14. Solving Triangles Using the Law of Cosines
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Chapter 2
14.
Solving Triangles Using the Law of Cosines
| Student Learning Objectives: |
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Why
The study of triangles is enriched with the law of cosine, a pivotal concept in trigonometry. This law offers a relationship between a triangle's side lengths and the cosine of its angles. It becomes particularly useful when certain triangle measurements are known, and others are to be determined. For instance, when two side lengths and the angle between them are known, the law of cosine can help find the missing side. In real-world scenarios, such as determining distances or calculating areas using trigonometry, the law of cosine is invaluable. Moreover, in situations where the law of sines isn't suitable, the law of cosine becomes the go-to method. The lesson provides both theoretical insights and practical examples, ensuring a well-rounded understanding of the topic.
Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Triangles Using the Law of Cosines
Slide of 10
Trigonometric ratios are useful in solving right triangles. Furthermore, the Law of Sines is a fairly used law when solving non-right triangles. However, there are cases in which this law is not applicable. This lesson will explore those cases, where the relationship between the side lengths and an angle measure of different triangles.
Catch-Up and Review
Here is a bundle of recommended readings before getting started with this lesson.
Try your knowledge on these topics.
Consider the triangle with vertices A, B, and C.
a Find the measure of the missing angle.
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b Find the length of the longest side. If necessary, round the answer to the nearest integer.
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c If the altitude of the triangle is 6, find AD and DB. If necessary, round the answer to the nearest integer. 
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Challenge
Investigate Distances in Space Using Trigonometry
On a June night, Zosia observes an astronomical asterism, which is called the Summer Triangle.
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Explore
Limitations of the Law of Sines
The applet below shows two different cases for a triangle with vertices A, B, and C. Case I shows the lengths of two sides and the measure of their included angle. Case II shows the lengths of all three sides.
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Discussion
Law of Cosines
When the Law of Sines cannot be used to solve triangles, the Law of Cosines may be applied.
Consider △ ABC with sides of length a, b, and c, which are respectively opposite the angles with measures A, B, and C.
The following equations hold true with regard to △ ABC.
a^2=b^2+c^2-2bc cos(A)
b^2=a^2+c^2-2ac cos(B)
c^2=a^2+b^2-2ab cos(C)
Proof
For Acute AnglesThe first equation will be proven. The other two equations can be proven by following the same procedure.
a^2 = b^2+c^2 - 2bccos A
Begin by drawing the altitude from B to its opposite side AC.
By the definition of an altitude, both △ ADB and △ CDB are right triangles. By applying the Pythagorean Theorem to △ CDB and △ ADB, two equations can be obtained. Equation I: & a^2=(b-x)^2+h^2 [0.6em] Equation II: & c^2= x^2+h^2 The binomial in Equation I can be expanded.
Notice that Equation II says that x^2 + h^2 is equal to c^2. Therefore, the expanded form of Equation I can be rewritten by using the Substitution Property of Equality.
a^2 = b^2- 2bx + x^2 + h^2
Substitute
x^2+h^2= c^2
a^2 = b^2- 2bx + c^2
CommutativePropAdd
Commutative Property of Addition
a^2 = b^2 + c^2 - 2bx
Now, the x-term in this equation can be written using the cosine ratio.
In △ ADB, the cosine of A is the ratio of x to c. cos A = x/c ⇔ x = c cos A Finally, ccos A can be substituted for x into a^2 = b^2 + c^2 - 2bx. By doing so, the formula for the Law of Cosines is obtained. a^2 & = b^2+c^2 - 2b x a^2 & = b^2+c^2 - 2b ccos A
Proof
For Obtuse AnglesThe first equation will be proven for obtuse angles. The remaining equations can be proven similarly. a^2=b^2+c^2-2bccosA
Consider △ABC with side lengths of a, b, and c, respectively opposite the angles with measures A, B, and C, such that m∠A is greater than 90^(∘).
The altitude of the triangle is the perpendicular segment from B to the extension of the base AC. Let D be the endpoint of this segment and x be the distance from D to A.
From the definition of an altitude, it follows that △BDA and △BDC are right triangles. Two equations can be obtained by applying the Pythagorean Theorem to both triangles. Equation I: & a^2=h^2+(b+x)^2 [0.6em] Equation II: & c^2= h^2+x^2 Expand the binomial in Equation I.
a^2=h^2+(b+x)^2
ExpandPosPerfectSquare
(a+b)^2=a^2+2ab+b^2
a^2=h^2+b^2+2bx+x^2
CommutativePropAdd
Commutative Property of Addition
a^2=h^2+x^2+b^2+2bx
From Equation II, h^2+x^2 is equal to c^2. Therefore, the expanded form of Equation I can be rewritten by using the Substitution Property of Equality.
Note that A and 180^(∘)-A are supplementary angles. Using the cosine ratio of 180^(∘)-A then gives an expression for the x-term.
In △BDA, the cosine of 180^(∘)-A is the ratio of x to c. cos(180^(∘)-A) = x/c ⇕ x = c cos(180^(∘)-A) By the Sine and Cosine of Supplementary Values Angles, cos(180^(∘)-A) and cosA have opposite values. x=ccos(180^(∘)-A) ⇕ x=- ccosA Finally, by substituting - ccosA for x into a^2=b^2+c^2+2bx, the Law of Cosines is obtained.
a^2=b^2+c^2+2bx
Substitute
x= - ccosA
a^2=b^2+c^2+2b( - ccosA)
MultPosNeg
a(- b)=- a * b
a^2=b^2+c^2-2bccosA
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Example
Knowing Two Sides and the Included Angle of a Triangle
In a triangle, when the lengths of two sides and the measure of their included angle are known, the missing side length can be found by applying the Law of Cosines.
Kriz wants to determine the distance between two trees on the other side of the river. Kriz uses a tool that measures the distances to objects. The tool is able to find the distances to each tree as 4 meters and 3.5 meters.
External credits: @kdekiara
The angle between these sides, from where the Kriz stands with the measuring tool, measures 67^(∘). Find the distance between the trees, and round the answer to the nearest tenth of a meter.
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Hint
Start by naming the vertices and sides of the triangle.
Solution
For simplicity, the vertices and sides of the triangle will be named.
The Law of Cosines relates the lengths of the sides and the cosine of one of the angles. Therefore, the missing side length of the triangle can be found by using this law. a^2 = b^2 + c^2 - 2 bc cos A The next step is to substitute b=4, c=3.5, and A= 67^(∘) into the equation. Then, solve for a.
a^2 = b^2 + c^2 - 2 bc cos A
SubstituteValues
Substitute values
a^2 = 4^2 + 3.5^2 - 2 ( 4)( 3.5)cos 67^(∘)
▼
Solve for a
CalcPow
Calculate power
a^2 = 16 + 12.25 - 2 (4)(3.5) cos 67^(∘)
Multiply
Multiply
a^2 = 16 + 12.25 -28 cos 67^(∘)
AddTerms
Add terms
a^2 = 28.25 -28 cos 67^(∘)
UseCalc
Use a calculator
a^2 = 17.309528 ...
SqrtEqn
sqrt(LHS)=sqrt(RHS)
a =sqrt(17.309528 ...)
UseCalc
Use a calculator
a = 4.160472 ...
RoundDec
Round to 1 decimal place(s)
a ≈ 4.2
Note that only the principal root was considered because a side length cannot be negative. Therefore, the distance between the trees is about 4.2 meters.
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Example
Knowing Three Sides of a Triangle
When all the three side lengths of a triangle are known, the Law of Cosines can be used to find the measure of the angles.
Ramsha lives near a lighthouse. As she likes to observe the landscape, she notices that the light rays coming out of the lighthouse create an angle. She decides to ask the lighthouse keeper, but he insists on not telling her the measure of the angle. She sees a blueprint on the desk behind him, and quickly writes the lengths shown in the diagram before the grumpy keeper blocks her view!
