Law of Sines – edMe Learning

The Rule: Law of Sines

In a triangle with angles and opposite sides shown below, the ratio of the the 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 written as:

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

Simply put, the ratio of the sine of an angle and the length of the opposite side is equal for each pair in the triangle. We use greek letters for the angle measures. If this is confusing just use capital letters when writing down the rule above and this triangle.

Example 1: Solving for Two Unknown Sides

An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.

The three angles must add up to 180 degrees. From this, we can determine that

$\Large\beta = 180\degree - 50\degree - 30\degree$
$\Large\beta = 100\degree$

Now we can put the information into these relationships:

$\LARGE\frac{sin 50\degree}{10} = \frac{sin 100\degree}{b} = \frac{sin 30\degree}{c}$

Then, you can solve them two at a time:

$\LARGE\frac{sin 50\degree}{10} = \frac{sin 100\degree}{b}$

Then, multiply by 10b

$\LARGE\frac{(10b)(sin 50\degree)}{10} = \frac{(10b)(sin 100\degree)}{b}$

$\Large(b)(sin 50\degree) = (10)(sin 100\degree)$

$\Large b = 12.9$

Then, you can solve the last ratio:

$\LARGE\frac{sin 50\degree}{10} = \frac{sin 30\degree}{c}$

Then, multiply by 10c

$\LARGE\frac{(10c)(sin 50\degree)}{10} = \frac{(10c)(sin 30\degree)}{c}$

$\Large(c)(sin 50\degree) = (10)(sin 30\degree)$

$\Large c = 6.5$

Checking your answer quickly: The shortest side should be opposite the smallest angle. The longest side should be opposite the largest angle. You can quickly check each answer to make sure you get all the points.

Application: Two radar towers are used to locate a plane. How far is distance a?

A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.

Solution:

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side𝑎, and then use right triangle relationships to find the height of the aircraft,ℎ .

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

$\LARGE\frac{sin(130\degree)}{20}=\frac{sin(35\degree)}{𝑎}$
$\text{[math]}$ \Largeasin(130\degree)=20sin(35\degree)[\latex]
𝑎=20sin(35°)sin(130°) 𝑎≈14.98

Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 2. Figure 2
AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 3. Figure 3
SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 4. Figure 4

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5.

Figure 5

Using