Toby draws a triangle that has two side lengths of 8 inches and 5 inches. He measures the included angle with a protractor and gets \begin{align*}75^\circ\end{align*}. What is the area of this triangle?

### Area of a Triangle

Recall the non right triangle for which we derived the law of sine.

We are most familiar with the area formula: \begin{align*}A=\frac{1}{2}bh\end{align*} where the base, \begin{align*}b\end{align*}, is the side length which is perpendicular to the altitude. If we consider angle \begin{align*}C\end{align*} in the diagram, we can write the following trigonometric expression for the altitude of the triangle, \begin{align*}h\end{align*}:

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

No we can replace \begin{align*}h\end{align*} in the formula with \begin{align*}b \sin C\end{align*} and the side perpendicular to \begin{align*}h\end{align*} is the base, \begin{align*}a\end{align*}. Our new area formula is thus:

\begin{align*}A=\frac{1}{2}ab \ \sin C.\end{align*}

It is important to note that \begin{align*}C\end{align*} is the angle between sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and that any two sides and the included angle can be used in the formula.

Let's find the area of the following triangles.

We are given two sides and the included angle so let \begin{align*}a=6\end{align*}, \begin{align*}b=9\end{align*} and \begin{align*}C=62^\circ\end{align*}. Now we can use the formula to find the area of the triangle:

\begin{align*}A=\frac{1}{2}(6)(9)\sin(62^\circ) \approx 23.8 \ square \ units\end{align*}

In this triangle we do not have two sides and the included angle. We must first find another side length using the Law of Sines. We can find the third angle using the triangle sum: \begin{align*}180^\circ - 51^\circ - 41^\circ = 88^\circ\end{align*}. Use the Law of Sines to find the side length opposite \begin{align*}41^\circ\end{align*}:

\begin{align*}\frac{\sin 88^\circ}{17}&=\frac{\sin 41^\circ}{x} \\ x&=\frac{17 \sin 41^\circ}{\sin 88^\circ} \approx 11.2\end{align*}

We now have two sides and the included angle and can use the area formula:

\begin{align*}A=\frac{1}{2}(11.2)(17)\sin(51^\circ) \approx 74 \ square \ units\end{align*}

Finally, given \begin{align*}c=25 \ cm\end{align*}, \begin{align*}a=31 \ cm\end{align*} and \begin{align*}B=78^\circ\end{align*}, find the area of \begin{align*}\Delta ABC\end{align*}.

Here we are given two sides and the included angle. We can adjust the formula to represent the sides and angle we are given: \begin{align*}A=\frac{1}{2}ac \ \sin B\end{align*}. It really doesn’t matter which “letters” are in the formula as long as they represent **two sides and the included angle** (the angle *between* the two sides.) Now put in our values to find the area: \begin{align*}A=\frac{1}{2}(31)(25)\sin(78^\circ)\approx 379 \ cm^2\end{align*}.

### Examples

#### Example 1

Earlier, you were asked to find the area of the triangle that has two side lengths of 8 inches and 5 inches and an included angle of \begin{align*}75^\circ\end{align*}.

We are given two sides and the included angle so let \begin{align*}a=8\end{align*}, \begin{align*}b=5\end{align*} and \begin{align*}C=75^\circ\end{align*}. Now we can use the formula to find the area of the triangle:

\begin{align*}A=\frac{1}{2}(8)(5)\sin(75^\circ) \approx 19.4 \ square \ inches\end{align*}

**Find the area of each of the triangles below. Round answers to the nearest square unit.**

#### Example 2

Two sides and the included angle are given so \begin{align*}A=\frac{1}{2}(20)(23)\sin 105^\circ \approx 222 \ sq \ units\end{align*}.

#### Example 3

Find side \begin{align*}a\end{align*} first: \begin{align*}\frac{\sin 70^\circ}{8}=\frac{\sin 60^\circ}{a}\end{align*}, so \begin{align*}a=\frac{8 \sin 60^\circ}{\sin 70^\circ}\approx 7.4\end{align*}. Next find \begin{align*}m \angle C=180^\circ - 60^\circ - 70^\circ = 50^\circ\end{align*}.

Using the area formula, \begin{align*}A=\frac{1}{2}(7.4)(8)\sin 50^\circ \approx 22.7 \ sq \ units\end{align*}.

#### Example 4

Find \begin{align*}m \angle C=180^\circ - 80^\circ - 41^\circ = 59^\circ\end{align*}. Find a second side: \begin{align*}\frac{\sin 59^\circ}{50}=\frac{\sin 80^\circ}{a}\end{align*}, so \begin{align*}a=\frac{50 \sin 80^\circ}{\sin 59^\circ} \approx 57.4\end{align*}.

Using the area formula, \begin{align*}A=\frac{1}{2}(57.4)(50)\sin 41^\circ \approx 941 \ sq \ units\end{align*}.

### Review

Find the area of each of the triangles below. Round your answers to the nearest square unit.

- \begin{align*}m \angle A=71^\circ, b=15, c=19\end{align*}
- \begin{align*}m \angle C=120^\circ, b=22, a=16\end{align*}
- \begin{align*}m \angle B=60^\circ, a=18, c=12\end{align*}
- \begin{align*}m \angle A=28^\circ, m \angle C=73^\circ, b=45\end{align*}
- \begin{align*}m \angle B=56^\circ, m \angle C=81^\circ, c=33\end{align*}
- \begin{align*}m \angle A=100^\circ, m \angle B=30^\circ, a=100\end{align*}
- The area of \begin{align*}\Delta ABC\end{align*} is 66 square units. If two sides of the triangle are 11 and 21 units, what is the measure of the included angle? Is there more than one possible value? Explain.
- A triangular garden is bounded on one side by a 20 ft long barn and a second side is bounded by a 25 ft long fence. If the barn and the fence meet at a \begin{align*}50^\circ\end{align*} angle, what is the area of the garden if the third side is the length of the segment between the ends of the fence and the barn?
- A contractor is constructing a counter top in the shape of an equilateral triangle with side lengths 3 ft. If the countertop material costs $25 per square foot, how much will the countertop cost?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 13.14.