Select your language

Suggested languages for you:
Log In Start studying!
StudySmarter - The all-in-one study app.
4.8 • +11k Ratings
More than 3 Million Downloads
Free
|
|

All-in-one learning app

  • Flashcards
  • NotesNotes
  • ExplanationsExplanations
  • Study Planner
  • Textbook solutions
Start studying

Law of Sines

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.

Real-world problem, Aishah Amri - StudySmarter 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 - StudySmarter Originals

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

sin θ=oppositehypotenuse

Finding the Area of a Triangle

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

Area =12×height×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:

Area ABC=12ch.

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=hb

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=12bc 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.

Area ABC=12ac sin BArea ABC=12ab sin C

Let us look at an example.

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

Solution

Let us begin by sketching this triangle.

Example 1, Aishah Amri - StudySmarter Originals

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

Area ABC=12bc sin AArea ABC=12×22×18× sin (31)Area ABC101.98 cm2 (correct to two decimal places)

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 - 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.

12bc sin A=12ac sin B=12ab sin C

Now, by dividing this entire expression by 12abc and simplifying it, we obtain the Law of Sines.

12bc sin A12abc=12ac sin B12abc=12ab sin C12abcsin Aa=sin Bb=sin Cc

Law of Sines: For any triangle ABC, sin Aa=sin Bb=sin Cc.

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

sin Aa=sin Bbsin Bb=sin Ccsin Aa=sin Cc

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

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

Example 2, Aishah Amri - StudySmarter Originals

Solution

We are given the angles B = 80o and C = 40o 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 180o.

A+B+C=180oA=180o-B-CA=180o-80o-40oA=60o

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

sin Aa=sin Bbsin (60)a=sin (80)14a=14 sin (60)sin (80)a12.31 cm (correct to two decimal places)sin Cc=sin Bbsin (40)c=sin (80)14c=14 sin (40)sin (80)c9.14 cm (correct to two decimal places)

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 - StudySmarter Originals

Solution

We are given the angle A = 27o 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.

sin Aa=sin Ccsin (27)10=sin C12sin C=12 sin (27)10C=sin-112 sin (27)10C 33.01o (correct to two decimal places)

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

A+B+C=180oB=180o-A-CB180o-27o-33.01oB119.99o correct to two decimal places)

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

sin Aa=sin Bbsin (27)10=sin (119.99)bb=10 sin (119.99)sin (27)b 19.08 cm (correct to two decimal places)

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:

  1. No solutions = No triangle exists

  2. One solution = Exactly one triangle exists

  3. 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 < 90o
A is Right or Obtuse
A ≥ 90o
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

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 = 95o, 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 - StudySmarter Originals

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

sin Bb=sin Ccsin (95)19=sin C12sin C=12 sin (95)19C=sin-112 sin (95)19C 38.99o (correct to two decimal places)

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

A=180o-B-CA180o-95o-38.99oA46.01o (correct to two decimal places)

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

sin Aa=sin Bbsin (46.01)a=sin (95)19a=19 sin (46.01)sin (95)a 13.72 (correct to two decimal places)

Thus, the length of a is approximately 13.72 units.

No Solutions

For triangle ABC where B = 95o, 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)b sin A13.2 (correct to one decimal place)

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 - StudySmarter 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

sin Aa=sin B1bsin (44)14=sin B119sin B1=19 sin (44)14B1=sin-119 sin (44)14B1 70.52o (correct to two decimal places)

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

C1=180o- A-B1C1180o- 44o - 70.52oC165.48o (correct to two decimal places)

Again, applying the Law of Sines

sin Aa=sin C1c1sin (44)14=sin (65.48)c1c1=14 sin (65.48)sin (44)c1 18.34 (correct to two decimal places)

Thus, the length of c1 is approximately 18.34 units.

Case 2: B is Obtuse, B2

Notice that the triangle B1 C3 B3 is an isosceles triangle. Recall that the base angles of an isosceles triangle are always congruent. Therefore, angles B1 and B3 are approximately 70.52o.

