# Lesson Explainer: Applications on Sine and Cosine Laws Mathematics

In this explainer, we will learn how to use the laws of sines and cosines to solve real-world problems.

The applications of these two laws are wide-ranging. They may be applied to problems within the field of engineering to calculate distances or angles of elevation, for example, when constructing bridges or telephone poles. In navigation, pilots or sailors may use these laws to calculate the distance or the angle of the direction in which they need to travel to reach their destination.

Let us begin by recalling the two laws. Consider triangle , with corresponding sides of lengths , , and .

### Definition: The Law of Sines

In a triangle as described above, the law of sines states that

The reciprocal is also true:

We can recognize the need for the law of sines when the information given consists of opposite pairs of side lengths and angle measures in a non-right triangle. In practice, we usually only need to use two parts of the ratio in our calculations.

### Definition: The Law of Cosines

In a triangle as described above, the law of cosines states that

The law of cosines can be rearranged to

We can recognize the need for the law of cosines in two situations:

• We use the first form when we have been given the lengths of two sides of a non-right triangle and the measure of the included angle, and we wish to calculate the length of the third side.
• We use the rearranged form when we have been given the lengths of all three sides of a non-right triangle and we wish to calculate the measure of any angle.

It is best not to be overly concerned with the letters themselves, but rather what they represent in terms of their positioning relative to the side length or angle measure we wish to calculate. For example, in our second statement of the law of cosines, the letters and represent the lengths of the two sides that enclose the angle whose measure we are calculating and a represents the length of the opposite side.

Provided we remember this structure, we can substitute the relevant values into the law of sines and the law of cosines without the need to introduce the letters , , and in every problem.

We should already be familiar with applying each of these laws to mathematical problems, particularly when we have been provided with a diagram. The focus of this explainer is to use these skills to solve problems which have a real-world application. It will often be necessary for us to begin by drawing a diagram from a worded description, as we will see in our first example.

### Example 1: Using the Law of Cosines to Calculate an Unknown Length in a Triangle in a Word Problem

A farmer wants to fence off a triangular piece of land. The lengths of two sides of the fence are 72 metres and 55 metres, and the angle between them is . Find the perimeter of the fence giving your answer to the nearest metre.

We begin by sketching the triangular piece of land using the information given, as shown below (not to scale).

In order to find the perimeter of the fence, we need to calculate the length of the third side of the triangle. We identify from our diagram that we have been given the lengths of two sides and the measure of the included angle. We can, therefore, calculate the length of the third side by applying the law of cosines:

We may find it helpful to label the sides and angles in our triangle using the letters corresponding to those used in the law of cosines, as shown below.

However, this is not essential if we are familiar with the structure of the law of cosines. If we recall that and represent the two known side lengths and represents the included angle, then we can substitute the given values directly into the law of cosines without explicitly labeling the sides and angles using letters.

Substituting , , and into the law of cosines, we obtain

Evaluating and simplifying gives

We solve for by square rooting. We can ignore the negative solution to our equation as we are solving to find a length:

Finally, we recall that we are asked to calculate the perimeter of the triangle. Summing the three side lengths and rounding to the nearest metre as required by the question, we have the following:

The perimeter of the field, to the nearest metre, is 212 metres.

Other problems to which we can apply the laws of sines and cosines may take the form of journey problems. We may be given a worded description involving the movement of an object or the positioning of multiple objects relative to one another and asked to calculate the distance or angle between two points. In more complex problems, we may be required to apply both the law of sines and the law of cosines. We will now consider an example of this.

### Example 2: Determining the Magnitude and Direction of the Displacement of a Body Using the Law of Sines and the Law of Cosines

A person rode a bicycle km east, and then he rode for another 21 kmΒ  south of east. Determine the magnitude and direction of the displacement, rounding the direction to the nearest minute.

We begin by sketching the journey taken by this person, taking north to be the vertical direction on our screen. His start point is indicated on our sketch by the letter , and the dotted line represents the continuation of the easterly direction to aid in drawing the line for the second part of the journey. An angle south of east is an angle measured downward (clockwise) from this line.

We are asked to calculate the magnitude and direction of the displacement. The magnitude is the length of the line joining the start point and the endpoint . The direction of displacement of point from point is southeast, and the size of this angle is the measure of angle .

Let us consider triangle , in which we are given two side lengths. We can calculate the measure of their included angle, angle , by recalling that angles on a straight line sum to . Subtracting from gives

As we now know the lengths of two sides and the measure of their included angle, we can apply the law of cosines to calculate the length of the third side:

Substituting , , and gives

We solve for by square rooting:

We add the information we have calculated to our diagram.

To calculate the measure of angle , we have a choice of methods:

• We could apply the law of cosines using the three known side lengths.
• We could apply the law of sines using the opposite length of 21 km and the side angle pair shown in red.

We will apply the law of sines, using the version that has the sines of the angles in the numerator:

Substituting , , and gives

Multiplying each side of this equation by 21 leads to

We solve for by applying the inverse sine function:

Recall that we are asked to give our answer to the nearest minute, so using our calculator function to convert between an answer in degrees and an answer in degrees and minutes gives Β .

The magnitude of the displacement is km and the direction, to the nearest minute, is Β  south of east.

We saw in the previous example that, given sufficient information about a triangle, we may have a choice of methods. Applying the law of sines and the law of cosines will of course result in the same answer and neither is particularly more efficient than the other. It is also possible to apply either the law of sines or the law of cosines multiple times in the same problem.

