=7 \sin 34^\circ\end{align*}. For the height that has been drawn, BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} is the base.

AAAA=12bh=12(12)(7sin34)=12(12)(7)(sin34)23.45 un2\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: A=12(12)(7)(sin34)\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 B\begin{align*}\angle B\end{align*}, the area will be Area=12(a)(c)sinB\begin{align*}Area = \frac{1}{2} (a)(c) \sin B\end{align*}. If you know C\begin{align*}\angle C\end{align*}, the area will be Area=12(a)(b)sinC\begin{align*}Area=\frac{1}{2} (a)(b) \sin C\end{align*}.

#### Finding the Area Given Angle Measurements

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 h\begin{align*}h\end{align*}. Then, find the area of the triangle.

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

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

sin55h=h7=7sin55\begin{align*}\sin 55^\circ &= \frac{h}{7}\\ \rightarrow h &= 7 \sin 55^\circ\end{align*}

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

AAAA=12bh=12(12)(7sin55)=12(12)(7)(sin55)34.4 un2\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.

### Examples

#### Example 1

Earlier, you were asked why is it helpful to have another formula for calculating the area of a triangle.

You have previously learned that the area of a triangle is A=12bh\begin{align*}A=\frac{1}{2}bh\end{align*} where b\begin{align*}b\end{align*} is a base of the triangle and h\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. A=12(a)(b)sinC\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.

#### Example 2

Find the area of the triangle.

The 65\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.

AA=12(7)(10)sin6531.72 un2\begin{align*}A &= \frac{1}{2} (7)(10) \sin 65^\circ\\ A & \approx 31.72 \ un^2\end{align*}

#### Example 3

Find the area of the triangle.

The sine ratio is actually a function that takes any angle measure as an input (not just angles between 0\begin{align*}0^\circ\end{align*} and 90\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!

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

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

AA=12(9)(15)sin30=33.75 un2\begin{align*}A &= \frac{1}{2} (9)(15) \sin 30^\circ\\ A &= 33.75 \ un^2\end{align*}

#### Example 4

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

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

AA=12(9)(15)sin150=33.75 un2\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*}

### Review

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?

To see the Review answers, open this PDF file and look for section 7.5.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Included Angle

The included angle in a triangle is the angle between two known sides.

sine

The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.