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Law of Sines

Let me paint you a scenario. Say you are flying a kite that is exactly 9 feet off the ground. Meanwhile, you are standing 12 feet away from where the kite is flying. You are curious to know what is the angle of elevation of the kite you are flying. Drawing this out on paper, you find that this problem creates a triangle as shown below.

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Fact Checked Content
Last Updated: 05.05.2022
12 min reading time

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Real-world problem, Aishah Amri - Vaia Originals

So, how would you determine this angle? Now, this is where the Law of Sines come into play! We can indeed use this concept to find this angle. Not only that, but we could even find the diagonal distance between you and the kite. Isn't that pretty nifty? In this topic, we shall explore the Law of Sines and observe its applications when solving triangles.

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 - Vaia Originals

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

$\sin θ = \frac{o p p o s i t e}{h y p o t e n u s e}$

Finding the Area of a Triangle

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

$A r e a △ = \frac{1}{2} \times h e i g h t \times b a s e$

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 - Vaia 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:

A r e a △ A B C = \frac{1}{2} c h

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.

$A r e a △ A B C = \frac{1}{2} b c \sin A$

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.

$A r e a △ A B C = \frac{1}{2} a c \sin B A r e a △ A B C = \frac{1}{2} a b \sin C$

Let us look at an example.

Find the area of a triangle if A = 31^o, b = 22 cm and c = 18 cm.

Solution

Let us begin by sketching this triangle.

Example 1, Aishah Amri - Vaia Originals

By the given formula, the area of this triangle is given by:

$A r e a △ A B C = \frac{1}{2} b c \sin A \Rightarrow A r e a △ A B C = \frac{1}{2} \times 22 \times 18 \times \sin (31) \Rightarrow A r e a △ A B C \approx 101.98 c m^{2} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, the area of this triangle is approximately 101.98 cm².

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 - Vaia 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.

$\frac{1}{2} b c \sin A = \frac{1}{2} a c \sin B = \frac{1}{2} a b \sin C$

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

$\Rightarrow \frac{\frac{1}{2} b c \sin A}{\frac{1}{2} a b c} = \frac{\frac{1}{2} a c \sin B}{\frac{1}{2} a b c} = \frac{\frac{1}{2} a b \sin C}{\frac{1}{2} a b c} \Rightarrow \frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c}$

Law of Sines: For any triangle ABC, $\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c}$ .

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

$\frac{s i n A}{a} = \frac{s i n B}{b} \frac{s i n B}{b} = \frac{s i n C}{c} \frac{s i n A}{a} = \frac{s i n C}{c}$

Application of the Law of Sines

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

The value of two angles and any side
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

Given the triangle below, find the angle of A and the length of a and c.

Example 2, Aishah Amri - Vaia Originals

Solution

We are given the angles B = 80^o and C = 40^o and the side b = 14 cm.

Here, we have the value of two angles and a side.

We begin by finding angle A. Recall that the sum of the interior angles of any triangle is equal to 180^o.

$A + B + C = 180^{o} \Rightarrow A = 180^{o} - B - C \Rightarrow A = 180^{o} - 80^{o} - 40^{o} \Rightarrow A = 60^{o}$

Thus, angle A is equal to 60^o. We shall now use the Law of Sines to find the lengths of a and c.

$\frac{\sin A}{a} = \frac{\sin B}{b} \Rightarrow \frac{\sin (60)}{a} = \frac{\sin (80)}{14} \Rightarrow a = \frac{14 \sin (60)}{\sin (80)} \Rightarrow a \approx 12.31 c m (c o r r e c t t o t w o d e c i m a l p l a c e s) \frac{\sin C}{c} = \frac{\sin B}{b} \Rightarrow \frac{\sin (40)}{c} = \frac{\sin (80)}{14} \Rightarrow c = \frac{14 \sin (40)}{\sin (80)} \Rightarrow c \approx 9.14 c m (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, a and c are approximately 12.31 cm and 9.14 cm respectively.

Solving a Triangle Given Two Sides and an Angle

Given the triangle below, find the angle of B and C and the length of b.

Example 3, Aishah Amri - Vaia Originals

Solution

We are given the angle A = 27^o and sides a = 10 cm and c = 12 cm.

Here, we have the value of two sides and an angle opposite one of them.

We begin by evaluating angle C using the Law of Sines below.

$\frac{\sin A}{a} = \frac{\sin C}{c} \Rightarrow \frac{\sin (27)}{10} = \frac{\sin C}{12} \Rightarrow \sin C = \frac{12 \sin (27)}{10} \Rightarrow C = \sin^{- 1} (\frac{12 \sin (27)}{10}) \Rightarrow C \approx 33 . 01^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, angle C is approximately 33.01^o. To find angle B, we shall subtract the value of angles A and C from 180^o.

$A + B + C = 180^{o} \Rightarrow B = 180^{o} - A - C \Rightarrow B \approx 180^{o} - 27^{o} - 33 . 01^{o} \Rightarrow B \approx 119 . 99^{o} c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, angle B is approximately 119.99^o. Finally, we can find the length of b by using the Law of Sines.

$\frac{\sin A}{a} = \frac{\sin B}{b} \Rightarrow \frac{\sin (27)}{10} = \frac{\sin (119.99)}{b} \Rightarrow b = \frac{10 \sin (119.99)}{\sin (27)} \Rightarrow b \approx 19.08 c m (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, b is approximately 19.08 cm.

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:

No solutions = No triangle exists
One solution = Exactly one triangle exists
Two solutions = Two triangles exist

If two solutions for a triangle exist, it is called an ambiguous case.

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.

Number of Solutions	A is Acute A < 90^o	A is Right or ObtuseA ≥ 90^o
No Solutions	Glencoe McGraw-Hill, Algebra 2 (2008) a < b sin A	Glencoe McGraw-Hill, Algebra 2 (2008) a ≤ b
One Solution	Glencoe McGraw-Hill, Algebra 2 (2008) a = b sin A	Glencoe McGraw-Hill, Algebra 2 (2008) a > b
One Solution	Glencoe McGraw-Hill, Algebra 2 (2008) a ≥ b	Glencoe McGraw-Hill, Algebra 2 (2008) a > b
Two Solutions	Glencoe McGraw-Hill, Algebra 2 (2008) b > a > b sin A

One Solution

For triangle ABC where B = 95^o, b = 19, and c = 12, identify whether ABC has no solution, one solution, or two solutions.

Solution

Since angle B is obtuse and b > c, we must have one solution. The triangle ABC is sketched below.

Example 4, Aishah Amri - Vaia Originals

Let us now use this to find angles A and C and side a. Using the Law of Sines

$\frac{\sin B}{b} = \frac{\sin C}{c} \Rightarrow \frac{\sin (95)}{19} = \frac{\sin C}{12} \Rightarrow \sin C = \frac{12 \sin (95)}{19} \Rightarrow C = \sin^{- 1} (\frac{12 \sin (95)}{19}) \Rightarrow C \approx 38 . 99^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, angle C is approximately 38.99^o. We can thus find angle A as

$A = 180^{o} - B - C \Rightarrow A \approx 180^{o} - 95^{o} - 38 . 99^{o} \Rightarrow A \approx 46 . 01^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Angle A is approximately 46.01^o. Again, applying the Law of Sines

$\frac{\sin A}{a} = \frac{\sin B}{b} \Rightarrow \frac{\sin (46.01)}{a} = \frac{\sin (95)}{19} \Rightarrow a = \frac{19 \sin (46.01)}{\sin (95)} \Rightarrow a \approx 13.72 (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, the length of a is approximately 13.72 units.

No Solutions

For triangle ABC where B = 95^o, b = 10, and c = 12, determine whether ABC has no solution, one solution, or two solutions.

Solution

Since angle B is obtuse and b < c, there are no solutions. More precisely, no such triangle exists for these given dimensions.

Two Solutions

For triangle ABC where A = 44°, b = 19, and a = 14, specify whether ABC has no solution, one solution, or two solutions.

Solution

Here, angle A is acute. Furthermore, evaluating b sin A, we obtain:

$b \sin A = 19 \sin (44) \Rightarrow b \sin A \approx 13.2 (c o r r e c t t o o n e d e c i m a l p l a c e)$

Now compare this with the given value of a and b. Notice that b > a > b sin A and so we must have two solutions. The sketch of this triangle is shown below.

Example 5, Aishah Amri - Vaia Originals

Let us now use this to find the unknown angles and sides. There are two cases to consider here.

Case 1: B is Acute, B₁

Using the Law of Sines

$\frac{\sin A}{a} = \frac{\sin B_{1}}{b} \Rightarrow \frac{\sin (44)}{14} = \frac{\sin B_{1}}{19} \Rightarrow \sin B_{1} = \frac{19 \sin (44)}{14} \Rightarrow B_{1} = \sin^{- 1} (\frac{19 \sin (44)}{14}) \Rightarrow B_{1} \approx 70 . 52^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, angle B₁ is approximately 70.52^o. We can thus find angle C₁ as

$C_{1} = 180^{o} - A - B_{1} \Rightarrow C_{1} \approx 180^{o} - 44^{o} - 70 . 52^{o} \Rightarrow C_{1} \approx 65 . 48^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Again, applying the Law of Sines

$\frac{\sin A}{a} = \frac{\sin C_{1}}{c_{1}} \Rightarrow \frac{\sin (44)}{14} = \frac{\sin (65.48)}{c_{1}} \Rightarrow c_{1} = \frac{14 \sin (65.48)}{\sin (44)} \Rightarrow c_{1} \approx 18.34 (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, the length of c₁ is approximately 18.34 units.

Case 2: B is Obtuse, B₂

Notice that the triangle B₁C₃B₃is an isosceles triangle. Recall that the base angles of an isosceles triangle are always congruent. Therefore, angles B₁ and B₃are approximately 70.52^o.

Furthermore, triangles B₁C₁B₃ and AB₂C₂ are supplementary. Thus, B₂ can be found by 180^o - 70.52^o ≈ 109.48^o.

Another way to find B₂ is to identify an obtuse angle whose sine is also

\frac{19 \sin (44)}{14} \approx 0.94

We do this by subtracting the angle of B₁= 70.52^o found in Case 1 from 180^o.

Thus, B₂ is approximately 180^o - 70.52^o ≈ 109.48^o, as before. We can thus find angle C₂ as

$C_{2} = 180^{o} - A - B_{2} \Rightarrow C_{1} \approx 180^{o} - 44^{o} - 109 . 48^{o} \Rightarrow C_{1} \approx 26 . 52^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Using the Law of Sines once more

$\frac{\sin A}{a} = \frac{\sin C_{2}}{c_{2}} \Rightarrow \frac{\sin (44)}{14} = \frac{\sin (26.52)}{c_{2}} \Rightarrow c_{2} = \frac{14 \sin (26.52)}{\sin (44)} \Rightarrow c_{2} \approx 9.0 (c o r r e c t t o t w o d e c i m a l p l a c e s)$

Thus, the length of c₂ is approximately 9.0 units.

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 - Vaia Originals

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

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

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

$\frac{\sin X}{1200} = \frac{\sin (24)}{575} \Rightarrow \sin X = \frac{1200 \sin (24)}{575} \Rightarrow X = \sin^{- 1} (\frac{1200 \sin (24)}{575}) \Rightarrow X \approx 58 . 09^{o} (c o r r e c t t o t w o d e c i m a l p l a c e s)$

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

$X + A = 90^{o} (a n g l e s X a n d A a r e c o m p l e m e n t a r y) \Rightarrow 58 . 09^{o} + (Y + 24^{o}) \approx 90^{o} (\sin c e A = Y + 24^{o}) \Rightarrow Y \approx 90^{o} - 58 . 09^{o} - 24^{o} \Rightarrow Y \approx 7 . 91^{o}$

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

$\cos Y = \frac{a d j a c e n t}{h y p o t e n u s e} = \frac{A B}{A D} \Rightarrow \cos (7.91) \approx \frac{d}{1200} \Rightarrow d \approx 1200 \cos (7.91) \Rightarrow d \approx 1188.58 (c o r r e c t t o t w o d e c i m a l p l a c e s)$

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

Law of Sines - Key takeaways

The area of a triangle can be found by

$A r e a △ A B C = \frac{1}{2} b c \sin A A r e a △ A B C = \frac{1}{2} a c \sin B A r e a △ A B C = \frac{1}{2} a b \sin C$

The Law of Sines states that

$\frac{s i n A}{a} = \frac{s i n B}{b} \frac{s i n B}{b} = \frac{s i n C}{c} \frac{s i n A}{a} = \frac{s i n C}{c}$

We can use the Law of Sines to solve triangles when given
- The value of two angles and any side
- The value of two sides and an angle opposite one of them
Solutions of a Triangle
Number of Solutions A is Acute A is Right or Obtuse
0 a < b sin A
a ≤ b
1
a = b sin A
a ≥ b
a > b
2 b > a > b sin A

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Frequently Asked Questions about Law of Sines

What is an example of the law of sines?

The law of sines relates the sides of a triangle with the sine of their opposite angles

What is the concept behind the law of sines?

The concept behind the law of sines stems from the formula for the area of a triangle

What is the Law of Sines?

Given a triangle ABC, the law of sines states that Sin A/a = Sin B/b = Sin C/c

When can we use the law of sines?

We can use the law of sines given the value of two angles and any side or the value of two sides and an angle opposite one of them

What is the formula when solving the laws of sines?

The formula for the law of sines is Sin A/a = Sin B/b = Sin C/c

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