Help Ramsha calculate the measure of each angle of the triangle formed by the light rays. Round the answers to the nearest degree.
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Hint
To find m ∠ L, use the Law of Cosines. Then, the measures of other two angles can be found by using either the Law of Cosines or the Law of Sines.
Solution
As the diagram indicates, the light rays form a triangle. In this triangle, the three side lengths are known.
The measure of the three angles will be found one at a time.
Finding m ∠ L
To find the measure of ∠ L, the Law of Cosines can be used. l^2 = m^2 + k^2 -2 mk cos L Substitute 35 for l, 105 for m, and 130 for k into the equation.
l^2 = k^2 + m^2 -2km cos L
SubstituteValues
Substitute values
35^2 = 105^2 + 130^2 -2( 105)( 130) cos L
▼
Solve for cos L
CalcPow
Calculate power
1225 = 11 025 + 16 900 - 2(105)(130) cos L
Multiply
Multiply
1225 = 11 025 + 16 900 - 27 300 cos L
AddTerms
Add terms
1225 = 27 925 - 27 300 cos L
SubEqn
LHS-27 925=RHS-27 925
- 26 700 = - 27 300 cos L
DivEqn
.LHS /(- 27 300).=.RHS /(- 27 300).
- 26 700/- 27 300= cos L
DivNegNeg
- a/- b=a/b
26 700/27 300= cos L
RearrangeEqn
Rearrange equation
cos L = 26 700/27 300
ReduceFrac
a/b=.a /300./.b /300.
cos L = 89/91
Now, take the inverse cosine of both sides of the equation.
cos L = 89/91
cos^(-1)(LHS) = cos^(-1)(RHS)
L = cos ^(- 1) (89/91 )
L ≈ 12^(∘)
The angle L measures about 12^(∘).
Finding m ∠ M
Since the ratio of the sine of an angle to the length of its opposite side is constant, the following proportion can be written. sin L/l=sin M/m ⇒ sin 12^(∘)/35=sin M/105 The equation can be solved for M.
sin 12^(∘)/35=sin M/105
▼
Solve for M
MultEqn
LHS * 105=RHS* 105
105* sin 12^(∘)/35= sin M
RearrangeEqn
Rearrange equation
sin M = 105* sin 12^(∘)/35
sin^(-1)(LHS) = sin^(-1)(RHS)
M =sin ^(- 1)(105* sin 12^(∘)/35)
UseCalc
Use a calculator
M=38.589405 ... ^(∘)
RoundInt
Round to nearest integer
M≈ 39^(∘)
The measure of ∠ M is about 39^(∘).
Finding m∠ K
By the Triangle Angle Sum Theorem, the sum of the interior angles of a triangle is 180^(∘). m ∠ K + 12 ^(∘) + 39^(∘) = 180^(∘) m ∠ K = 129^(∘) Therefore, all measures of △ KLM were found.
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Investigating the Cosine Ratio of Obtuse Angles
In △ ABC, all three side lengths and the measure of the angle at C are given. Examine how the length of AB changes as the measure of ∠ C varies.
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Discussion
Relationship Between the Angles of a Triangle
Consider △ ABC, in which all three sides lengths and the measure of the angle at C are known. Let a, b, and c be the lengths of the sides opposite A, B, and C, respectively. In the following applet, the values of a^2+b^2 and c^2 are shown. Move the slider to change the measure of ∠ C.
