Lesson Explainer: Law of Sines Mathematics

In this explainer, we will learn how to apply the law of sines to find lengths and angle measures in non-right triangles.

We should already be familiar with how to apply the sine, cosine, and tangent ratios in right triangles. The law of sines is an extension of these trigonometric techniques to include a wider class of triangles.

Definition: The Law of sines

Consider triangle 𝐴𝐵𝐶, with corresponding sides of lengths 𝑎, 𝑏, and 𝑐. The convention when labeling such triangles is that each side is labeled using the lowercase letter corresponding to its opposite angle.

The law of sines states that 𝑎𝐴=𝑏𝐵=𝑐𝐶.sinsinsin

The reciprocal is also true: sinsinsin𝐴𝑎=𝐵𝑏=𝐶𝑐.

In other words, the law of sines tells us that the ratio between any side length and the sine of its opposite angle is the same for all three pairs of sides and angles within any triangle. In practice, we usually use only two parts of the ratio in our calculations, rather than all three. The first version of the law of sines, where the sides appear in the numerator, is most useful when calculating the length of a side, while the reciprocal version, where the sines of the angles appear in the numerator, is most useful when calculating an angle simply because less algebraic rearrangement is required to reach an answer.

As the law of sines describes the ratio between sides and the sines of their opposite angles, we can recognize the need for the law of sines by identifying that the information given in a question consists of pairs of sides and opposite angles in a non-right triangle.

Let us now demonstrate how to apply the law of sines to calculate an unknown length in a triangle.

Example 1: Using the Law of sines to Calculate an Unknown Length

For the given figure, 𝐴𝐵=3 and 𝐵𝐶=𝑎. Use the law of sines to work out 𝑎. Give your answer to two decimal places.

Answer

We will find it helpful to add the given side lengths to the diagram.

We can now see that the side length of 3 units is opposite the angle of 31, and the side length we wish to calculate, 𝑎, is opposite the angle of 64. We can therefore apply the law of sines: 𝑎𝐴=𝑐𝐶.sinsin

Substituting the values from the diagram and solving for 𝑎 gives 𝑎64=331𝑎=36431.sinsinsinsin

Evaluating and then rounding our answer to two decimal places as required by the question gives 𝑎=5.2355.24.

The length of 𝑎, to two decimal places, is 5.24 units.

In our previous example, we demonstrated how to calculate the length of a missing side using the law of sines. We will now consider how to apply a similar process to calculate the measure of an unknown angle.

Example 2: Using the Law of sines to Calculate an Unknown Angle in a Triangle

𝐴𝐵𝐶 is a triangle, where 𝑎=9, 𝑏=6, and 𝑚𝐴=58.1. Find 𝑚𝐵 to the nearest tenth of a degree.

Answer

When approaching this type of problem, we should begin by sketching the triangle using the known information as shown below (not to scale).

We see that the side of length 9 units is opposite the angle of 58.1, and the side of length 6 units is opposite the angle we need to calculate. As we are working with opposite pairs of sides and angles, we can apply the law of sines. since we are calculating an angle, we will use the version that has the sines of the angles in the numerator: sinsin𝐵𝑏=𝐴𝑎.

Substituting the values from our diagram into the law of sines and rearranging we obtain sinsinsinsin𝐵6=58.19𝐵=658.19.

We solve this equation for 𝐵 by applying the inverse sine function: 𝐵=658.19=(0.565)=34.47034.5.sinsinsin

The measure of angle 𝐵, to the nearest tenth of a degree, is 34.5.

We have now seen examples of how to calculate both the length of a side and the measure of an angle using the law of sines. We can summarize this process in the following steps.

How To: Calculating an Unknown Length or Angle Using the Law of Sines

  1. Make sure you have a diagram labeled clearly with all known sides and angles.
  2. Identify the opposite pairs of sides and angles in the diagram.
  3. Substitute the known sides and angles into the law of sines, ideally using the version that has the unknown you need to calculate in the numerator.
  4. Rearrange the equation to solve for the unknown.

