Solving Triangles Using The Law of Sines

Previous concepts explained how to use trigonometry to find the measures of the angles and sides of right triangles. We will now discuss the law of sines, which allows us to solve for the angles and side lengths of any triangle. A right triangle contains a $90^{\circ}$ angle, while any other triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides.

The law of sines states that:

$\displaystyle{\frac{ \sin \alpha}{a} = \frac{ \sin \beta}{b} = \frac{ \sin \gamma}{c}}$

where $\alpha, \beta,$ and $\gamma$ are angles and $a$, $b$, and $c$ are the lengths of the sides opposite them, respectively.

Oblique triangle

The sides of this oblique triangle are labeled a, b, and c, and the angles are labeled $\alpha$, $\beta$, and $\gamma$. 

Note the standard way of labeling triangles: angle $\alpha$ (alpha) is opposite side $a$; angle $\beta$ (beta) is opposite side $b$; and angle $\gamma$ (gamma) is opposite side $c$.

To solve an oblique triangle, use any pair of applicable ratios from the law of sines formula. While calculating angles and sides, be sure to carry the exact values through to the final answer. 

Example

Solve the triangle shown in the figure, with final answers rounded to the nearest tenth.

Oblique triangle with unknown sides and angles

In this triangle, $\alpha = 50\degree$, $\gamma = 30\degree$, and $a=10$. The angle $\beta$ and the side-lengths $b$ and $c$ are unknown.

First, notice that two of the three angles are already identified. We can subtract these from $180^{\circ}$ to find the measure of angle $\beta$:

$\begin{aligned} \beta &= 180^{\circ} - 50^{\circ} - 30^{\circ} \\ &= 100^{\circ} \end{aligned}$

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle $\alpha = 50^{\circ}$ and its corresponding side is $a = 10$. Let's find the length of side $c$, which corresponds to angle $\gamma = 30^{\circ}$. We can use the following proportion from the law of sines: 

$\displaystyle{\frac{\sin{\left(50^{\circ}\right)}}{10} = \frac{\sin{\left(30^{\circ}\right)}}{c}}$

Multiply both sides by $c$:

$\displaystyle{ c \cdot \frac{\sin{\left(50^{\circ}\right)}}{10} = \sin{\left(30^{\circ}\right)} }$

Multiply by the reciprocal of $\displaystyle{\frac{\sin{\left(50^{\circ}\right)}}{10}}$ to isolate $c$:

$\displaystyle{ c = \sin{\left(30^{\circ}\right)} \cdot \frac{10}{\sin{\left(50^{\circ}\right)}} }$

Solving this with a calculator, we obtain: 

$c \approx 6.5$

The last unknown side is $b$, and we will follow a similar process for this. Set up another proportion:

$\displaystyle{ \frac{\sin{\left(50^{\circ}\right)}}{10} = \frac{\sin{\left(100^{\circ}\right)}}{b} }$

Solve for $b$:

$\displaystyle{ \begin{aligned} \frac{b \cdot \sin{\left(50^{\circ}\right)}}{10} &= \sin{\left(100^{\circ}\right)} \\ b \cdot \sin{\left(50^{\circ}\right)} &= 10 \cdot \sin{\left(100^{\circ}\right)} \\ b &= \frac{10 \cdot \sin{\left(100^{\circ}\right)}}{\sin{\left(50^{\circ}\right)}} \\ b &\approx 12.9 \end{aligned} }$

Therefore, the complete set of sides and angles is:

$\alpha = 50^{\circ} \quad \quad \quad a = 10 \\ \beta = 100^{\circ} \quad \quad \space \space b \approx 12.9 \\ \gamma = 30^{\circ} \quad \quad \quad c \approx 6.5 $