15. Application of the Laws of Sine and Cosine
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Chapter 2
15.
Application of the Laws of Sine and Cosine
Understanding mathematical principles and their applications can be intriguing. The lesson dives deep into how the law of cosine and law of sine play an essential role in solving real-world problems. These laws, especially when combined, can address intricate challenges that arise in fields like physics, engineering, and even everyday life. Recognizing their potential allows one to approach problems with a broader toolkit, facilitating more effective solutions. Whether you're trying to figure out the dimensions of a triangle in a construction project or understanding the relationships between angles and sides, these laws are indispensable.
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| | 14 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Application of the Laws of Sine and Cosine
Slide of 14
This lesson will explore the applications of the Law of Sines and the Law of Cosines to solve different types of triangles. Furthermore, real-life applications of these ideas will also be explored.
Catch-Up and Review
Here are a some recommended readings before getting started with this lesson.
Explore
Which Law Should be Applied?
Take a look at the following triangles. Think whether they can be solved by using the Law of Sines or the Law of Cosines.
Challenge
Area of a Circle Circumscribed Around a Non-Right Triangle
The following figure shows a circle circumscribed around a non-right triangle. Notice that none of the triangle's sides correspond to the diameter of the circle. What is the area of the circle?
Discussion
Solving Triangles Using the Law of Sines and the Law of Cosines
Trigonometric ratios are often used to solve right triangles, but they cannot be used to solve non-right, or oblique, triangles. For these triangles, the Law of Sines and the Law of Cosines are particularly useful because they can be used to solve any triangle, right or oblique. To solve a triangle using the Law of Sines or the Law of Cosines, three pieces of information must be known.
| Case | Given Information | Law | Strategy |
|---|---|---|---|
| 1 | Two angles and a side length | Law of Sines | The Triangle Angle Sum Theorem can be used to find the missing angle measure. Then the Law of Sines can be used to find the unknown side lengths. |
| 2 | Two side lengths and a non-included angle | Law of Sines | The Law of Sines can be used to solve for one of the unknown angle measures. Then the Triangle Angle Sum Theorem can be used to find the third angle measure. Finally, the Law of Sines can be applied one more time to find the unknown side length. |
| 3 | Three side lengths | Law of Cosines | The Law of Cosines can be used to find any of the unknown angle measures. Then, either the Law of Sines or the Law of Cosines can be used to find another missing angle measure. Finally, the Triangle Angle Sum Theorem can be used to find the third angle measure. |
| 4 | Two side lengths and their included angle | Law of Cosines | The Law of Cosines can be used to find the missing side length. Then, either the Law of Cosines or the Law of Sines can be used used to find a missing angle measure. Finally, the Triangle Angle Sum Theorem can be used to find the last angle measure. |
Pop Quiz
Analyzing Different Triangles to Determine Which Law Can Be Used
Example
Calculating the Height of the Leaning Tower of Pisa
Zain is vacationing in Italy. They were in Pisa to see the famous Leaning Tower when a question came across their mind. What would the tower's height be if it was not a leaning tower? Zain's distance to the tower is 80 meters and they can measure an angle of elevation of 37∘ to the tower's top. Furthermore, the guidebook states that the tower's inclination is about 4∘.
Remembering a Geometry lesson, they realize that the situation can be modeled using a non-right triangle. Help Zain calculate the height of the upright tower. Write the answer rounded to one decimal place.
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Hint
The inclination angle and its adjacent angle are complementary angles.
Solution
The situation can be modeled using a non-right triangle.
4∘+m∠B=90∘⇕m∠B=86∘
Now that two angles and their included side are known, it is possible to use the Law of Sines. To do so, the measure of ∠A must be found. This will be done by using the Triangle Angle Sum Theorem. m∠A+86∘+37∘=180∘⇕m∠A=57∘
Finally, the Law of Sines will be used to write a proportion. This equation will be solved for c, the height of the upright tower.
sinAa=sinCc
SubstituteValues
Substitute values
sin57∘80=sin37∘c
▼
Solve for c
MultEqn
LHS⋅sin37∘=RHS⋅sin37∘
sin57∘80(sin37∘)=c
UseCalc
Use a calculator
57.406571…=c
RearrangeEqn
Rearrange equation
c=57.406571…
RoundDec
Round to 1 decimal place(s)
c≈57.4
Pop Quiz
Measuring Trees Using the Law of Sines
Magdalena is doing some research in the forest for a biology project. She has a device that allows her to measure angles. She can also measure the distance from a tree to her device. To help Magdalena complete her research project, find the lengths of different trees, rounded to one decimal place.
