Law of Cosines: How and when to use Formula, examples problems …

# Law of Cosines: How and when to use Formula, examples problems …

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4. Use the Law of Cosines for SAS

# Use the Law of Cosines for SAS

Related Book

## Trigonometry For Dummies®

By Mary Jane Sterling

When you have two sides of a triangle and the angle between them, otherwise known as SAS (side-angle-side), you can use the law of cosines to solve for the other three parts. Consider the triangle ABC where a is 15, c is 20, and angle B is 124 degrees. The following figure shows what this triangle looks like.

A sample triangle that allows for the law of cosines.

Now, to solve for the measure of the missing side and angles:

1. Find the measure of the missing side by using the law of cosines.

Use the law that solves for side b.

You end up with the value for b2. Take the square root of each side and just use the positive value (because a negative length doesn’t exist).

The length of side b is about 31.

2. Find the measure of one of the missing angles by using the law of cosines.

Using the law that solves for a, fill in the values that you know.

Solve for cos A by simplifying and moving all the other terms to the left.

Using a scientific calculator to find angle A, you find that A = cos–1(0.916) = 23.652, or about 24 degrees.

You can also switch to the law of sines to solve for this angle. Don’t be afraid to mix and match when solving these triangles.

1. Find the measure of the last angle.

Determine angle B by adding the other two angle measures together and subtracting that sum from 180.

180 – (124 + 24) = 180 – 148 = 32. Angle B measures 32 degrees.

How about an application that uses this SAS portion of the law of cosines? Consider the situation: A friend wants to build a stadium in the shape of a regular pentagon (five sides, all the same length) that measures 920 feet on each side. How far is the center of the stadium from the corners? The left part of figure shows a picture of the stadium and the segment you’re solving for.

You can divide the pentagon into five isosceles triangles. The base of each triangle is 920 feet, and the two sides are equal, so call them both a. Refer to the right-hand picture in the preceding figure. Use the law of cosines to solve for a, because you can get the angle between those two congruent sides, plus you already know the length of the side opposite that angle.

1. Determine the measure of the angle at the center of the pentagon.

A circle has a total of 360 degrees. Divide that number by 5, and you find that the angle of each triangle at the center of the pentagon is 72 degrees.

2. Use the law of cosines with the side measuring 920 feet being the side solved for.

Because the other two sides are the same measure, write them both as a in the equation.

3. Solve for the value of a.

The distance from the center to a corner is between 782 and 783 feet.

The computations here involve using rounded values. It’s usually best to hold off doing the rounding until you’re ready to report your final answer. In these cases, it didn’t really matter, but you want to be cautious if more accuracy is needed.

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# The Law of Cosines

For any triangle:

 a, b and c are sides. C is the angle opposite side c

The Law of Cosines (also called the Cosine Rule) says:

c2 = a2 + b2 − 2ab cos(C)

It helps us solve some triangles. Let’s see how to use it.

### Example: How long is side "c" … ?

We know angle C = 37º, and sides a = 8 and b = 11

The Law of Cosines says:c2 = a2 + b2 − 2ab cos(C)
Put in the values we know:c2 = 82 + 112 − 2 × 8 × 11 × cos(37º)
Do some calculations:c2 = 64 + 121 − 176 × 0.798…
More calculations:c2 = 44.44…
Take the square root:c = √44.44 = 6.67 to 2 decimal places

Answer: c = 6.67

## How to Remember

How can you remember the formula?

Well, it helps to know it’s the Pythagoras Theorem with something extra so it works for all triangles:

Pythagoras Theorem:
(only for Right-Angled Triangles)
a2 + b2 = c2
Law of Cosines:
(for all triangles)
a2 + b2 − 2ab cos(C) = c2

So, to remember it:

• think "abc": a2 + b2 = c2,
• then a 2nd "abc": 2ab cos(C),
• and put them together: a2 + b2 − 2ab cos(C) = c2

## When to Use

The Law of Cosines is useful for finding:

• the third side of a triangle when we know two sides and the angle between them (like the example above)
• the angles of a triangle when we know all three sides (as in the following example)

### Example: What is Angle "C" …?

The side of length "8" is opposite angle C, so it is side c. The other two sides are a and b.

Now let us put what we know into The Law of Cosines:

Start with:c2 = a2 + b2 − 2ab cos(C)
Put in a, b and c:82 = 92 + 52 − 2 × 9 × 5 × cos(C)
Calculate:64 = 81 + 25 − 90 × cos(C)

Now we use our algebra skills to rearrange and solve:

Subtract 25 from both sides:39 = 81 − 90 × cos(C)
Subtract 81 from both sides:−42 = −90 × cos(C)
Swap sides:−90 × cos(C) = −42
Divide both sides by −90:cos(C) = 42/90
Inverse cosine:C = cos−1(42/90)
Calculator:C = 62.2° (to 1 decimal place)

## In Other Forms

### Easier Version For Angles

We just saw how to find an angle when we know three sides. It took quite a few steps, so it is easier to use the "direct" formula (which is just a rearrangement of the c2 = a2 + b2 − 2ab cos(C) formula). It can be in either of these forms:

cos(C) =
a2 + b2 − c2
2ab

cos(A) =
b2 + c2 − a2
2bc

cos(B) =
c2 + a2 − b2
2ca

### Example: Find Angle "C" Using The Law of Cosines (angle version)

In this triangle we know the three sides:

• a = 8,
• b = 6 and
• c = 7.

Use The Law of Cosines (angle version) to find angle C :

cos C= (a2 + b2 − c2)/2ab
= (82 + 62 − 72)/2×8×6
= (64 + 36 − 49)/96
= 51/96
= 0.53125
C= cos−1(0.53125)
= 57.9° to one decimal place

### Versions for a, b and c

Also, we can rewrite the c2 = a2 + b2 − 2ab cos(C) formula into a2= and b2= form.

Here are all three:

a2 = b2 + c2 − 2bc cos(A)

b2 = a2 + c2 − 2ac cos(B)

c2 = a2 + b2 − 2ab cos(C)

But it is easier to remember the "c2=" form and change the letters as needed !

As in this example:

### Example: Find the distance "z"

The letters are different! But that doesn’t matter. We can easily substitute x for a, y for b and z for c

Start with:c2 = a2 + b2 − 2ab cos(C)
x for a, y for b and z for cz2 = x2 + y2 − 2xy cos(Z)
Put in the values we know:z2 = 9.42 + 6.52 − 2×9.4×6.5×cos(131º)
Calculate:z2 = 88.36 + 42.25 − 122.2 × (−0.656…)
z2 = 130.61 + 80.17…
z2 = 210.78…
z = √210.78… = 14.5 to 1 decimal place.

Answer: z = 14.5

Did you notice that cos(131º) is negative and this changes the last sign in the calculation to + (plus)? The cosine of an obtuse angle is always negative (see Unit Circle ).

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