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Law of Sines

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Law of Sines with AAS and ASA

A triangle has two angles that measure 60^\circ and 45^\circ . The length of the side between these two angles is 10. What are the lengths of the other two sides?

Guidance

Consider the non right triangle below. We can construct an altitude from any one of the vertices to divide the triangle into two right triangles as show below.

from the diagram we can write two trigonometric functions involving h :

\sin C&=\frac{h}{b} \qquad and \qquad \sin B=\frac{h}{c}\\b \sin C&=h \qquad \qquad \quad \ c \sin B=h

Since both are equal to h , we can set them equal to each other to get:

b \sin C=c \sin B and finally divide both sides by bc to create the proportion:

\frac{\sin C}{c}=\frac{\sin B}{b}

If we construct the altitude from a different vertex, say B , we would get the proportion: \frac{\sin A}{a}=\frac{\sin C}{c} . Now, the transitive property allows us to conclude that \frac{\sin A}{a}=\frac{\sin B}{b} . We can put them all together as the Law of Sines: \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} . In the examples that follow we will use the Law of Sines to solve triangles.

Example A

Solve the triangle.

Solution: Since we are given two of the three angles in the triangle, we can find the third angle using the fact that the three angles must add up to 180^\circ . So, m \angle A=180^\circ - 45^\circ - 70^\circ = 650^\circ . Now we can substitute the known values into the Law of Sines proportion as shown:

\frac{\sin 65^\circ}{a}=\frac{\sin 70^\circ}{15}=\frac{\sin 45^\circ}{c}

Taking two ratios at a time we can solve the proportions to find a and c using cross multiplication.

To find a :

\frac{\sin 65^\circ}{a}&=\frac{\sin 70^\circ}{15} \\a&=\frac{15 \sin65^\circ}{\sin 70^\circ} \approx 14.5

To find c :

\frac{\sin 70^\circ}{15}&=\frac{\sin 45^\circ}{c} \\c&=\frac{15 \sin45^\circ}{\sin 70^\circ} \approx 11.3

This particular triangle is an example in which we are given two angles and the non-included side or AAS (also SAA).

Example B

Solve the triangle.

Solution: In this example we are given two angles and a side as well but the side is between the angles. We refer to this arrangement as ASA. In practice, in doesn’t really matter whether we are given AAS or ASA. We will follow the same procedure as Example A. First, find the third angle: m \angle A=180^\circ - 50^\circ - 80^\circ = 50^\circ .

Second, write out the appropriate proportions to solve for the unknown sides, a and b .

To find a :

\frac{\sin 80^\circ}{a}&=\frac{\sin 50^\circ}{20} \\a&=\frac{20 \sin80^\circ}{\sin 50^\circ} \approx 25.7

To find b :

\frac{\sin 50^\circ}{b}&=\frac{\sin 50^\circ}{20} \\b&=\frac{20 \sin50^\circ}{\sin 50^\circ} = 20

Notice that c=b and m \angle C=m \angle B . This illustrates a property of isosceles triangles that states that the base angles (the angles opposite the congruent sides) are also congruent.

Example C

Three fishing ships in a fleet are out on the ocean. The Chester is 32 km from the Angela. An officer on the Chester measures the angle between the Angela and the Beverly to be 25^\circ . An officer on the Beverly measures the angle between the Angela and the Chester to be 100^\circ . How far apart, to the nearest kilometer are the Chester and the Beverly?

Solution: First, draw a picture. Keep in mind that when we say that an officer on one of the ships is measuring an angle, the angle she is measuring is at the vertex where her ship is located.

Now that we have a picture, we can determine the angle at the Angela and then use the Law of Sines to find the distance between the Beverly and the Chester.

The angle at the Angela is 180^\circ - 100^\circ - 25^\circ = 55^\circ .

Now find x ,

\frac{\sin 55^\circ}{x}&=\frac{\sin 100^\circ}{32} \\x&=\frac{32 \sin 55^\circ}{\sin 100^\circ} \approx 27

The Beverly and the Chester are about 27 km apart.

Concept Problem Revisit The measure of the triangle's third angle is 180^\circ - 60^\circ -45^\circ =75^\circ

\frac{\sin 45^\circ}{x}&=\frac{\sin 75^\circ}{10}, \ \text{so} \ x=\frac{10 \sin 45^\circ}{\sin 75^\circ} \approx 7.29 \\\frac{\sin 60^\circ}{y}&=\frac{\sin 75^\circ}{10}, \ \text{so} \ y=\frac{10 \sin 60^\circ}{\sin 75^\circ} \approx 8.93

Guided Practice

Solve the triangles.

1.

2.

3. A surveying team is measuring the distance between point A on one side of a river and point B on the far side of the river. One surveyor is positioned at point A and the second surveyor is positioned at point C , 65 m up the riverbank from point A . The surveyor at point A measures the angle between points B and C to be 103^\circ . The surveyor at point C measures the angle between points A and B to be 42^\circ . Find the distance between points A and B .

Answers

1. m \angle A=180^\circ - 82^\circ -24^\circ =74^\circ

\frac{\sin 24^\circ}{b}&=\frac{\sin 74^\circ}{11}, \ \text{so} \ b=\frac{11 \sin 24^\circ}{\sin 74^\circ} \approx 4.7 \\\frac{\sin 82^\circ}{c}&=\frac{\sin 74^\circ}{11}, \ \text{so} \ c=\frac{11 \sin 82^\circ}{\sin 74^\circ} \approx 11.3

2. m \angle C=180^\circ - 110^\circ -38^\circ =32^\circ

\frac{\sin 38^\circ}{a}&=\frac{\sin 110^\circ}{18}, \ \text{so} \ a=\frac{18 \sin 38^\circ}{\sin 110^\circ} \approx 11.8 \\\frac{\sin 32^\circ}{c}&=\frac{\sin 110^\circ}{18}, \ \text{so} \ c=\frac{18 \sin 32^\circ}{\sin 110^\circ} \approx 10.2

3.

m \angle B &= 180^\circ - 103^\circ - 42^\circ = 35^\circ \\\frac{\sin 35^\circ}{65} &= \frac{\sin 42^\circ}{c} \\c &= \frac{65 \sin 42^\circ}{\sin 35^\circ} \approx 75.8 \ m

Vocabulary

Law of Sines
For any triangle, the ratio of the sine of an angle over its opposite side is equal to the sine of any other angle in the triangle over its opposite side. \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}

Explore More

Solve the triangles. Round your answers to the nearest tenth.

Using the given information, solve \Delta ABC .

  1. .
m \angle A&=85^\circ \\m \angle C&=40^\circ \\a&=12
  1. .
m \angle B&=60^\circ \\m \angle C&=25^\circ \\a&=28
  1. .
m \angle B&=42^\circ \\m \angle A&=36^\circ \\b&=8
  1. .
m \angle B&=30^\circ \\m \angle A&=125^\circ \\c&=45

Use the Law of Sines to solve the following world problems.

  1. A surveyor is trying to find the distance across a ravine. He measures the angle between a spot on the far side of the ravine, X , and a spot 200 ft away on his side of the ravine, Y , to be 100^\circ . He then walks to Y the angle between X and his previous location to be 20^\circ . How wide is the ravine?
  2. A triangular plot of land has angles 46^\circ and 58^\circ . The side opposite the 46^\circ angle is 35 m long. How much fencing, to the nearest half meter, is required to enclose the entire plot of land?

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