The laws of sines and cosines can also be applied to problems involving other geometric shapes such as quadrilaterals, as these can be divided up into triangles. Let us now consider an example of this, in which we apply the law of cosines twice to calculate the measure of an angle in a quadilateral.

### Example 3: Using the Law of Cosines to Find the Measure of an Angle in a Quadrilateral

is a quadrilateral where , , , , and . Find giving the answer to the nearest degree.

We begin by sketching quadrilateral as shown below (not to scale).

We can also draw in the diagonal and identify the angle whose measure we are asked to calculate, angle .

The diagonal divides the quadrilaterial into two triangles. We see that angle is one angle in triangle , in which we are given the lengths of two sides. If we knew the length of the third side, , we could apply the law of cosines to calculate the measure of any angle in this triangle.

The side is shared with the other triangle in the diagram, triangle , so let us now consider this triangle. We are given two side lengths ( and ) and their included angle, so we can apply the law of cosines to calculate the length of the third side. For this triangle, the law of cosines states that

Substituting , , and gives

We solve for by square rooting, ignoring the negative solution as represents a length:

We add the length of to our diagram.

We now know the lengths of all three sides in triangle , and so we can calculate the measure of any angle. Recall the rearranged form of the law of cosines: where and are the side lengths which enclose the angle we wish to calculate and is the length of the opposite side. In our figure, the sides which enclose angle are of lengths 40 cm and cm, and the opposite side is of length 43 cm. Substituting these values into the law of cosines, we have

We solve for angle by applying the inverse cosine function:

The measure of angle , to the nearest degree, is .

We have now seen examples of calculating both the lengths of unknown sides and the measures of unknown angles in problems involving triangles and quadrilaterals, using both the law of sines and the law of cosines. Another application of the law of sines is in its connection to the diameter of a triangleβs circumcircle. Recall that the circumcircle of a triangle is the circle that passes through all three of the triangleβs vertices, as shown in the figure below.

### Definition: The Law of Sines and Circumcircle Connection

For any triangle , the diameter of its circumcircle is equal to the law of sines ratio:

We will now see how we can apply this result to calculate the area of a circumcircle given the measure of one angle in a triangle and the length of its opposite side.

### Example 4: Finding the Area of a Circumcircle given the Measure of an Angle and the Length of the Opposite Side

is a triangle where and . Find the area of the circumcircle giving the answer to the nearest square centimetre.

We recall first that the circumcircle of a triangle is the circle passing through all three of the triangleβs vertices. To calculate the area of any circle, we use the formula , so we need to consider how we can determine the radius of this circle.

The information given in the question consists of the measure of an angle and the length of its opposite side. We know this because the length given is for the side connecting vertices and , which will be opposite the third angle of the triangle, angle . An alternative way of denoting this side is .

We recall the connection between the law of sines ratio and the radius of the circumcircle:

Substituting and into the first part of this ratio and ignoring the middle two parts that are not required, we have

We solve this equation to determine the radius of the circumcircle:

We are now able to calculate the area of the circumcircle:

The area of the circumcircle, to the nearest square centimetre, is 431 cm2.

We can also combine our knowledge of the laws of sines and co sines with other results relating to non-right triangles. We should recall the trigonometric formula for the area of a triangle where and represent the lengths of two of the triangleβs sides and represents the measure of their included angle. In our final example, we will see how we can apply the law of sines and the trigonometric formula for the area of a triangle to a problem involving area.

### Example 5: Using the Law of Sines and Trigonometric Formula for Area of Triangles to Calculate the Areas of Circular Segments

Find the area of the green part of the diagram, given that , , and . Give the answer to the nearest square centimetre.

We begin by adding the information given in the question to the diagram. We may also find it helpful to label the sides using the letters , , and .

The shaded area can be calculated as the area of triangle subtracted from the area of the circle:

We recall the trigonometric formula for the area of a triangle, using two sides and the included angle:

In order to compute the area of triangle , we first need to calculate the length of side . We already know the length of a side in this triangle (side ) and the measure of its opposite angle (angle ). We can determine the measure of the angle opposite side by subtracting the measures of the other two angles in the triangle from :

As the information we are working with consists of opposite pairs of side lengths and angle measures, we recognize the need for the law of sines:

Substituting , , and , we have

We solve this equation to find by multiplying both sides by :

We are now able to substitute , , and into the trigonometric formula for the area of a triangle:

To find the area of the circle, we need to determine its radius. This circle is in fact the circumcircle of triangle as it passes through all three of the triangleβs vertices. We recall the connection between the law of sines ratio and the radius of the circumcircle:

Using the length of side and the measure of angle ,we can form an equation:

Solving for gives

Hence, the area of the circle is as follows:

Finally, we subtract the area of triangle from the area of the circumcircle:

The shaded area, to the nearest square centimetre, is 187 cm2.

Let us finish by recapping some key points from this explainer.

### Key Points

• For a triangle , as shown in the figure below, the law of sines states that The law of cosines states that
• The law of sines and the law of cosines can be applied to problems in real-world contexts to calculate unknown lengths and angle measures in non-right triangles. These questions may take a variety of forms including worded problems, problems involving directions, and problems involving other geometric shapes.
• If we are not given a diagram, our first step should be to produce a sketch using all the information given in the question.
• The law we use depends on the combination of side lengths and angle measures we are given. We may have a choice of methods or we may need to apply both the law of sines and the law of cosines or the same law multiple times within the same problem.
• We can combine our knowledge of the laws of sines and cosines with other geometric results, such as the trigonometric formula for the area of a triangle,
• The law of sines is related to the diameter of a triangleβs circumcircle. For any triangle , the diameter of its circumcircle is equal to the law of sines ratio: