Lesson 10.1 : Non-right Triangles: Law of Sines

New Concept

Moving beyond right triangles, the Law of Sines is a powerful tool for solving any oblique triangle. It establishes a proportional relationship, $\frac{\sin \alpha}{a} = \frac{\sin \beta}{b} = \frac{\sin \gamma}{c}$, allowing you to find unknown sides, angles, and even the triangle's area.

What’s next

This card is just the beginning. Next, you'll work through interactive examples of solving oblique triangles and tackle practice problems on the ambiguous SSA case.

Property

Given a triangle with angles and opposite sides labeled, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

To solve an oblique triangle, use any pair of applicable ratios.

Examples

• Solve for side $b$ in a triangle where $\alpha = 40^\circ, \beta = 75^\circ, a = 10$. First, find side $b$ using the proportion $\frac{b}{\sin(75^\circ)} = \frac{10}{\sin(40^\circ)}$. This gives $b = \frac{10 \sin(75^\circ)}{\sin(40^\circ)} \approx 15.0$.

• In a triangle, angles are $\alpha = 55^\circ, \gamma = 80^\circ$ and the included side is $b = 12$. First, find $\beta = 180^\circ - 55^\circ - 80^\circ = 45^\circ$. Then, find side $a$ using $\frac{a}{\sin(55^\circ)} = \frac{12}{\sin(45^\circ)}$, so $a \approx 13.9$.

Property

Triangles classified as SSA, where we know two sides and the angle opposite one of them, may result in one, two, or no solutions. This is the ambiguous case. Given angle $\alpha$ and sides $a$ and $b$:

1. No triangle exists if $a < b \sin \alpha$.
2. One right triangle exists if $a = b \sin \alpha$.
3. Two triangles exist if $b \sin \alpha < a < b$.
4. One triangle exists if $a \ge b$.

Examples

• Given $\alpha = 40^\circ, a = 9, b = 12$. The height is $h = 12 \sin(40^\circ) \approx 7.71$. Since $7.71 < 9 < 12$, there are two possible triangles. One angle is $\beta_1 \approx 59.4^\circ$ and the other is $\beta_2 = 180^\circ - 59.4^\circ = 120.6^\circ$.

• Given $\alpha = 50^\circ, a = 5, b = 8$. The height is $h = 8 \sin(50^\circ) \approx 6.13$. Since side $a$ is shorter than the height ($5 < 6.13$), it cannot form a triangle, so there is no solution.

Property

An oblique triangle is any triangle that is not a right triangle. Solving an oblique triangle means finding all three angles and all three sides. To do so, we need at least three of these values, including at least one side. The Law of Sines is used for:

1. ASA (angle-side-angle): Knowing two angles and the included side.
2. AAS (angle-angle-side): Knowing two angles and a non-included side.

Examples

• AAS Case: Given $\alpha = 45^\circ, \beta = 65^\circ, a = 10$. First find $\gamma = 180^\circ - 45^\circ - 65^\circ = 70^\circ$. Then find side $b$ with the Law of Sines: $\frac{b}{\sin(65^\circ)} = \frac{10}{\sin(45^\circ)}$, which gives $b \approx 12.8$.

• ASA Case: Given $\alpha = 50^\circ, \gamma = 30^\circ, b = 15$. First find $\beta = 180^\circ - 50^\circ - 30^\circ = 100^\circ$. Then find side $a$ with the Law of Sines: $\frac{a}{\sin(50^\circ)} = \frac{15}{\sin(100^\circ)}$, which gives $a \approx 11.7$.

Property

The formula for the area of an oblique triangle is given by

This is equivalent to one-half of the product of two sides and the sine of their included angle.

Examples

• Find the area of a triangle with sides $a = 80$, $b = 60$, and included angle $\gamma = 110^\circ$. The area is $\operatorname{Area} = \frac{1}{2}(80)(60)\sin(110^\circ) \approx 2255$ square units.

• A triangular park has two sides measuring 150 feet and 200 feet, with an included angle of $70^\circ$. Its area is $\operatorname{Area} = \frac{1}{2}(150)(200)\sin(70^\circ) \approx 14095$ square feet.