Use the law of cosines to find the side opposite an angle #7-12
Use the law of cosines to find an angle #13-20
Use the law of cosines to find a side adjacent to an angle #21-26
Decide which law to use #27-34
Solve a triangle #35-42
Solve problems using the law of cosines #43-56
Law of Cosines
If we know two angles and one side of a triangle, we can use the law of sines to solve the triangle. We can also use the law of sines when we know two sides and the angle opposite one of them. But the law of sines is not helpful for the problem that opened this chapter, finding the distance from Avery to Clio. In this case we know two sides of the triangle, [latex]a[/latex] and [latex]c{,}[/latex] and the included angle, [latex]B{.}[/latex]
To solve a triangle when we know two sides and the included angle, we will need a generalization of the Pythagorean theorem known as the law of cosines.
In a right triangle, with [latex]C = 90°{,}[/latex] the Pythagorean theorem tells us that
[latex]c^{2} = a^{2} + b^{2}[/latex]
If we allow angle [latex]C[/latex] to vary but keep [latex]a[/latex] and [latex]b[/latex] the same length, the side [latex]c[/latex] will grow or shrink, depending on whether we increase or decrease the angle [latex]C{,}[/latex] as shown below.
The Pythagorean theorem is actually a special case of a more general law that applies to all triangles, no matter what the size of angle [latex]C{.}[/latex] The equation relating the three sides of a triangle is
[latex]c^{2} = a^{2} + b^{2} - 2ab \cos C[/latex]
You can see that when [latex]C[/latex] is a right angle, [latex]\cos 90° = 0{,}[/latex] so the equation reduces to the Pythagorean theorem.
We can write similar equations involving the angles or [latex]A[/latex] or [latex]B{.}[/latex] The three equations are all versions of the law of cosines.
Law of Cosines.
If the angles of a triangle are [latex]A, B{,}[/latex] and [latex]C{,}[/latex] and the opposite sides are, respectively, [latex]a, b,[/latex] and [latex]c{,}[/latex] then
For a proof of the law of cosines, see Homework Problems 57 and 58.
Finding a Side
Now we can solve the problem of the distance from Avery to Clio. Here is the figure from Section 3.1 showing the location of the three towns.
Example 3.39.
Example 3.39.
How far is it from Avery to Clio?
Solution
The angle [latex]\angle ABC = 35° + 90° = 125°{.}[/latex] Thus, in [latex]\triangle ABC[/latex] we have [latex]a = 34,~ c = 48[/latex] and [latex]B = 125°{.}[/latex] The distance from Avery to Clio is represented by [latex]b[/latex] in the figure.
We know two sides and the included angle, and we choose the version of the law of cosines that uses our known angle, [latex]B{.}[/latex]
[latex]b^{2} = a^{2} + c^{2} - 2ac \cos B \qquad \text{Substitute the known values.}[/latex]
[latex]b^{2} = 34^{2} + 48^{2} - 2(34)(48) \cos 125° \qquad \text{Simplify the right side}[/latex]
[latex]b^{2} = 3460 - 3264 \cos 125° = 5332.153 \qquad \text{Take Square roots}[/latex]
[latex]b = 73.02[/latex]
Avery is about [latex]73[/latex] miles from Clio.
Caution 3.40.
When simplifying the law of cosines, be careful to follow the order of operations. In the previous example, the right side of the equation
[latex]b^{2} = 34^{2} + 48^{2} - 2(34)(48) \cos 125°[/latex]
Note that [latex]3264[/latex] is the coefficient of [latex]\cos 125°{,}[/latex] so it would be incorrect to subtract [latex]3264[/latex] from [latex]3460[/latex]. If you are using a graphing calculator, you can enter the right side of the equation exactly as it is written.
Checkpoint 3.41.
In [latex]\triangle ABC{,}[/latex] [latex]a = 11,~c = 23{,}[/latex] and [latex]B = 87°{.}[/latex] Find [latex]b[/latex] and round your answer to two decimal places.