Discussion
Extending the Definition of Sine and Cosine to Obtuse Angles
As mentioned before, the Law of Sines and the Law of Cosines are valid for all types of triangles, including both right and non-right triangles. However, the definitions of the sine and cosine of an angle are given in terms of the ratios of a right triangle's sides.
Therefore, these definitions do not seem to be compatible with obtuse angles. Nevertheless, they can be extended to deal with obtuse angles by considering the following identities.
Rule
Sine and Cosine of Supplementary Angles
The sine of supplementary angles are equal. Conversely, the cosine of supplementary angles are opposite values.
sin(180∘−θ)cos(180∘−θ)=sinθ=-cosθ
sin(π−θ)cos(π−θ)=sinθ=-cosθ
Example
Calculating the Perimeter of a Piece of Land
Ignacio's grandparent wants to construct a fence for a quadrilateral piece of land. To find the perimeter of the land, he starts measuring its sides using an old trundle wheel. Unfortunately, after measuring just two sides, the trundle wheel breaks.
Ignacio wants to help his grandparent and, in an attempt to simplify the problem, he divides the land into two triangles. Then, by using a compass, he is able to measure the angles of these triangles.
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Hint
Since two side lengths and the measure of their included angle are known in △ABC, the Law of Cosines can be used to solve for the missing side length.
Solution
Note that two side lengths and all the angle measurements are known in △ABC.
Now, △ACD will be considered.
P=457 m+560 m+400 m+260 m⇕P=1677 m
The perimeter of the piece of land is about 1677 meters.
Pop Quiz
Law of Cosines in Soccer Practice
Kriz is setting up for a free shots on an empty goal. When considering their distance to both goal posts, they realize that the Law of Cosines can be used to calculate the top measure of the angle in which they must kick the ball in order to score. They are practicing with a standard 7.3 meter net. Help Kriz calculate this angle and score the goal! Write the answer rounded to one decimal place.
External credits: @kdekiara
Example
Locating of a Smartphone Using Triangulation
A burglar robbed a store and took the cashier's smartphone. In an attempt to outsmart the police, the burglar turned off the phone's GPS. Once at their secret location, the burglar felt safe and made a call to plan their next move. However, the smartphone signal was detected by two nearby towers, estimating their distance to the phone in use.
With the distance from the towers to the location of the smartphone and knowing that the towers are 6 blocks apart, the police think they have enough information to locate the burglar. Help the police find the burglar's location! Write the horizontal and vertical distance in blocks with respect to the phone tower at the left. Round the answers to the nearest integer. 
Even if no angle is known at first, knowing the three side lengths allows the use of the Law of Cosines to solve for any angle. In this case, it is convenient to solve for ∠A, which corresponds to the vertex represented by the left tower.
Now that the measure of ∠A is known, the horizontal and vertical distance from the left tower to the smartphone, H and V, respectively, can be calculated. To do this, a right triangle formed by the left tower and the phone will be considered. Then, trigonometric ratios will be used. 
The vertical distance to the smartphone V is the opposite leg to the angle whose measure is 26∘. Therefore, the length of the leg can be found by using the sine ratio.
Similarly, the horizontal distance H is the adjacent leg to the angle whose measure is 26∘. Therefore, the cosine ratio can be used to find the value of H.
The horizontal distance from the left tower to the burglar's location is about 4 blocks, and the vertical distance is about 2 blocks.
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Hint
The situation can be modeled using a triangle with three known side lengths.
Solution
The distance from the tower at the left to the smartphone is about 4.5 blocks. The distance from the tower at the right to the phone is about 2.8 blocks. The towers are 6 blocks apart. Therefore, the situation can be modeled using a triangle with three known sides lengths.
