Finding the Area of a Triangle Using Sine

Example 1:

Find the area of $\Delta PQR$ .

You have the lengths of two sides and the measure of the included angle. So, you can use the formula $R=\frac{1}{2}pr\sin \left(Q\right)$ where $p$ and $r$ are the lengths of the sides opposite to the vertices $P$ and $R$ respectively.

Using the formula the area, $R=\frac{1}{2}\left(3\right)\left(4\right)\sin \left(145°\right)$ .

Simplify.

$\begin{array}{l}R=6\sin \left(145°\right)\\ \approx 6\left(0.5736\right)\\ \approx 3.44\end{array}$

Therefore, the area of $\Delta PQR$ is about $3.44$ sq.cm.

Example 2:

The area of the right $\Delta XYZ$ with the right angle at the vertex $Y$ is $39$ sq. units. If $YZ=12$ and $XZ=13$ , solve the triangle.

First, draw a figure with the given measures.

Use the Pythagorean Theorem to find the length of the third side of the triangle.

$\begin{array}{l}XY=\sqrt{{\left(XZ\right)}^2-{\left(YZ\right)}^2}\\ =\sqrt{{13}^2-{12}^2}\\ =\sqrt{169-144}\\ =\sqrt{25}\\ =5\end{array}$

Now, you have lengths of the three sides and the area of the triangle.

Substitute in the area formula.

$\begin{array}{l}\text{Area}=\frac{1}{2}\times \left(YZ\right)\times \left(XZ\right)\times \sin \left(Z\right)\\ 39=\frac{1}{2}\left(12\right)\left(13\right)\sin \left(Z\right)\end{array}$

Solve for $Z$ .

$\begin{array}{l}\sin \left(Z\right)=\frac{\left(39\right)\left(2\right)}{\left(12\right)\left(13\right)}\\ =0.5\end{array}$

Taking the inverse,

$\begin{array}{l}Z={\sin }^{-1}\left(0.5\right)\\ =30°\end{array}$

That is, $m\angle Z=30°$ .

Given that the angle at the vertex $Y$ is a right angle. Therefore, $m\angle Y=90°$ .

Using the Triangle Angle Sum Theorem , the measure of the third angle is,

$\begin{array}{l}m\angle X=180-\left(m\angle Y+m\angle Z\right)\\ =180-\left(90+30\right)\\ =60\end{array}$

Therefore, the measure of $\angle X$ is $60°$ .