Solution
[latex]24.97[/latex]
Finding an Angle
We can also use the law of cosines to find an angle when we know all three sides of a triangle. Pay close attention to the algebraic steps used to solve the equation in the next example.
Example 3.42.
In the triangle at right, [latex]a = 6,~ b = 7{,}[/latex] and [latex]c = 11{.}[/latex] Find angle [latex]C{.}[/latex]
Solution
We choose the version of the law of cosines that uses angle [latex]C{.}[/latex]
[latex]c^{2} = a^{2} + b^{2} - 2ab \cos C \qquad \text{Substitute the known values.}[/latex]
[latex]11^{2} = 6^{2} + 7^{2} - 2 (6) (7) \cos \qquad \text{Simplify each side.}[/latex]
[latex]121 = 36 + 49 - 84 \cos C \qquad \text{Isolate the cosine term.}[/latex]
[latex]36 = -84 \cos C \qquad \text{Solve for cos C.}[/latex]
[latex]\dfrac{-3}{7} = \cos C \qquad \text{Solve for C.}[/latex]
[latex]C = \cos^{-1} (\dfrac{-3}{7}) = 115.4°[/latex]
Angle [latex]C[/latex] is about [latex]115.4°{.}[/latex]
Checkpoint 3.43.
In [latex]\triangle ABC{,}[/latex] [latex]a = 5.3,~b = 4.7{,}[/latex] and [latex]c = 6.1{.}[/latex] Find angle [latex]B[/latex] and round your answer to two decimal places.
Solution
[latex]48.07°[/latex]
Once we have calculated one of the angles in a triangle, we can use either the law of sines or the law of cosines to find a second angle. Here is how we would use the law of sines to find angle [latex]A[/latex] in the previous example.
[latex]\dfrac{\sin A}{a} = \dfrac{\sin C}{c} \qquad \text{Substitute the known values.}[/latex]
[latex]\dfrac{\sin A}{6} = \dfrac{\sin 115.4°}{11} \qquad\text{Solve for sin A.}[/latex]
[latex]\sin A = 6 \cdot \dfrac{\sin 115.4°}{11} \approx 0.4928[/latex]
Thus, [latex]A = \sin^{-1}(0.4928) = 29.5°{.}[/latex] (We know that [latex]A[/latex] is an acute angle because it is opposite the shortest side of the triangle.) Finally,
[latex]B = 180° - (A + C) \approx 35.1°[/latex]
Alternatively, we can use the law of cosines to find angle [latex]A{.}[/latex]
[latex]a^{2} = b^{2} + c^{2} - 2bc \cos A \qquad \text{Substitute the known values.}[/latex]
[latex]6^{2} = 7^{2} + 11^{2} - 2(7) (11) \cos A \qquad \text{Simplify each side.}[/latex]
[latex]36=49+121-154\cos A \qquad \text{Isolate the cosine term.}[/latex]
[latex]-134 = -154 \cos A \qquad \text{Solve for cos A.}[/latex]
[latex]\dfrac{67}{77} = \cos A \qquad \text{Solve for A.}[/latex]
[latex]A = \cos^{-1} (\dfrac{67}{77}) = 29.5°[/latex]
Note 3.44.
Using the law of sines requires fewer calculations than the law of cosines, but the law of cosines uses only the original values instead of the results of our previous calculations and approximations.
Whenever we round off a number, we introduce inaccuracy into the calculations, and these inaccuracies grow with each additional calculation. Thus, for the sake of accuracy, it is best to use given values in preference to calculated values whenever possible.
Using the Law of Cosines for the Ambiguous Case
In Section 3.2 we encountered the ambiguous case:
If we know two sides [latex]a[/latex] and [latex]b[/latex] of a triangle and the acute angle [latex]\alpha[/latex] opposite one of them, there may be one solution, two solutions, or no solution, depending on the size of [latex]a[/latex] in relation to [latex]b[/latex] and [latex]\alpha{,}[/latex] as shown below.
The Ambiguous Case.
No solution: [latex]a \lt b \sin \alpha[/latex][latex]\alpha[/latex] is too short to make a triangle.
One solution: [latex]a = b \sin \alpha[/latex][latex]\alpha[/latex] is exactly the right \length to make a right triangle.
Two solutions: [latex]b \sin \alpha \lt a \lt b[/latex]
One solution: [latex]a \gt b[/latex]
If [latex]\alpha[/latex] is an obtuse angle, things are simpler: there is one solution if [latex]a \gt b[/latex] and no solution if [latex]a \le b{.}[/latex]
Because there are always two angles with a given sine, if we use the law of sines for the ambiguous case, we must check whether both possible angles result in a triangle. But here is another approach: We can apply the law of cosines to find the third side first. With that method, we’ll need the quadratic formula.
Quadratic Formula.
The solutions of the quadratic equation [latex]ax^{2} + bx + c = 0,~~a not= 0,[/latex] are given by [latex]x = \dfrac{-b^ {\pm} \sqrt{b^{2} - 4ac}}{2a}[/latex]
A quadratic equation can have one solution, two solutions, or no solution, depending on the value of the discriminant, [latex]b^{2} - 4ac{.}[/latex] If we use the law of cosines to find a side in the ambiguous case, the quadratic formula will tell us how many triangles have the given properties.
If the quadratic equation has one positive solution, there is one triangle.
If the quadratic equation has two positive solutions, there are two triangles.
If the quadratic equation has no positive solutions, there is no triangle with the given properties.
Example 3.45.
In [latex]\triangle ABC,~ B = 14.4°,~ a = 8{,}[/latex] and [latex]b = 3{.}[/latex] Solve the triangle.
Solution
We begin by finding the third side of the triangle. Using the law of cosines, we have
[latex]b^{2} = a^{2} + c^{2} - 2ac \cos B \qquad \text{Substitute the known values.}[/latex]
[latex]3^{2} = 8^{2} + c^{2} - 2(8)c \cos 14.4° \qquad \text{Simplify.}[/latex]
[latex]9 = 64 + c^{2} - 16c (0.9686) \qquad \text{Write the quadratic equation in standard form.}[/latex]
[latex]0 = c^{2}-15.497c + 55 \qquad \text{Apply the quadratic formula.}[/latex]
Because there are two positive solutions for side [latex]c{,}[/latex] either [latex]c = 5.503[/latex] or [latex]c = 9.994{,}[/latex] there are two triangles with the given properties. We apply the law of cosines again to find angle [latex]C[/latex] in each triangle.
For the triangle with [latex]c = 5.503{,}[/latex] we have
[latex]c^{2} = a^{2} + b^{2} - 2ab \cos C \qquad \text{Substitute the known values.}[/latex]
[latex]5.503^{2} = 8^{2} + 3^{2} - 2(8)(3) \cos C \qquad \text{Solve for cos C}[/latex]
[latex]\cos C = \dfrac{5.503^{2} - 64 - 9}{-48} = 0.889876[/latex]
[latex]C = \cos^{-1} 0.889876 = 27.1°[/latex]
and [latex]A = 180° - (14.4° + 27.1°) = 138.5°{.}[/latex]
For the triangle with [latex]c = 9.994{,}[/latex] we have
[latex]c^{2} = a^{2} + b^{2} - 2ab \cos C \qquad \text{Substitute the known values.}[/latex]
[latex]9.994^{2} = 8^{2} + 3^{2} - 2(8)(3) \cos C. \text{Solve for cos C.}[/latex]
[latex]\cos C = \dfrac{9.994^{2} - 64 - 9}{-48} = -0.560027[/latex]
[latex]C = \cos^{-1} (-0.560027) = 124.1°[/latex]
and [latex]A = 180° - (14.4° + 124.1°) = 41.5°{.}[/latex] Both triangles are shown below.
Note 3.46.
In the previous example, notice that we used the law of cosines instead of the law of sines to find a second angle in the triangle because there is only one angle between [latex]0°[/latex] and [latex]180°[/latex] with a given cosine. We don’t have to check the results, as we would if we used the law of sines.
Checkpoint 3.47.
Use the law of cosines to find all triangles [latex]\triangle ABC[/latex] with [latex]\angle A = 48°,~ a = 10{,}[/latex] and [latex]b = 15{.}[/latex]
Solution
No solution
Navigation
Even with the aid of GPS (global positioning system) instruments, aircraft pilots and ship captains need to understand navigation based on trigonometry.
Example 3.48.
The sailing club leaves the marina on a heading [latex]15°[/latex] east of north and sails for [latex]18[/latex] miles. They then change course, and after traveling for [latex]12[/latex] miles on a heading [latex]35°[/latex] east of north, they experience engine trouble and radio for help. The marina sends a speedboat to rescue them. How far should the speedboat go, and on what heading?
Solution
We’d like to find the distance [latex]x[/latex] and the angle [latex]\theta[/latex] shown in the figure. In [latex]\triangle ABC{,}[/latex] we can calculate the angle at point [latex]B[/latex] where the sailing club changed course:
[latex]B = 180° - 35°+ 15° = 160°[/latex]
We know [latex]a = 12[/latex] and [latex]c = 18{.}[/latex] We use the law of cosines to find [latex]b[/latex] and [latex]\angle A{.}[/latex]
Next, we apply the law of cosines again to find [latex]\angle A{.}[/latex]
[latex]a^{2} = b^{2} + c^{2} - 2bc \cos A \qquad \text{Substitute the known values.}[/latex]
[latex]12^{2} = 29.56^{2} + 18^{2} - 2(29.56)(18) \cos A \qquad \text{Solve for cos A.}[/latex]
[latex]\cos A = \dfrac{12^{2} - 29.56^{2} - 18^{2}}{-2(29.56)(18)} = 0.9903[/latex]
[latex]C = \cos^{-1} (0.9903) = 8°[/latex]
Thus, [latex]x = 29.56[/latex] and [latex]\theta = 8° + 15° = 23°{.}[/latex] The speedboat should travel 29.56 miles on a heading [latex]23°[/latex] east of north.
Checkpoint 3.49.
Howard wants to fly from Anchorage to Nome, Alaska, a distance of 540 miles on a heading [latex]57°[/latex] west of north. After flying for some time, he discovers that his heading is in error, and he is actually flying [latex]47°[/latex] west of north. Howard corrects his flight plan and changes course when he is exactly 200 miles from Anchorage. What is his new heading, and how far is he from Nome?
Solution
[latex]62.8°[/latex] west of north, 344.8 miles
Which Law to Use
How can we decide which law, the law of sines or the law of cosines, is appropriate for a given problem?
If we are solving a right triangle, we don’t need the Laws of Sines and cosines; all we need are the definitions of the trigonometric ratios.
But for oblique triangles, we can identify the following cases:
How to Solve an Oblique Triangle.
If we know:
We can use:
1. One side and two angles [latex]SAA[/latex]
1. Law of sines to find another side
2. Two sides and the angle opposite
one of them (SSA, the ambiguous case)
2. Law of sines to find another angle,
or law of cosines to find another
side
3. Two sides and the included angle
[latex]SAS[/latex]
3. Law of cosines to find the third
side
4. Three sides [latex]SSS[/latex]
4. Law of cosines to find an angle
Note 3.50.
In the ambiguous case, SSA, the law of sines is easier to apply, but there will be two possible angles, and we must check each angle to see if it produces a solution. Using the law of cosines involves solving a quadratic equation, but each positive solution of the equation yields a solution of the triangle.