Furthermore, triangles B1 C1 B3 and A B2 C2 are supplementary. Thus, B2 can be found by 180o - 70.52o ≈ 109.48o.

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

19 sin (44)140.94We do this by subtracting the angle of B1 = 70.52o found in Case 1 from 180o.

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

C2=180o-A-B2C1180o- 44o - 109.48oC126.52o (correct to two decimal places)

Using the Law of Sines once more

sin Aa=sin C2c2sin (44)14=sin (26.52)c2c2=14 sin (26.52)sin (44)c2 9.0 (correct to two decimal places)

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 - StudySmarter Originals

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

460(360o)=24o

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

sin X1200=sin (24)575sin X=1200 sin (24)575X=sin-11200 sin (24)575X58.09o (correct to two decimal places)

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

X+A=90o (angles X and A are complementary)58.09o+(Y+24o)90o (since A=Y+24o)Y90o-58.09o-24oY7.91o

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

cos Y=adjacenthypotenuse=ABADcos (7.91)d1200d1200 cos (7.91)d1188.58 (correct to two decimal places)

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

Area ABC=12bc sin AArea ABC=12ac sin BArea ABC=12ab sin C

  • The Law of Sines states that

sin Aa=sin Bbsin Bb=sin Ccsin Aa=sin Cc

  • 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 SolutionsA is AcuteA is Right or Obtuse
    0a < b sin A

    a ≤ b

    1

    a = b sin A

    a ≥ b

    a > b
    2b > a > b sin A

Frequently Asked Questions about Law of Sines

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

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 

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

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

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

Final Law of Sines Quiz

Question

When can you apply the Law of Sines?

Show answer

Answer

We can use the Law of Sines to solve triangles when given  

  1. The value of two angles and any side
  2. The value of two sides and an angle opposite one of them

Show question

Question

Given a triangle ABC with angle A and sides a and b, if angle A is acute and a < b sin A, how many solutions does the triangle have?

Show answer

Answer

No solutions

Show question

Question

Given a triangle ABC with angle A and sides a and b, if angle A is acute and a = b sin A, how many solutions does the triangle have?

Show answer

Answer

One solution

Show question

Question

Given a triangle ABC with angle A and sides a and b, if angle A is acute and a ≥  b, how many solutions does the triangle have?

Show answer

Answer

One solution

Show question

Question

Given a triangle ABC with angle A and sides a and b, if angle A is acute and b > a > b sin A, how many solutions does the triangle have? 

Show answer

Answer

Two solutions

Show question

Question

Given a triangle ABC with angle A and sides a and b, if angle A is obtuse and a ≤ b, how many solutions does the triangle have? 

Show answer

Answer

No solution

Show question

Question

Given a triangle ABC with angle A and sides a and b, if angle A is obtuse and a > b, how many solutions does the triangle have? 

Show answer

Answer

One solution

Show question

Question

What do you call a triangle with two solutions?

Show answer

Answer

An ambiguous case

Show question

60%

of the users don't pass the Law of Sines quiz! Will you pass the quiz?

Start Quiz

Discover the right content for your subjects

No need to cheat if you have everything you need to succeed! Packed into one app!

Study Plan

Be perfectly prepared on time with an individual plan.

Quizzes

Test your knowledge with gamified quizzes.

Flashcards

Create and find flashcards in record time.

Notes

Create beautiful notes faster than ever before.

Study Sets

Have all your study materials in one place.

Documents

Upload unlimited documents and save them online.

Study Analytics

Identify your study strength and weaknesses.

Weekly Goals

Set individual study goals and earn points reaching them.

Smart Reminders

Stop procrastinating with our study reminders.

Rewards

Earn points, unlock badges and level up while studying.

Magic Marker

Create flashcards in notes completely automatically.

Smart Formatting

Create the most beautiful study materials using our templates.

Sign up to highlight and take notes. It’s 100% free.