When working with triangles, it is important to be aware that the longest side of a triangle is always opposite the angle with the largest measure and the shortest side is always opposite the angle with the smallest measure. We will now consider an example in which this knowledge is helpful.

Example 3: Using the Law of sines to Calculate the Shortest Side in a Triangle

𝐴𝐵𝐶 is a triangle, where 𝑚𝐴=461117, 𝑚𝐵=27446, and length 𝑎=21.4cm. Find the length of the shortest side of 𝐴𝐵𝐶 giving the answer to one decimal place.

Answer

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

The question asks for the length of the shortest side of this triangle. We do not need to calculate the lengths of both missing sides and then answer with whichever is shorter. Instead, we should recall that the shortest side in any triangle is opposite its smallest angle.

We know the measures of angles 𝐴 and 𝐵, and we can calculate the measure of angle 𝐶 using the angle sum of a triangle: 𝑚𝐶=18046111727446=1064357.

It may have been clear already, but we can now see explicitly that the smallest angle in the triangle is angle 𝐵, and therefore it is the length of side 𝑏 that we are looking to calculate. The information we are given consists of a side, 𝑎, and its opposite angle, 𝐴, together with the angle opposite the side we wish to calculate. We can therefore apply the law of sines: 𝑎𝐴=𝑏𝐵.sinsin

Substituting the two known angles and the one known side, we have 21.4(461117)=𝑏(27446).sinsin

Solving for 𝑏 gives 𝑏=21.4(27446)(461117)=13.50013.5.sinsin

To one decimal place, the length of the shortest side of triangle 𝐴𝐵𝐶 is 13.5 cm.

In some problems, we may need to combine the law of sines with other geometric properties, such as angle rules or right angle trigonometry. Let us consider an example in which we apply all of these techniques together.

Example 4: Using the Law of sines to Calculate the Length of a Side in a Composite Figure

In the given figure, 𝐵𝐶𝐷𝑌 is a rectangle and 𝐵 is a point on the straight line 𝐴𝐶. 𝐵𝐶=405m, 𝑚𝐷𝐴𝐶=21, and 𝑚𝑌𝐴𝐶=59. Find the length of 𝐷𝐶 giving the answer to the nearest metre.

Answer

Let us begin by adding the information given in the question to the diagram. We can separate the angle of 59 into 21, which is the measure of angle 𝐷𝐴𝐶, and the remaining 38.

As 𝐵𝐶𝐷𝑌 is a rectangle, all of its interior angles are 90. The length we want to calculate, 𝐷𝐶, is part of a right triangle 𝐷𝐴𝐶, none of whose full side lengths we know. We need at least one length in order to be able to calculate 𝐷𝐶, so let us consider instead the other triangle in the figure, triangle 𝐴𝑌𝐷. This triangle shares side 𝐴𝐷 with our right triangle, so we may be able to use it to work out this shared length.

We know or can find by calculation some more information about triangle 𝐴𝑌𝐷. Firstly, using alternate angles in parallel lines, we see that the measure of angle 𝑌𝐷𝐴 is the same as the measure of angle 𝐷𝐴𝐶, so it is 21. We can then calculate the measure of angle 𝐴𝑌𝐷 using the angle sum of a triangle: 𝑚𝐴𝑌𝐷=1803821=121.

Because 𝐵𝐶𝐷𝑌 is a rectangle, its opposite sides are equal. So, 𝑌𝐷=𝐵𝐶=405m.

We can now use the law of sines to calculate the length of side 𝐴𝐷. 𝐴𝐷 is opposite the angle measuring 121, and we have the length of 405 m opposite the angle of 38. Applying the law of sines gives 𝐴𝐷121=40538.sinsin

We can then solve for 𝐴𝐷 by multiplying by sin121 and evaluating: 𝐴𝐷=40512138=563.869.sinsin

We have now calculated the length of the shared side 𝐴𝐷, which is the hypotenuse of the right triangle 𝐴𝐶𝐷. In relation to the angle of 21, 𝐷𝐶 is the opposite side and we can therefore apply right triangle trigonometry using the sine ratio: sin21=𝐷𝐶𝐴𝐷=𝐷𝐶563.869.