The ambiguous case can be avoided on the SSA triangle if you use the law of cosines to find the LARGEST angle first, then any method to find the remaining acute angle.
Example 3.51.
In the triangle at right, which law should you use to find [latex]\angle B[/latex]? Which law should you use to find [latex]c[/latex]?
Solution
We know two sides of the triangle and the angle opposite one of them. We can use the law of sines to find [latex]\angle B{.}[/latex]
[latex]\dfrac{\sin B}{6} = \dfrac{\sin 40°}{10}[/latex]
[latex]\sin B = 0.3857[/latex]
This is the ambiguous case; there are two angles between [latex]0°[/latex] and [latex]180°[/latex] with sine [latex]0.3857{,}[/latex] and we must check each angle to see if it produces a solution.
If instead we start by finding side [latex]c{,}[/latex] we use the law of cosines.
[latex]10^{2} = 6^{2} + c^{2} - 2(6)c \cos 40°[/latex]
[latex]c^{2} - (12 \cos 40°)c - 64 = 0[/latex]
[latex]c = \dfrac {12 \cos 40°^ {\pm} \sqrt {(-12 \cos 40°)^{2} - 4(1)(-64)}}{2(1)}[/latex] You can check that the quadratic equation has only one positive solution for [latex]c[/latex] (or notice that because [latex]a \gt b{,}[/latex] angle [latex]B[/latex] must be acute).
Checkpoint 3.52.
In the triangle at right, which part of the triangle can you find, and which law should you use?
Solution
We first find side [latex]c[/latex] using the law of cosines
The law of sines is not helpful when we know two sides of the triangle and the included angle. In this case we need the law of cosines.
Law of cosines: If the angles of a triangle are [latex]A, B{,}[/latex] and [latex]C{,}[/latex] and the opposite sides are, respectively, [latex]a, b,[/latex] and [latex]c{,}[/latex] then
[latex]a^{2} = b^{2} + c^{2} - 2bc \cos A[/latex]
[latex]b^{2} = a^{2} + c^{2} - 2ac \cos B[/latex]
[latex]c^{2} = a^{2} + b^{2} - 2ab \cos C[/latex]
We can also use the law of cosines to find an angle when we know all three sides of a triangle.
We can use the law of cosines to solve the ambiguous case.
How to solve an oblique triangle:
If we know:
We can use:
1. One side and two angles [latex]SAA[/latex]
1. Law of sines to find another side
2. Two sides and the angle opposite
one of them [latex]SSA[/latex], the ambiguous
case
2. Law of sines to find another angle,
or law of cosines to find another
side
3. Two sides and the included angle
[latex]SAS[/latex]
3. Law of cosines to find the third
side
4. Three sides [latex]SSS[/latex]
4. Law of cosines to find an angle
Study Questions
The law of cosines is really a generalization of what familiar theorem?
If you know all three sides of a triangle and one angle, what might be the advantage in using the law of cosines to find another angle instead of the law of sines?
State the quadratic formula from memory. Try to sing the quadratic formula to the tune of “Pop Goes the Weasel.”
Francine is solving a triangle in which [latex]a = 20,~ b = 16{,}[/latex] and [latex]\angle A = 26°{.}[/latex] She finds that [latex]\sin B = 0.3507{.}[/latex] How does she know that [latex]\angle B = 20.5°{,}[/latex] and not [latex]159.5°{?}[/latex]
definition
A polynomial equation with degree 2 is called quadratic. The graph of a quadratic is called a parabola.
It is simply written in the form \[ ax^2+bx + c = 0 \], where [latex]a \neq 0[/latex]
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Discriminant \[ (D) = b^2 - 4ac \]
If the discriminant is greater than zero (D > 0), then the quadratic equation has two distinct real solutions. This means the graph of the equation intersects the x-axis at two different points.
If the discriminant is equal to zero (D = 0), then the quadratic equation has one real solution.
If the discriminant is less than zero (D < 0), then the quadratic equation has no real solutions.