| m∠ C | c^2 = a^2 + b^2- 2ab cos C | Relationship between a^2 + b^2 and c^2 | Conclusion |
|---|---|---|---|
| m∠ C < 90^(∘) | c^2 = a^2 + b^2- 2ab cos C_(> 0) | c^2 < a^2 + b^2 | If ∠ C is acute, not too many conclusions can be made. The opposite side to ∠ C can be the largest side, the shortest side, or none. |
| m∠ C = 90^(∘) | c^2 = a^2 + b^2- 2ab cos 90^(∘)_(= 0) | c^2 = a^2 + b^2 | The cosine of 90^(∘) is 0. In this case, the Law of Cosines becomes the Pythagorean Theorem. This means that the opposite side to ∠ C is the largest side of the triangle. |
| 90^(∘) < m∠ C < 180^(∘) | c^2 = a^2 + b^2- 2ab cos C_(< 0) | c^2 > a^2 + b^2 | If ∠ C is obtuse, its measure is greater than the measures of ∠ A and ∠ C. Therefore, its opposite side is the largest side of the triangle. |
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Example
Solving Problems Using the Law of Cosines
Once Diego completed a grueling month-long shift as a lighthouse keeper, he decided to fly from San Juan to New York. After flying for 3 hours on a straight path, he felt that the pilot made a course correction, then continued to fly for about 2 more hours on a path still toward New York. On Diego's return flight, the pilot flew on a straight path, without any change in direction, from New York to San Juan.
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Hint
Use the speed formula to calculate the distance traveled before the change in direction SD, and the distance traveled after the change DN.
Solution
First, SD and DN will be found. Then, the Law of Cosines will be used to find NS, the distance between New York and San Juan.
Finding SD and DN
The average speed is the distance traveled divided by the amount of time spent traveling. Speed = Distance/Time The flight from S to D takes 3 hours and the speed of the plane is 330 miles per hour. Substitute these values into the formula and solve for SD.
The distance covered in the first 3 hours of the flight is 990 miles. Similarly, the distance covered in the next 2 hours can be calculated.
| Speed = Distance/Time | ||
|---|---|---|
| Length | SD | DN |
| Substitution | 330 = SD/3 | 330 = DN/2 |
| Calculation | SD = 990 mi | DN = 660 mi |
Finding SD
Since the plane was deflected 10^(∘) from the first route, the measure of the angle SDN is 170^(∘).
Now, in △ SDN, the lengths of two sides and the measure of their included angle are known. Therefore, the Law of Cosines can be used. Let d, s, and n be the lengths of the sides opposite D, S, and N, respectively. d^2 = s^2 + n^2 -2sn cos D Substitute 660 for s, 990 for n, and 170^(∘) for D.
d^2 = s^2 + n^2 -2sn cos D
SubstituteValues
Substitute values
d^2 = 660^2 + 990^2 -2( 660)( 990)cos 170^(∘)
▼
Solve for d
CalcPow
Calculate power
d^2 = 435 600 + 980 100 - 2 (660)(990)cos 170^(∘)
AddTerms
Add terms
d^2 = 1 415 700 - 2 (990)(660)cos 170^(∘)
UseCalc
Use a calculator
d^2 = 2 702 646.7716204...
SqrtEqn
sqrt(LHS)=sqrt(RHS)
d = sqrt(2 702 646.7716204...)
UseCalc
Use a calculator
d = 1643.972862 ...
Approximate to nearest hundred
d ≈ 1600
Since a length cannot be negative, only the principal root is considered here. Therefore, the distance from San Juan to New York is about 1600 miles.
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Closure
Calculating Distances in Space Using Trigonometry
In this course, the use of the Law of Cosines in solving any type of triangle has been studied. By using this law, the challenge presented at the beginning can be solved.
Zosia knows the lengths of two sides of a triangle and the measure of their included angle. Let A, V, and D denote the vertices of the Summer Triangle.
What is the angular distance between Deneb and Vega? Round the answer to the nearest integer.
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Hint
The Law of Cosines states the relationship between the side lengths of a triangle and the cosine of one of the angles.
Answer
Let a, v, and d be the lengths of the sides opposite A, V, and D, respectively. By the Law of Cosines, the following equations holds true for △ AVD. a^2 &= v^2 + d^2 -2vd cos A v^2 &= d^2 + a^2 -2da cos V d^2 &= a^2 + v^2 -2av cos D
Since the values of v, d, and A are given, the first equation will give the length of the third side.