a2=b2+c2−2bccosA
SubstituteValues
Substitute values
2.82=62+4.52−2(6)(4.5)cosA
▼
Solve for A
CalcPow
Calculate power
7.84=36+20.25−2(6)(4.5)cosA
AddTerms
Add terms
7.84=56.25−2(6)(4.5)cosA
Multiply
Multiply
7.84=56.25−54cosA
SubEqn
LHS−56.25=RHS−56.25
-48.41=-54cosA
DivEqn
LHS/(-54)=RHS/(-54)
0.896481…=cosA
RearrangeEqn
Rearrange equation
cosA=0.896481…
cos-1(LHS)=cos-1(RHS)
A=cos-10.896481…
UseCalc
Use a calculator
A=26.300639…∘
RoundInt
Round to nearest integer
A≈26∘
sinθ=HypotenuseOpposite Leg
SubstituteValues
Substitute values
sin26∘=4.5V
▼
Solve for V
MultEqn
LHS⋅4.5=RHS⋅4.5
(4.5)sin26∘=V
UseCalc
Use a calculator
1.972670…=V
RoundInt
Round to nearest integer
2≈V
RearrangeEqn
Rearrange equation
V≈2
cosθ=HypotenuseAdjacent Leg
SubstituteValues
Substitute values
cos26∘=4.5H
▼
Solve for H
MultEqn
LHS⋅4.5=RHS⋅4.5
(4.5)cos26∘=H
UseCalc
Use a calculator
4.044573…=H
RoundInt
Round to nearest integer
4≈H
RearrangeEqn
Rearrange equation
H≈4
Example
Finding the Altitude of a Helicopter
Two radar stations located 30 kilometers apart detect a passing helicopter. The first station measures an angle of elevation to the helicopter of 40∘, while the second station measures an angle of elevation of 45∘. At what altitude, rounded to one decimal place, is the helicopter flying?
To do this, the Triangle Angle Sum Theorem can be used. The measure of ∠A will now be found.
The distance from Station 1 to the helicopter is about 21.3 kilometers. This distance is the hypotenuse of the right triangle formed by considering the helicopter and Station 1 as vertices. Furthermore, in this right triangle, the opposite leg to the angle that measures 40∘ represents the helicopter's altitude. 
Finally, the sine ratio can be used to find the altitude of the helicopter.
The helicopter is flying at an altitude of about 13.7 kilometers.
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Hint
The missing angle can be found by using the Triangle Angle Sum Theorem.
Solution
Because two angles and the included side are known, this problem can be approached by using the Law of Sines to find the distance between the helicopter and one of the radar stations. Once one of the missing side lengths is determined, trigonometric ratios can be used to find the altitude, or height, of the helicopter. The first thing to do is to find the unknown angle.
m∠A+40∘+45∘=180∘⇕m∠A=95∘
Now that the opposite angle to the known side length a has been found, the Law of Sines can be used to write a proportion and solve for any of the missing sides. For instance, b can be calculated.
sinAa=sinBb
SubstituteValues
Substitute values
sin95∘30=sin45∘b
▼
Solve for b
MultEqn
LHS⋅sin45∘=RHS⋅sin45∘
sin95∘30(sin45∘)=b
UseCalc
Use a calculator
21.294234…=b
RearrangeEqn
Rearrange equation
b=21.294234…
RoundDec
Round to 1 decimal place(s)
b≈21.3
sinθ=hypotenuseopposite leg
SubstituteValues
Substitute values
sin40∘=21.3h
▼
Solve for h
MultEqn
LHS⋅21.3=RHS⋅21.3
sin40∘(21.3)=h
UseCalc
Use a calculator
13.691376…=h
RearrangeEqn
Rearrange equation
h=13.691376…
RoundDec
Round to 1 decimal place(s)
h≈13.7
Discussion
Connection Between the Law of Sines and the Circumcircle of a Triangle
The Law of Sines states that for any triangle, the ratio of the sine of an angle to the length of its opposite side is constant. However, this is not just any constant. In fact it has an important geometrical interpretation.
Rule
Extended Form of the Law of Sines
The following goes for any triangle. The diameter of a triangle's circumcircle is equal to the ratio of a side length to the sine of its opposite angle.
Now, consider the above figure and let D be the diameter of the circle. With the given information, the following equation holds true.
sinAa=sinBb=sinCc=D
Proof
The proof can be completed in two parts. Part I shows how the sides of a triangle are proportional to the sines of the opposite angles. Part II shows this proportion is equal to the diameter of the circle circumscribed around the triangle.
Part I: Proportional Sides
Consider a triangle ABC with side lengths a, b, and c, and angle measures A, B, and C.
sinAa=sinBb=sinCc
Part II: Diameter of the Circumscribed Circle
Draw the circumcircle of the triangle ABC with its circumcenter O.
By the Inscribed Angle Theorem, the measure of A is half of the measure of the intercepted arc BC.
Now, draw the central angle with the intercepted arc BC. Recall that a central angle and its intercepted arc have the same measure. Therefore, the measure of BC is equal to the measure of m∠COB.
⎩⎪⎨⎪⎧m∠A=21mBCmBC=m∠COB⇓m∠A=21m∠COB
Let m∠A be α. Then m∠COB becomes 2α. 
