13. Solving Triangles Using the Law of Sines
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Chapter 2
13.
Solving Triangles Using the Law of Sines
| Student Learning Objectives: |
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Why
Grasping how to solve triangles with the law of sines is essential for numerous real-world applications. This mathematical principle shines when we possess a triangle with a known angle and side length, or when two angles are known without the side between them. By employing the law of sines, we can unveil the unknown sides or angles. This becomes invaluable in professions like architecture, engineering, and navigation. Whether you're determining a building's height with a given angle and distance or setting a sailing course, the law of sines proves to be a crucial asset.
Lesson Settings & Tools
| | 11 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Triangles Using the Law of Sines
Slide of 11
In this lesson, a new relation between side lengths and angle measures of a triangle will be derived and used.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
- Triangle Angle Sum Theorem
- Right triangles
- Pythagorean Theorem
- Trigonometric ratios
- Area of a triangle and area of a triangle using sine
Try your knowledge on these topics.
Find the value of x. If needed, write the answer to the nearest tenth.
a 
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b 
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c 
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Calculate the area of the following triangles. If needed, write the answer correct to the nearest integer.
d 
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e 
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Challenge
Investigate Triangles With Two Sides and an Angle Known
Can the measure of ∠ B be found with the given information? If so, what is its value?
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Explore
Relationship Between a Triangle's Angles and Their Opposite Sides
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Discussion
Law of Sines
One conclusion obtained in the previous exploration can be generalized to all triangles.
For any triangle, the ratio of the sine of an angle to its opposite side is constant.
sin A/a=sin B/b=sin C/c
An alternative way to write the Law of Sines is involving the ratio of a side length to the sine of its opposite angle.
a/sin A=b/sin B=c/sin C
Proof
Acute Triangles
Consider an acute triangle with height h drawn from one of its vertices. Because h is perpendicular to the base, the original triangle is split into two right triangles.
In these two right triangles, ∠ A and ∠ B are both opposite angles to h. Therefore, by applying the definition of the sine ratio to ∠ A and ∠ B, it is possible to relate the sine of these angles with the side lengths a and b.
Next, h can be isolated and written in terms of the corresponding side length and angle, for both right triangles. sin A = h/b &⇔ h = bsin A [1em] sin B = h/a &⇔ h = asin B By the Transitive Property of Equality, it can be stated that bsin A and a sin B are equal. h = bsin A h = asin B ⇒ bsin A=asin B Finally, the obtained equation can be rearranged to obtain the desired result.
bsin A = a sin B
▼
Simplify
DivEqn
.LHS /ab.=.RHS /ab.
bsin A/ab = asin B/ab
CancelCommonFac
Cancel out common factors
bsin A/ab = asin B/ab
SimpQuot
Simplify quotient
sin A/a = sin B/b
By following the same procedure but drawing the height from vertex A, it can be shown that sin Bb = sin Cc. Putting these two results together, the Law of Sines is obtained.
sin A/a=sin B/b=sin C/c
Obtuse Triangles
An obtuse triangle will now be considered.
This proof is very similar to the proof for acute triangles, but it uses an interior and an exterior height. First, the height h_1 from the vertex where the obtuse angle is located will be drawn. Just as before, this generates two right triangles.
The sine ratio will be written for these right triangles.
In the equations, h_1 can be isolated. sin A = h_1/c &⇔ h_1 = csin A [1em] sin C = h_1/a &⇔ h_1 = asin C By the Transitive Property of Equality, it can be stated that csin A and a sin C are equal. h_1 = csin A h_1 = asin C ⇒ csin A=asin C The obtained equation can be rearranged to obtain the desired result.
csin A=asin C
▼
Simplify
DivEqn
.LHS /ac.=.RHS /ac.
csin A/ac = asin C/ac
CancelCommonFac
Cancel out common factors
csin A/ac = asin C/ac
SimpQuot
Simplify quotient
sin A/a = sin C/c
Next, the exterior height h_2 from vertex C will be drawn. Let E be the point of intersection of this height and the extension of AB.
Here, ∠ ABC and ∠ CBE form a linear pair and are therefore supplementary angles. Because the sine of supplementary angles is the same, the sine of ∠ ABC equals the sine of ∠ CBE. Also, using the sine ratio on △ BCE, it can be stated that the sine of ∠ CBE is the ratio of h_2 to a. sin ABC = sin CBE and sin CBE = h_2/a By the Transitive Property of Equality, the sine of ∠ ABC can be written in terms of h_2 and a. sin ABC = sin CBE sin CBE = h_2/a ⇓ sin ABC=h_2/a Now △ ACE will be considered. By using the sine ratio, it follows that the sine of ∠ A is the ratio of h_2 to b.
Now, h_2 can be written in terms of sin ABC and a, and in terms of sin A and b. sin ABC = h_2/a &⇔ h_2 = asin ABC [1em] sin A = h_2/b &⇔ h_2 = bsin A The Transitive Property of Equality can be used one more time. h_2 = asin ABC h_2 = bsin A ⇓ asin ABC = bsin A If only △ ABC is considered, then ∠ ABC can be named as ∠ B.
This allows sin ABC to be written as sin B. Therefore, asin B=bsin A, and this equation can be rearranged to obtain the desired formula.
asin B=bsin A
▼
Simplify
DivEqn
.LHS /ab.=.RHS /ab.
asin B/ab=bsin A/ab
CancelCommonFac
Cancel out common factors
asin B/ab=bsin A/ab
SimpQuot
Simplify quotient
sin B/b=sin A/a
Finally, this result can be combined with the previous determination sin Aa = sin Cc to derive the Law of Sines.
sin A/a=sin B/b=sin C/c
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Example
Using the Law of Sines To Find the Side of a Triangle
As previously stated, the Law of Sines can be used to find side lengths of a triangle.
Magdalena will go golfing after school. The locations of her school, the golf course, and her house form a triangular shape. She knows the measures of two of the triangle's angles, and she knows the distance from her house to the school is 3 kilometers.
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Hint
Start by labeling the sides and the angles of the triangle.
Solution
The sides of the triangle will be labeled a, b, and c. Similarly, the angles will be labeled A, B, and C.
By the Law of Sines, the ratio of a side length to the sine of its opposite angle is constant. With this information, a proportion that relates A, a, C, and c can be written. a/sin A=c/sin C The values a=3, A=55^(∘), and C=60^(∘) can be substituted into this formula. Then the resulting equation can be solved for c, the distance between Magdalena's house and the golf course.
a/sin A=c/sin C
SubstituteValues
Substitute values
3/sin 55^(∘)=c/sin 60^(∘)
▼
Solve for c
MultEqn
LHS * sin 60^(∘)=RHS* sin 60^(∘)
3/sin 55^(∘) (sin 60^(∘))=c
RearrangeEqn
Rearrange equation
c=3/sin 55^(∘) (sin 60^(∘))
UseCalc
Use a calculator
c=3.171665...
RoundSigDig
Round to 3 significant digit(s)
c≈ 3.17
It was found that the distance between the golf course and Magdalena's house is 3.17 kilometers, rounded to three significant figures.
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Pop Quiz
Practice Finding Sides Using the Law of Sines
In the following applet, x represents the side length of a triangle. By using the Law of Sines and, if needed, the Triangle Angle Sum Theorem, find the value of x. Round the answer to two decimal places.
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Example
Using the Law of Sines To Find Angles of a Triangle
The Law of Sines can also be used to find angle measures of a triangle.
Emily will go backpacking across South America! She will visit Buenos Aires, Santiago, and Asunción, among other cities. Emily knows that the distances from Buenos Aires to Santiago and Asunción are 1140 and 1040 kilometers, respectively. She also knows that the angle whose vertex is located at Santiago and whose sides pass through Buenos Aires and Asunción measures 44^(∘).
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Hint
Start by labeling the sides and the angles of the triangle.
Solution
The angles of the triangle will be labeled A, B, and S. Similarly, the sides will be labeled a, b, and s.
By the Law of Sines, the ratio of the sine of an angle to the length of its opposite side is constant. With this information, a proportion that relates A, a, S, and s can be written. sin S/s=sin A/a In the above formula, a=1140, s=1040, and S=44^(∘) can be substituted. Then the resulting equation can be solved for A, the measure of the angle formed at Asunción.
sin S/s=sin A/a
SubstituteValues
Substitute values
sin 44^(∘)/1040=sin A/1140
▼
Solve for A
MultEqn
LHS * 1140=RHS* 1140
sin 44^(∘)/1040(1140)=sin A
RearrangeEqn
Rearrange equation
sin A = sin 44^(∘)/1040(1140)
sin^(-1)(LHS) = sin^(-1)(RHS)
A=sin ^(- 1)(sin 44^(∘)/1040(1140))
UseCalc
Use a calculator
A=49.592410... ^(∘)
RoundInt
Round to nearest integer
A≈ 50^(∘)
Rounded to the nearest degree, it was found that the angle formed at Asunción measures 50^(∘).
Finally, to find the measure of the angle at Buenos Aires, the Triangle Angle Sum Theorem can be used. 44^(∘)+ 50^(∘)+B=180^(∘) ⇕ B= 86^(∘) This information can be added to the diagram.
Finally, the Law of Sines can be used again to find the distance between Santiago and Asunción. To do so, recall that the ratio of the length of a side to the sine of its opposite angle is constant. b/sin B=s/sin S Next, s=1040, S=44^(∘), and B=86^(∘) will be substituted into the above equation.
Emily has all she needs to have a great journey!
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Pop Quiz
Practice Finding Angles Using the Law of Sines
In the following applet, x represents the measure of an angle of a triangle. By using the Law of Sines and, if needed, the Triangle Angle Sum Theorem, find the value of x. Write the answer as a single number rounded to the nearest degree, without the degree symbol.
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Example
Using the Law of Sines To Find Measurements of a Triangle
The Law of Sines can be used to find missing side lengths and angle measures of a triangle. When those side lengths and angle measures are known, the area and perimeter of a triangle can also be found.
Emily is planning to continue her travels. This time, she wants to visit some cities in the UK, as well as Ireland. She would really like to visit London, Edinburgh, and Dublin. It becomes clear to her that these three cities form a triangle. She is able to figure out that the angles at London and Edinburgh measure 41^(∘) and 59^(∘), respectively. She also knows that the distance between these two cities is 530 kilometers.
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Hint
The perimeter of a triangle is the sum of its three side lengths. The area of the triangle can be found by calculating half the product of two side lengths and the sine of their included angle.
Solution
The perimeter and the area of the triangle will be calculated one at a time.
Perimeter
The perimeter of a triangle is the sum of its three side lengths. Therefore, the missing side lengths need to be found. To make things clearer, the angles and sides will first be labeled.
Note that the missing angle is opposite to the side of the triangle whose length is known. Therefore, in order to use the Law of Sines to find the missing side lengths, the angle at Dublin must be calculated. To do this, the Triangle Angle Sum Theorem can be used. D+ 59^(∘)+ 41^(∘)=180^(∘) ⇕ D= 80^(∘) The measure of the angle at Dublin is 80^(∘).
The Law of Sines can now be used to find the missing side lengths. The law states that, for every triangle, the ratio of the length of a side to the sine of its opposite angle is constant. With this rule in mind, a proportion involving D, d, L and l can be written. l/sin L=d/sin D Next, L= 41^(∘), D= 80^(∘), and d= 530 can be substituted in the above equation, and l can be isolated.
The distance between Dublin and Edinburgh is about 353 kilometers. By following the same procedure, the distance between London and Dublin can be found.
| Law of Sines | Substitute | Simplify |
|---|---|---|
| l/sin L=d/sin D | l/sin 41^(∘)=530/sin 80^(∘) | l≈ 353 |
| e/sin E=d/sin D | e/sin 59^(∘)=530/sin 80^(∘) | e≈ 461 |
The distance between London and Dublin is about 461 kilometers.
The perimeter of the triangle formed by the three cities can now be calculated. Perimeter P= 530+ 353+ 461 ⇕ P=1344km
Area
The area of a triangle can be found by calculating half the product of two side lengths and the sine of their included angle. Since all angle measures and sides lengths are known, any of them can be used. In this case, E, d, and l will be arbitrarily used. Area: 1/2 d l sin E The respective values can be substituted to find the area of the triangle.
Area= 1/2 d l sin E
SubstituteValues
Substitute values
Area= 1/2( 530)( 353) sin 59^(∘)
▼
Evaluate right-hand side
Multiply
Multiply
Area= 1/2(187 090) sin 59^(∘)
MoveRightFacToNumOne
1/b* a = a/b
Area= 187 090/2 sin 59^(∘)
CalcQuot
Calculate quotient
Area= 93 545 sin 59^(∘)
UseCalc
Use a calculator
Area= 80 183.715144...
Approximate to the nearest thousand
Area≈ 80 000
The area of the triangle formed by London, Dublin, and Edinburgh is about 80 000 square kilometers. Thanks for helping Emily plan her travels!
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Discussion
The Ambiguous Case of the Law of Sines
The Law of Sines is useful to find missing angle measures and side lengths for any type of triangle.
The measure of the angle at A can be found by using the Law of Sines.
sin A/10=sin 34^(∘)/6
▼
Solve for A
MultEqn
LHS * 10=RHS* 10
sin A=sin 34^(∘)/6 (10)
CommutativePropMult
Commutative Property of Multiplication
sin A=10 (sin 34^(∘)/6)
MoveLeftFacToNum
a*b/c= a* b/c
sin A=10sin 34^(∘)/6
ReduceFrac
a/b=.a /2./.b /2.
sin A=5sin 34^(∘)/3
sin^(-1)(LHS) = sin^(-1)(RHS)
A=sin ^(- 1)(5sin 34^(∘)/3)
UseCalc
Use a calculator
A=68.746886... ^(∘)
RoundInt
Round to nearest integer
A≈ 69^(∘)
Is this the only possible measure for the angle at A? To find the measure of this angle, the Law of Sines was applied. That is, for any triangle, the ratio of the sine of an angle to the length of its opposite side is constant. sin 69^(∘)/10=sin 34^(∘)/6 Recall now that the sine of an angle is equal to the sine of the supplement of the angle. sin θ = sin (180^(∘)-θ) With this information, it can be stated that the sine of an angle whose measure is 69^(∘) is equal to the sine of an angle whose measure is 180^(∘)- 69^(∘)= 111^(∘). Therefore, by the Substitution Property of Equality, a new proportion can be derived. sin 69^(∘)/10=sin 34^(∘)/6 sin 69^(∘)= sin 111^(∘) ⇓ sin 111^(∘)/10=sin 34^(∘)/6 The obtained proportion implies a new triangle that satisfies the initial conditions.
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Closure
Calculating an Angle of a Triangle Using the Law of Sines
The challenge presented at the beginning of this lesson can be solved by using the Law of Sines.
Find the measure of the angle at B. Write the answer rounded to the nearest degree.
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Solution
Since the ratio of the sine of an angle to the length of its opposite side is constant, a proportion can be written.
sin B/AC=sin A/BC ⇒ sin B/15=sin 70^(∘)/20
The above equation can be solved for B.
sin B/15=sin 70^(∘)/20
B≈ 45^(∘)
The measure of the angle at B is about 45^(∘).
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Solving Triangles Using the Law of Sines
Exercise 3.1
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Let's draw a possible triangle that can be created by the flag, the support beam, and the wall. We will label the point where the support beam meets the flagpole as B.
We can use the Law of Sines to determine the length of PA. To first determine that length, we need to know the measure of m∠ B. For that angle measure, we need to know the measure of m∠ A. Then we will have enough information to use the Law of Sines. Now, let's solve for m∠ A.
Now we can use the Interior Angles Theorem to determine the unknown angle. m∠ B+30^(∘)+36.2215...^(∘) =180^(∘) ⇓ m∠ B=113.7784...^(∘) With this information, we can use the Law of Sines to calculate the length of PA.
As we can see, when PA equals 6.04 meters, the angle at P is 30^(∘).
Notice that sin θ=sin(180^(∘)-θ). Therefore, ∠ A can have a second measure.
m∠ A = 180^(∘)-36.2215...^(∘) = 143.7784...^(∘)
Since AB
Now we can use the Law of Sines to calculate this second possible length of PA.
As we can see, PA could also have a length of 0.72 meters.
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Hint & Answer
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Solution
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In the figure, we can identify two right triangles inside of the greater triangle ABC. Notice that we know the measures of two angles of △ ABC which means we can find the measure of ∠ C by using the Interior Angles Theorem. 83^(∘)+87^(∘)+m∠ C = 180^(∘) ⇓ m∠ C = 10^(∘) Let's add this finding to the triangle.
With the given information, we can find the length of either AC or BC by using the Law of Sines. Let's find BC.
Let's keep the side length in exact form for precision purposes and add it to the diagram. At the same time, we will focus on the right triangle BDC.
Let's substitute the hypotenuse, the given acute angle, and its opposite side of △ BDC in the sine ratio and solve for h.
The height from the ground to the northern lights is about 171 kilometers.
Instead of numerical values, we have A and B. This means we can express the angle at C as 180^(∘)-(A+B).
Now we can use the Law of Sines to write a general expression to describe the length of BC.
As in Part A, we will use the sine ratio to find an expression for h.
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Solving Triangles Using the Law of Sines
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