﻿ Law of Sines formula, how and when to use , examples and practice problems

# Law of Sines, Trigonometry of Triangles

## Formula to find angle and side lengths

#### When to use the law of sines formula

When you know 2 sides and the non-included angle or when you know 2 angles and the non-included side .

What You Know What You Can Find
2 sides and the non-included angle angle opposite to known side

We can use the formula because we are working with 2 opposite pairs of sides/angles .

What You Know What You Can Find
2 angles and the non-included side side opposite to known angle

We can use the formula because we are working with 2 opposite pairs of sides/angles .

#### Two different Cases? Or really just one?

There are two different situations when you use this formula. But really, there is just one case .

Just look at it: You can always immediately look at a triangle and tell whether or not you can use the Law of Sines -- you need 3 measurements: either 2 sides and the non-included angle or 2 angles and the non-included side.

It's all about opposites: To use the law of sines, you need to know one opposite angle/side pair measurements.

#### Cases when you can not use the Law of Sines

The picture below illustrates a case not suited for the law of sines. Since we do not know an opposite side and angle, we cannot employ the formula.

Or, just look at it: Remember when you have 2 sides, the angle must be non-included. By the way, we could use the law of cosines to find the length of the side opposite the 115° angle.

##### Do you know when to use the formula?

In which triangle(s) below, can we use the formula? Both triangles below have 3 known measurements.

Triangle 1 has only one opposite pair that we are dealing with, but that does not help us because we need to know both the angle and itsopposite side.

Or, just look at it: Remember when you have 2 sides, the angle must be non-included.

Triangle 2 does have 2 opposite pairs that we are dealing with, and we do know the measurements of one angle and its opposite side

Or, just look at it: We can use the formula, when we have 2 sides and the non-included angle.

In the full practice problems below, we will solve for triangle 2's unknown angle measure.

### Practice Problems

##### Problem 1
Step 1

Set up proportion with 2 pairs of opposite sides/sines of angles

$$\\ \frac{ sin( \red b)}{ 16} = \frac{ sin(115)} {123}$$

Step 2

Solve for unknown

$$\frac{ sin( \red b)}{ 16} = \frac{ sin(115)} {123} \\ sin( \red b ) = \frac{ 16 \cdot sin(115)} {123} \\ sin( \red b ) = 0.11789369587468619 \\ \red b = sin^{-1} ( 0.11789369587468619 ) \\ \red b = 6.770557323410266 \\ \red b \approx 6.8$$

##### Problem 2

No, because we need to know the measure of 1 opposite side and angle.

We can not use side with length 20 because we don't know its opposite angle. And we can't use 66 ° angle because we don't know its opposite side. And, of course, we do not know the measure of the angle opposite of the side of length 13 because... well, because that's the very thing we are solving for!

Just look at it: Remember when you have 2 sides, the angle must be non-included.

##### Problem 3

Yes, because we need to know the measures of one opposite side and angle which we have with the 29 ° angle and the side of length 11. And we know the angle (118 °) opposite the side length that we are solving for.

Or just look at it: Remember when you have 2 angles, the side must be non-included.

##### Problem 4
Step 1

Set up proportion with 2 pairs of opposite sides/sines of angles

$$\frac{ \red b}{ sin(118)} = \frac{ 11 } {sin(29)}$$

Step 2

Solve for unknown

$$\frac{ \red b}{ sin(118)} = \frac{ 11 } {sin(29)} \\ \red b = \frac { sin(118) \cdot 11 }{ sin(29)} \\ \red b = 20.033 \\ \red b \approx 20.0$$

##### Problem 5
Step 1

Set up proportion with 2 pairs of opposite sides/sines of angles

$$\frac{ \red e}{ sin(67)} = \frac{ 7 } {sin(54)}$$

Step 2

Solve for unknown

$$\frac{ \red e}{ sin(67)} = \frac{ 7 } {sin(54)} \\ \red e = \frac { sin(67) \cdot 7 }{ sin(54)} \\ \red e = 7.9646460 \\ \red e \approx 7.96$$

### Challenge Problems

##### Challenge Problem 1
Step 1

Yes, first you must remember that the sum of the interior angles of a triangle is 180 in order to calculate the measure of the angle opposite of the side of length 19.

Now that we have the measure of that angle, use the law of sines to find the value of x

Step 2

Solve for unknown

Set up the equation:

$$\frac { \color{red}{x} }{ sin(116)} = \frac{19}{sin (34) }$$

Then cross multiply to calculate x:

$$\red x = 30.5$$

##### Challenge Problem 2
Step 1

Use the fact the sum of the interior angles of a triangle is 180° to calculate all of the angles inside the triangles.

Step 2

Now, use the law of sines formula to set up an equation.

Step 3

From here you have several options

Step 4

The easiest one is to use sohcahtoa to solve for a side length to arrive at the answer.