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# Law of Sines and Cosines Help

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Introduction to Law of Sines and Cosines

We can solve other triangles using inverse trigonometric functions and the Law of Sines and/or the Law of Cosines. Although all triangles can be solved, sometimes we are given information that is true about more than one triangle or about a triangle that cannot exist. In the following problems, we will use the labels in the following triangles.

Fig. 13.55

Fig. 13.56

The angles are A , B , and C . The sides opposite these angles are a , b , and c , respectively.

We cannot solve a triangle if all we know are all three angles. Two triangles can be different sizes but have the same angles. Also, we might be given an angle with the side opposite the angle and another side that makes two triangles true. For example, suppose we are told to find a triangle where ∠ A = 21°, a = 3, and b = 8. There are two triangles that satisfy these conditions.

 Triangle 1 Triangle 2 ∠A = 21° ∠A = 21° ∠B ≈ 72.9° ∠B ≈ 107.1° ∠C ≈ 86.1° ∠C ≈ 52° a = 3 a = 3 b = 8 b = 8 c ≈ 8.4 c ≈ 6.6

There are two triangles when b sin A < a < b . If we have another number in addition to A , a , and b , then there will only be one triangle.

As an example of a triangle that cannot exist, let ∠ A = 20°, b = 10, and a = 2. As you can see in Figure 13.57, a is too short to close the triangle. This happens when a < b sin A .

Fig. 13.57

### The Law of Sines and Cosines

We can use the Law of Sines to solve a triangle if we know two sides and one of the angles opposite these sides or two angles and one side (if we know two angles, then we know all three because their sum is 180°). If do not have this information, the Law of Cosines works. We can use the Law of Cosines when we have two sides and any angle or when we have all three sides.

Here is the Law of Sines.

This is really three separate equations.

Here is the Law of Cosines.

a 2 = b 2 + c 2 − 2 bc cos A

b 2 = a 2 + c 2 − 2 ac cos B

c 2 = a 2 + b 2 − 2 ab cos C

#### Examples

Solve the triangle. When rounding is necessary, give your solutions accurate to one decimal place.

• A = 30°, B = 70°, and a = 5
• We will use the Law of Sines because we know an angle, A , and the side opposite it, a .

Now we will use (sin A )/ a = (sin C )/ c to find c . ( C = 180° − 30° − 70° = 80°)

• a = 5, b = 8, and c = 12
• There is not enough information to get one equation with one variable using the Law of Sines, so we will use the Law of Cosines.

We can use either the Law of Sines or the Law of Cosines to find B . The Law of Sines is a little easier.

C ≈ 180° − 17.6° − 28.9° ≈ 133.5°

•

Fig. 13.58

We will call the 120° angle A , then b = 10 and c = 6. (It does not matter which side is b and which side is c , as long as we do not label either one of them a .) There is not enough information to use the Law of Sines, so we will use the Law of Cosines.

We can use either the Law of Sines or the Law of Cosines to find B or C . We will use the Law of Sines to find B .

Find practice problems and solutions for these concepts at Law of Sines and Cosines Practice Problems.

More practice problems for this concept can be found at: Trigonometry Practice Test.

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