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 𝐴𝐡𝐢, with corresponding sides of lengths π‘Ž, 𝑏, and 𝑐.

Definition: The Law of Sines

In a triangle 𝐴𝐡𝐢 as described above, the law of sines states that π‘Žπ΄=𝑏𝐡=𝑐𝐢.sinsinsin

The reciprocal is also true: sinsinsinπ΄π‘Ž=𝐡𝑏=𝐢𝑐.

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 𝐴𝐡𝐢 as described above, the law of cosines states that π‘Ž=𝑏+π‘βˆ’2𝑏𝑐𝐴.cos

The law of cosines can be rearranged to cos𝐴=𝑏+π‘βˆ’π‘Ž2𝑏𝑐.

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 𝑏 and 𝑐 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 π‘Ž, 𝑏, and 𝑐 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 83∘. Find the perimeter of the fence giving your answer to the nearest metre.

Answer

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: π‘Ž=𝑏+π‘βˆ’2𝑏𝑐𝐴.cos

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 𝑏 and 𝑐 represent the two known side lengths and 𝐴 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 𝑏=55, 𝑐=72, and 𝐴=83∘ into the law of cosines, we obtain π‘Ž=55+72βˆ’(2Γ—55Γ—72)83.∘cos

Evaluating and simplifying gives π‘Ž=3025+5184βˆ’792083=7243.794….∘cos

We solve for π‘Ž by square rooting. We can ignore the negative solution to our equation as we are solving to find a length: π‘Ž=√7243.794…=85.110….mm

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: perimeter=72+55+(85.110…)=212.110β€¦β‰ˆ212.

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 7√2 km east, and then he rode for another 21 km 45∘ south of east. Determine the magnitude and direction of the displacement, rounding the direction to the nearest minute.

Answer

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 𝐴, 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 45∘ south of east is an angle measured 45∘ 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 𝐴 and the endpoint 𝐢. The direction of displacement of point 𝐢 from point 𝐴 is southeast, and the size of this angle is the measure of angle 𝐡𝐴𝐢.

Let us consider triangle 𝐴𝐡𝐢, in which we are given two side lengths. We can calculate the measure of their included angle, angle 𝐴𝐡𝐢, by recalling that angles on a straight line sum to 180∘. Subtracting 45∘ from 180∘ gives π‘šβˆ π΄π΅πΆ=180βˆ’45=135.∘∘∘

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: 𝐴𝐢=𝐴𝐡+π΅πΆβˆ’(2×𝐴𝐡×𝐡𝐢×𝐴𝐡𝐢).cos

Substituting 𝐴𝐡=7√2, 𝐡𝐢=21, and π‘šβˆ π΄π΅πΆ=135∘ gives 𝐴𝐢=ο€»7√2+21βˆ’ο€»2Γ—7√2Γ—21Γ—135=833.∘cos

We solve for 𝐴𝐢 by square rooting: 𝐴𝐢=√833=7√17.kmkm

We add the information we have calculated to our diagram.

To calculate the measure of angle 𝐡𝐴𝐢, 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 orange.

We will apply the law of sines, using the version that has the sines of the angles in the numerator: sinsinπ΄π‘Ž=𝐡𝑏.

Substituting π‘Ž=21, 𝑏=7√17, and 𝐡=135∘ gives sinsin𝐴21=1357√17.∘

Multiplying each side of this equation by 21 leads to sinsin𝐴=211357√17.∘

We solve for 𝐴 by applying the inverse sine function: 𝐴=ο€Ώ211357√17=30.963….sinsin∘∘

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 30∘ 58β€².

The magnitude of the displacement is 7√17 km and the direction, to the nearest minute, is 30∘ 58β€² 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

𝐴𝐡𝐢𝐷 is a quadrilateral where 𝐴𝐡=31cm, 𝐡𝐢=40cm, 𝐢𝐷=43cm, 𝐷𝐴=32cm, and π‘šβˆ π΄=64∘. Find π‘šβˆ πΆπ΅π· giving the answer to the nearest degree.

Answer

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

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

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

The side 𝐡𝐷 is shared with the other triangle in the diagram, triangle 𝐴𝐡𝐷, so let us now consider this triangle. We are given two side lengths (𝐴𝐷 and 𝐴𝐡) 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 𝐡𝐷=𝐴𝐡+π΄π·βˆ’(2×𝐴𝐡×𝐴𝐷×𝐴).cos

Substituting 𝐴𝐡=31, 𝐴𝐷=32, and 𝐴=64∘ gives 𝐡𝐷=31+32βˆ’(2Γ—31Γ—32Γ—64)=1985βˆ’198464=1115.271….∘∘coscos

We solve for 𝐡𝐷 by square rooting, ignoring the negative solution as 𝐡𝐷 represents a length: 𝐡𝐷=√(1115.271…)=33.395….cmcm

We add the length of 𝐡𝐷 to our diagram.

We now know the lengths of all three sides in triangle 𝐡𝐢𝐷, and so we can calculate the measure of any angle. Recall the rearranged form of the law of cosines: cos𝐴=𝑏+π‘βˆ’π‘Ž2𝑏𝑐, where 𝑏 and 𝑐 are the side lengths which enclose the angle we wish to calculate and π‘Ž is the length of the opposite side. In our figure, the sides which enclose angle 𝐡 are of lengths 40 cm and 33.395… cm, and the opposite side is of length 43 cm. Substituting these values into the law of cosines, we have cos(𝐢𝐡𝐷)=40+(33.395…)βˆ’432Γ—40Γ—(33.395…)=0.324….

