Law of Sines

Law of Sines

Before addressing the main topic of this article, let us first look at the area of a triangle in a new light. Following this, we aim to bridge the idea of the Law of Sines along with the area of a triangle and the Sine Ratio. You shall see that these concepts come together quite perfectly! Let us begin by recalling the Sine Ratio.

Recap: The Sine Ratio

Below is a right-angle triangle with an angle θ.

Right-angle triangle, Aishah Amri - StudySmarter Originals

Recall that the sine of an angle is found by dividing the opposite by the hypotenuse. This is the Sine Ratio.

$\sin \theta =\frac{opposite}{hypotenuse}$

Finding the Area of a Triangle

Furthermore, recall that the area of a triangle is given by the formula below.

$Area△=\frac{1}{2}\times height\times base$

Now that we have established the Sine Ratio and area of a triangle, let us observe their relationship. Consider the triangle below.

Area of a triangle, Aishah Amri - StudySmarter Originals

We can find the area of this triangle given its base and height. Using the formula, the area of the triangle above is given by:

Suppose that the height here was unknown. However, we are still told to find the area of this triangle. In fact, we can do so given the angle of A and the length of side b. To determine the height, we shall use the Sine Ratio for angle A.

$\sin A=\frac{h}{b}$

Hence, $h=b\sin A$.

Now that we have an expression for h, let us then substitute this into the initial formula for the area of this triangle.

$Area△ABC=\frac{1}{2}bc\sin A$

Similarly, we can find the area of this triangle using other variations of this formula. Now, why would there be different ways to express the formula above? Notice that the previous formula is written in terms of angle A and sides b and c. Say we have a different triangle and we are not given not the information on those measures. Instead, we are given the measures of other sides or angles. In this case, we need to solve them according to their accompanying pair of sides and corresponding angles. These other two variations are shown below.

$$\begin{gathered} Area△ABC=\frac{1}{2}ac\sin B \\ Area△ABC=\frac{1}{2}ab\sin C \end{gathered}$$

Let us look at an example.

The Law of Sines

So where does the Law of Sines come from? Given our ideas above, we have essentially built a foundation for deriving the Law of Sines. First of all, notice that all the area formulas for triangle ABC above describe the area of the same triangle.

Derivation of the Law of Sines, Aishah Amri - StudySmarter Originals

This means that the right-hand side for all three of these expressions equates to the same value. With that in mind, let us set these areas equal to each other.

$\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B=\frac{1}{2}ab\sin C$

Now, by dividing this entire expression by $\frac{1}{2}abc$ and simplifying it, we obtain the Law of Sines.

$$\begin{gathered} \Rightarrow \frac{\frac{1}{2}bc\sin A}{\frac{1}{2}abc}=\frac{\frac{1}{2}ac\sin B}{\frac{1}{2}abc}=\frac{\frac{1}{2}ab\sin C}{\frac{1}{2}abc} \\ \Rightarrow \frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c} \end{gathered}$$

The Law of Sines can be written as three separate equations, as below.

$$\begin{gathered} \frac{sinA}{a}=\frac{sinB}{b} \\ \frac{sinB}{b}=\frac{sinC}{c} \\ \frac{sinA}{a}=\frac{sinC}{c} \end{gathered}$$

Application of the Law of Sines

We can apply the Law of Sines for any triangle given the measures of two cases:

1. The value of two angles and any side

2. The value of two sides and an angle opposite one of them

Let us look at some worked examples that apply the Law of Sines.

Solving a Triangle Given Two Angles and a Side

Solving a Triangle Given Two Sides and an Angle

Solutions of a Triangle

A triangle may or may not have a unique solution given the value of two sides and an angle opposite one of them. To identify the number of solutions a triangle may have, it is important to analyze the angle provided and the lengths of the given sides of a triangle before applying the Law of Sines. In regards to the given angle, we must identify whether it is a right angle, an acute angle or an obtuse angle. There are three cases to consider for a triangle given the value of two sides and an angle opposite one of them:

1. No solutions = No triangle exists

2. One solution = Exactly one triangle exists

3. Two solutions = Two triangles exist

Suppose we are provided with the values of angle A and sides a and b for a given triangle. The table below summarizes the criteria for each case of a given triangle.

One Solution

No Solutions

Two Solutions

Real-World Example Involving Law of Sines

The light from a beacon of a vessel revolves clockwise at a steady rate of one revolution per minute. The beam strikes a point on the shore that is 1200 feet from the vessel. Four seconds later, the light strikes a point 575 feet further down the shore. How far is the vessel from the shore?

Solution

Let us begin by sketching the layout of this problem.

Boat example, Aishah Amri - StudySmarter Originals

The beacon makes one revolution every 60 seconds. Thus, the angle through which the light revolves in 4 seconds is

$\frac{4}{60}\left({360}^o\right)={24}^o$

To find angle X, we shall apply the Law of Sines.

$$\begin{gathered} \frac{\sin X}{1200}=\frac{\sin \left(24\right)}{575} \\ \Rightarrow \sin X=\frac{1200\sin \left(24\right)}{575} \\ \Rightarrow X={\sin }^{-1}\left(\frac{1200\sin \left(24\right)}{575}\right) \\ \Rightarrow X\approx 58.{09}^o(correcttotwodecimalplaces) \end{gathered}$$

Now that we have angle X, we can use this to find angle Y.

$$\begin{gathered} X+A={90}^o(anglesXandAarecomplementary) \\ \Rightarrow 58.{09}^o+(Y+{24}^o)\approx {90}^o(\sin ceA=Y+{24}^o) \\ \Rightarrow Y\approx {90}^o-58.{09}^o-{24}^o \\ \Rightarrow Y\approx 7.{91}^o \end{gathered}$$

Finally, we can determine the distance from the beacon to the shore by calculating d. We will use the Cosine Ratio here.

$$\begin{gathered} \cos Y=\frac{adjacent}{hypotenuse}=\frac{AB}{AD} \\ \Rightarrow \cos (7.91)\approx \frac{d}{1200} \\ \Rightarrow d\approx 1200\cos (7.91) \\ \Rightarrow d\approx 1188.58(correcttotwodecimalplaces) \end{gathered}$$

Thus, the distance from the beacon to the shore is approximately 1188.58 feet.