Note that OB and OC are both equal to the radius R. Hence, △COB is an isosceles triangle with two congruent sides OB and OC.
Now, focus on △COB. Draw its altitude from the vertex angle O. Since △COB is an isosceles triangle, the altitude bisects both the vertex angle and the opposite side.
sinα=R2a=2Ra
Rearrange sinα by substituting α=A.
Since R represents the radius, 2R is equal to the diameter D of the circle.
sinAa=2R⇒sinAa=D
Conclusion
Combining the results of Part I and Part II, the extended form of the Law of Sines can be obtained.⎩⎪⎪⎪⎨⎪⎪⎪⎧sinAa=sinBb=sinCcsinAa=D⇓sinAa=sinBb=sinCc=D
Closure
Calculating the Area of a Circle Circumscribed Around a Non-Right Triangle
The challenge presented at the beginning of this lesson can be solved by using a combination of the Law of Sines, the Law of Cosines, and the extended form of the Law of Sines.
The question here was to find the area of the circle. This will be answered by first finding the ratio of the triangle's side lengths to the sine of their opposite angles.
a What is the ratio of a side length to the sine of its opposite angle? Write the answer rounded to one decimal place.
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b What is the area of the circumscribed circle? Write the answer rounded to one decimal place.
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Hint
a Two side lengths and the measure of the included angle are known.
b The area of a circle is A=πr2, where r is the radius.
Solution
Since two side lengths and the included angle are known, the Law of Cosines can be used to find the missing side length. Because the missing side is opposite to the known angle, finding the side length will allow to calculate the desired ratio.
Radius:25.2=2.6 units
With this information, the area of the circle can be calculated. A=πr2
Substitute
r=2.6
A=π(2.6)2
▼
Simplify right-hand side
CalcPow
Calculate power
A=π(6.76)
CommutativePropMult
Commutative Property of Multiplication
A=6.76π
UseCalc
Use a calculator
A=21.237166…
RoundDec
Round to 1 decimal place(s)
A≈21.2
Application of the Laws of Sine and Cosine
Exercises
A satellite orbiting the Earth emits a signal to calculate the distance between itself and Earth. It found that the distances between itself and two cities are 398km and 361 km, respectively, and that the angle between them is 5∘. Find the distance between the cities. Round the answer to the nearest kilometer.
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Because two sides lengths and the included angle measure are known, we can apply the Law of Cosines to solve for the missing side length representing the distance between the cities.
Let's use the Law of Cosines to solve for the distance.
When solving the equation, only the principal root was considered because side lengths are always positive. The distance between the two cities is about 50 kilometers.
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A security camera has been installed in one of the corners in a room. The diagram shown below indicates the area that is visible in the camera's recording. What is the angle of the camera's field of vision? Write the answer rounded to the nearest degree.
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Examining the diagram, we see that the field of vision can be modeled using a non-right triangle where all side lengths are known.
Because of this, the Law of Cosines can be used to find m∠ A, which corresponds to the desired field of vision angle.
Therefore, the field of vision angle is about 26^(∘).
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A solar panel was installed on the roof of a house and set to an angle that optimizes the energy produced. The length of the panel is 1.96 meters and the length of the leg supporting it is 0.68 meters. If the support leg makes an angle of 117∘ with the roof, what is the optimal inclination angle of the solar panel? Write the answer rounded to the nearest degree.
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The situation can be modeled by using a non-right triangle.
Because two side lengths and an angle measure are known, it is possible to apply the Law of Sines to solve for the required angle measure. Let's use the Law of Sines to set up a proportion and solve for m∠ A.
The optimal angle for the solar panel is about 18 ^(∘).
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Two cellphone towers detect a smartphone in use. One tower finds that the smartphone is about 650 meters away at an angle of 41∘. The second tower is at an approximate distance of 534 meters from the smartphone at an angle of 53∘. Find the distance between the cellphone towers. Write the answer rounded to the nearest meter.
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The situation can be modeled by a non-right triangle.
The missing side length could be found by using the Law of Sines if the measure of ∠ C were known, so let's start by finding m∠ C. This can be done by using the Interior Angles Theorem. 41 ^(∘) + 53 ^(∘) + m∠ C = 180 ^(∘) ⇕ m∠ C = 86 ^(∘) Now that m∠ C is known, we can use the Law of Sines to write a proportion and solve for the distance between the towers c.
Therefore, the distance between the two towers is about 812 meters.
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Application of the Laws of Sine and Cosine
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