We solve for angle 𝐢𝐡𝐷 by applying the inverse cosine function: ∠𝐢𝐡𝐷=(0.324…)=71.080β€¦β‰ˆ71.cos∘

The measure of angle 𝐢𝐡𝐷, to the nearest degree, is 71∘.

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 triangle’s circumcircle. Recall that the circumcircle of a triangle is the circle that passes through all three of the triangle’s vertices, as shown in the figure below.

Definition: The Law of Sines and Circumcircle Connection

For any triangle 𝐴𝐡𝐢, the diameter of its circumcircle is equal to the law of sines ratio: π‘Žπ΄=𝑏𝐡=𝑐𝐢=2π‘Ÿ.sinsinsin

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

𝐴𝐡𝐢 is a triangle where π‘šβˆ π΄=152∘ and 𝐡𝐢=11cm. Find the area of the circumcircle giving the answer to the nearest square centimetre.

Answer

We recall first that the circumcircle of a triangle is the circle passing through all three of the triangle’s vertices. To calculate the area of any circle, we use the formula area=πœ‹π‘ŸοŠ¨, 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 𝐡 and 𝐢, which will be opposite the third angle of the triangle, angle 𝐴. An alternative way of denoting this side is π‘Ž.

We recall the connection between the law of sines ratio and the radius of the circumcircle: π‘Žπ΄=𝑏𝐡=𝑐𝐢=2π‘Ÿ.sinsinsin

Substituting π‘Ž=11 and 𝐴=152∘ into the first part of this ratio and ignoring the middle two parts that are not required, we have 11152=2π‘Ÿ.sin∘

We solve this equation to determine the radius of the circumcircle: π‘Ÿ=112152=11.715….sincm∘

We are now able to calculate the area of the circumcircle: area=πœ‹π‘Ÿ=πœ‹Γ—(11.715…)=431.178β€¦β‰ˆ431.

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 areasin=12π‘Žπ‘πΆ, where π‘Ž and 𝑏 represent the lengths of two of the triangle’s sides and 𝐢 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 π‘šβˆ πΆπ΄π΅=77∘, π‘šβˆ π΅πΆπ΄=57∘, and 𝐢𝐡=19cm. Give the answer to the nearest square centimetre.

Answer

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 π‘Ž, 𝑏, and 𝑐.

The shaded area can be calculated as the area of triangle 𝐴𝐡𝐢 subtracted from the area of the circle: shadedareaareaofcircleareaoftriangle=βˆ’π΄π΅πΆ.

We recall the trigonometric formula for the area of a triangle, using two sides and the included angle: areaoftrianglesin=12π‘Žπ‘πΆ.

In order to compute the area of triangle 𝐴𝐡𝐢, we first need to calculate the length of side 𝑏. We already know the length of a side in this triangle (side π‘Ž) and the measure of its opposite angle (angle 𝐴). We can determine the measure of the angle opposite side 𝑏 by subtracting the measures of the other two angles in the triangle from 180∘: π‘šβˆ π΅=180βˆ’77βˆ’57=46.∘∘∘∘

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: 𝑏𝐡=π‘Žπ΄.sinsin

Substituting π‘Ž=19, 𝐴=77∘, and 𝐡=46∘, we have 𝑏46=1977.sinsin∘∘

We solve this equation to find 𝑏 by multiplying both sides by sin46∘: 𝑏=194677=14.026….sinsincm∘∘

We are now able to substitute π‘Ž=19, 𝑏=14.026…, and 𝐢=57∘ into the trigonometric formula for the area of a triangle: areaoftrianglesincm𝐴𝐡𝐢=12Γ—19Γ—(14.026…)Γ—57=111.758….∘

To find the area of the circle, we need to determine its radius. This circle is in fact the circumcircle of triangle 𝐴𝐡𝐢 as it passes through all three of the triangle’s vertices. We recall the connection between the law of sines ratio and the radius of the circumcircle: π‘Žπ΄=𝑏𝐡=𝑐𝐢=2π‘Ÿ.sinsinsin

Using the length of side π‘Ž and the measure of angle 𝐴,we can form an equation: 1977=2π‘Ÿ.sin∘

Solving for π‘Ÿ gives π‘Ÿ=19277=9.749….sincm∘

Hence, the area of the circle is as follows: areaofcirclecm=πœ‹Γ—(9.749…)=298.640….

Finally, we subtract the area of triangle 𝐴𝐡𝐢 from the area of the circumcircle: shadedareaareaofcircleareaoftriangle=βˆ’π΄π΅πΆ=(298.640…)βˆ’(111.758…)=186.882β€¦β‰ˆ187.

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 𝐴𝐡𝐢, as shown in the figure below, the law of sines states that π‘Žπ΄=𝑏𝐡=𝑐𝐢.sinsinsin The law of cosines states that π‘Ž=𝑏+π‘βˆ’2𝑏𝑐𝐴.cos
  • 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, areasin=12π‘Žπ‘πΆ.
  • The law of sines is related to the diameter of a triangle’s circumcircle. For any triangle 𝐴𝐡𝐢, the diameter of its circumcircle is equal to the law of sines ratio: π‘Žπ΄=𝑏𝐡=𝑐𝐢=2π‘Ÿ.sinsinsin

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