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Finding the Area of a Triangle Using Sine

You are familiar with the formula $R = \frac{1}{2} b h$ to find the area of a triangle where $b$ is the length of a base of the triangle and $h$ is the height, or the length of the perpendicular to the base from the opposite vertex.

Suppose $Δ A B C$ has side lengths $a$ , $b$ , and $c$ . Let $h$ be the length of the perpendicular to the side of length $b$ from the vertex $B$ that meets the side $\bar{A C}$ at $D$ .

Math diagram

Then, the area $R$ of the triangle $A B C$ is $R = \frac{1}{2} b h$ .

Now, look at $Δ A D B$ . It is a right triangle with hypotenuse $\bar{A B}$ that has a length of $c$ units.

Consider the sine of $∠ A$ .

$\begin{array}{l} \sin (A) = \frac{Opposite Side}{Hypotenuse} \\ = \frac{h}{c} \\ \sin (A) = \frac{h}{c} \Rightarrow h = c \sin (A) \end{array}$

Substituting the value of $h$ in the formula for the area of a triangle, you get

$\begin{array}{l} R = \frac{1}{2} b (c \sin (A)) \\ = \frac{1}{2} b c \sin (A) \end{array}$

Similarly, you can write formulas for the area in terms of $\sin (B)$ or $\sin (C)$ .

$\begin{array}{l} R = \frac{1}{2} a b \sin (C) \\ R = \frac{1}{2} a c \sin (B) \end{array}$

Example 1:

Find the area of $Δ P Q R$ .

Math diagram

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

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

Simplify.

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

Therefore, the area of $Δ P Q R$ is about $3.44$ sq.cm.

Example 2:

The area of the right $Δ X Y Z$ with the right angle at the vertex $Y$ is $39$ sq. units. If $Y Z = 12$ and $X Z = 13$ , solve the triangle.

First, draw a figure with the given measures.

Math diagram

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

$\begin{array}{l} X Y = \sqrt{{(X Z)}^{2} - {(Y Z)}^{2}} \\ = \sqrt{13^{2} - 12^{2}} \\ = \sqrt{169 - 144} \\ = \sqrt{25} \\ = 5 \end{array}$

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

Substitute in the area formula.

$\begin{array}{l} Area = \frac{1}{2} \times (Y Z) \times (X Z) \times \sin (Z) \\ 39 = \frac{1}{2} (12) (13) \sin (Z) \end{array}$

Solve for $Z$ .

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

Taking the inverse,

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

That is, $m ∠ Z = 30 °$ .

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

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

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

Therefore, the measure of $∠ X$ is $60 °$ .