Substitute these values and solve for a.
a^2 = v^2 + d^2 -2vd cos A
SubstituteValues
Substitute values
a^2 = 38^2 + 34^2 -2( 38)( 34) cos 38 ^(∘)
▼
Solve for a
CalcPow
Calculate power
a^2 = 1444 + 1156 - 2(38)(34)cos 38^(∘)
Multiply
Multiply
a^2 = 1444 + 1156 -2584 cos 38^(∘)
AddTerms
Add terms
a^2 = 2600 -2584 cos 38^(∘)
UseCalc
Use a calculator
a^2 = 563.780212 ...
SqrtEqn
sqrt(LHS)=sqrt(RHS)
a = sqrt(563.780212 ...)
UseCalc
Use a calculator
a = 23.744056 ...
RoundInt
Round to nearest integer
a ≈ 24
The angular distance between Deneb and Vega is about 24 angular units.
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Solving Triangles Using the Law of Cosines
Exercise 2.1
Kevin wants to combine his favorite sports of ice skating and windsurfing into one activity. He calls it ice surfing.
Kevin has designed a new sail, specifically for windsurfing, with the following dimensions.
Before he buys the fabric to make the sail, he needs to know its area. What is the area of the sail? The fabric store only sells material by the meter and rounded to one decimal place.
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Examining the sail, consider what may help us find the total area. Well, treating the support beam as a line of reflection, we can see that the parts above and below are mirror images of each other.
That means if we can determine the area of one part of the sail, then we can double its area to find the total area of the sail. For now, let's work with the bottom half. We will add a diagonal to that part to split it into two triangles and find their areas.
Since one of the triangles is a right triangle and we know both of its leg measurements, we can find its area by using the formula for the area of a triangle. A=1/2( 3.30)( 0.85)= 1.4025 m^2 Next, we need to find the area of the second triangle. Again, because we know both of its leg measurements, we can find the length of the side labeled c by using the Pythagorean Theorem.
Now we can find either of its three angles by using the Law of Cosine.
Now let's apply the formula for calculating a triangle's area using sine to find its area.
Finally, we have the information we need to determine the total area of the sail. First, let's add the area of the two triangles. Recall that the two parts of the whole sail are a mirror image of each other. Therefore, we can multiply the sum of the two triangles by 2 to account for the entire sail. 2(1.4025+2.34340...)≈ 7.5m^2
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Solution
What is the perimeter of the triangle in centimeters? Round the answer to one decimal place.
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To determine the perimeter, we need to calculate the length of all sides. Currently, we do not know the length of x. Since we know two side lengths and an angle, we can use an equation from the Law of Cosines to determine x. Notice that x is not on the opposite side of the known angle A. Therefore, x goes on the right-hand side of the formula.
To solve this we will use the Quadratic Formula.
We cannot continue this simplification without a graphing calculator. Let's calculate 4.4cos 72^(∘).
Let's replace 4.4cos 72^(∘) by 1.35967... and continue.
When we know x, we can find the perimeter by adding all of the side lengths. P= 4.20576...+4.1+2.2≈ 10.5 cm
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Solution
What is the perimeter of the quadrilateral ABCD in inches? Give the answer in exact form.
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To determine the perimeter of ABCD we must know all of its sides. Notice that the figure consists of two triangles, △ ABD and △ BDC. In △ BDC we know two sides and the included angle. This is enough information to calculate the length of BD by using the Law of Cosine.
Notice that △ ABD is an equilateral triangle. Therefore, it has three congruent sides. Since BD is sqrt(4.75), we know that the remaining sides have the same length.
Finally, we will add AD, DC, CB, and AB to determine the perimeter of ABCD. Notice that BD is not part of the perimeter, as it is inside the shape. We will keep the answer in exact form. sqrt(4.75)+ 2.5+ 1+ sqrt(4.75) ⇓ 2sqrt(4.75)+3.5
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Solution
Solving Triangles Using the Law of Cosines
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