Rearranging this equation for 𝐷𝐶 and rounding to the nearest metre, we obtain 𝐷𝐶=563.86921=202.072202.sin

The length of 𝐷𝐶, to the nearest metre, is 202 m.

It is important to appreciate that when applying the law of sines, we are working with a ratio. The values we use for the side lengths do not necessarily need to be the side lengths themselves, but they must be in the same ratio as the side lengths. Let us consider an example in which we cannot calculate the side lengths explicitly but can calculate their ratio.

Example 5: Using the Law of Sines to Find the Ratio between the Side Lengths of a Triangle

𝐴𝐵𝐶 is a triangle where 8𝐴=11𝐵=16𝐶sinsinsin. Find the ratio 𝑎𝑏𝑐.

Answer

Recall that the law of sines tells us that the ratio between the length of each side and the sine of its opposite angle is the same within any triangle: sinsinsin𝐴𝑎=𝐵𝑏=𝐶𝑐.

This looks a little like the equation we have been given for triangle 𝐴𝐵𝐶, but our equation has coefficients for each of the sine terms and no denominators. We therefore need to manipulate the equation we have been given to write it in a format that mimics the law of sines. To do so, we can divide the equation through by the lowest common multiple of 8, 11, and 16. As 8 is a factor of 16 and 11 and 16 are coprime, the lowest common multiple is 11×16 (i.e., 176). Dividing through by 11×16 gives 8𝐴11×16=11𝐵11×16=16𝐶11×16.sinsinsin

We can then simplify each part of the ratio by cancelling common factors to give sinsinsin𝐴22=𝐵16=𝐶11.

Comparing this to the general form of the law of sines, in which the denominators represent the side lengths, we see that the ratio 𝑎𝑏𝑐 is 221611.

Note that this is the ratio between the side lengths as opposed to necessarily the side lengths themselves. If the side lengths were 44, 32, and 22 units, respectively, then the ratio would be sinsinsin𝐴44=𝐵32=𝐶22, which could be simplified by multiplying each part of the ratio by 2 to give the same equation of sinsinsin𝐴22=𝐵16=𝐶11 as we saw earlier.

We have seen several examples of how we can apply the law of sines to calculate both side lengths and angle measures in non-right triangles. In fact, the law of sines is also true in right triangles. Let us consider a triangle 𝐴𝐵𝐶 with a right angle at 𝐴.

Then considering just two of the angle-side pairs, we have 𝑎90=𝑏𝐵.sinsin

Recalling that sin90=1, it simplifies to 𝑎=𝑏𝐵.sin

We then rearrange by first multiplying by sin𝐵 and then dividing by 𝑎 to give 𝑎𝐵=𝑏𝐵=𝑏𝑎.sinsin

Returning to the original triangle, we see that side 𝑎 is the hypotenuse and, in relation to angle 𝐵, side 𝑏 is the opposite. So, we have that the sine of angle 𝐵 is equal to the length of the opposite divided by the length of the hypotenuse. Therefore, the law of sines has reduced to the sine ratio in a right triangle. It is perfectly possible then to apply the law of sines in right triangles, but it is unnecessarily complicated as our equations will simplify to equations we could have obtained using right triangle trigonometry.

Let us finish by recapping some key points.

Key Points

  • The law of sines allows us to calculate unknown lengths and angles in non-right triangles when the information we are given consists of opposite pairs of sides and angles.
  • Using lowercase letters to represent sides and uppercase letters to represent the opposite angles, the law of sines states that 𝑎𝐴=𝑏𝐵=𝑐𝐶.sinsinsin
  • When calculating a side length, we should use this version as the side lengths are in the numerator.
  • The reciprocal is also true, and we should use this version when calculating the measure of an unknown angle: sinsinsin𝐴𝑎=𝐵𝑏=𝐶𝑐.
  • We may need to use the law of sines in combination with other skills such as angle rules and right triangle trigonometry.

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