One Angle and the Law of Cosines - Sam Houston State ...kws006/Precalculus/5.4_Law_of_Cosines... ·...

Elementary FunctionsPart 5, Trigonometry

Lecture 5.4a, The Law of Cosines

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 22

Solving Triangles and the Law of Cosines

In this section we work out the law of cosines from our earlier identitiesand then practice applying this new identity.

c2 = a2 + b2 − 2ab cosC. (1)

Draw the triangle 4ABC on the Cartesian plane with the vertex C at theorigin.

In the drawing sinC = yb and cosC = x

b . We may relabel the x and ycoordinates of A(x, y) as x = b cosC and y = b sinC.



We get information if we compute c2. By the Pythagorean theorem,

c2 = (y2) + (a− x)2 = (b sinC)2 + (a− b cosC)2

= b2 sin2C + a2 − 2ab cosC + b2 cos2C.

We use the Pythagorean identity to simplify b2 sin2C + b2 cos2C = b2 andso

c2 = a2 + b2 − 2ab cosC


One Angle and the Law of Cosines

c2 = a2 + b2 − 2ab cosC.

It is straightforward to use the law of cosines when we know one angle andits two adjacent sides. This is the Side-Angle-Side (SAS) case, in whichwe may label the angle C and its two sides a and b and so we can solve forthe side c.

Or, if we have the Side-Side-Side (SSS) situation, in which we know allthree sides, we can label one angle C and solve for that angle in terms ofthe sides a, b and c, using the law of cosines.


A Worked Problem.

Solve the triangle C = 32◦, a = 100 feet, b = 150 feet.

Solution. If C = 32◦, a = 100 feet, b = 150 feet then, by the law ofcosines (c2 = a2 + b2 − 2ab cosC),

c2 = (100)2 + (150)2 − 2(100)(150) cos(32◦) ≈ 7059.

Soc ≈√7059 ≈ 84.018 feet.

Now we apply the law of sines:

sin 32◦

84.018=

sinA

100=

sinB

150This forces

sinA =100 sin 32◦

84.018≈ 0.6307

and

sinB =150 sin 32◦

84.018≈ 0.9461

If the sine of A is 0.6307 then either

A ≈ sin−1(0.63007) ≈ 39.1◦ or A ≈ 180− 39.1 = 140.9◦.Smith (SHSU) Elementary Functions 2013 5 / 22

A Worked Problem.

We solve the triangle C = 32◦, a = 100 feet, b = 150 feet.We found

A ≈ sin−1(0.63007) ≈ 39.1◦ or A ≈ 180− 39.1 = 140.9◦.

The second answer is too big, so A = 39.1◦.Since sinB = 0.9461 then either

B ≈ sin−1(0.94510) ≈ 71.1◦ or B = 180− 71.1 = 108.9◦.

The second answer, B = 108.9◦ , makes perfect sense because the anglesof the triangle need to add up to 180 degrees.


The law of cosines and SSS

Solve the triangle a = 30 feet, b = 20 feet, c = 15 feet.

Solution. If a = 30 feet, b = 20 feet, c = 15 feet then the law of cosinestells us that

152 − 302 − 202

−2(30)(20)= cosC

so−1075−1200

=43

48= cosC

Therefore

C = cos−1(43

48) ≈ 26.38◦ .

By the law of sines

sin 26.38◦

15=

sinA

30=

sinB

20,

sosinA = 2 sin 26.38◦.

and soA = 62.7◦ or A = 180◦ − 62.7◦ = 117.3◦Smith (SHSU) Elementary Functions 2013 7 / 22

The law of cosines and SSS

We solve the triangle a = 30 feet, b = 20 feet, c = 15 feet.We found

A = 62.7◦ or A = 180◦ − 62.7◦ = 117.3◦

Also

sinB =4

3sin 26.38◦

soB = 36.33◦ or 180◦ − 36.33◦ = 143.67◦

The second result here is too large.

Our final answer, after checking that angles add up to 180◦, is

A = 117.3◦, B = 36.3◦, C = 26.4◦.


Some Worked Problems.

A triangle has angle C = 20◦ and sides a = 10, b = 20. Use the law ofcosines to find the length of the side c.

Solution.

c2 = 102 + 202 − 2(10)(20) cos 20◦ = 500− 400 cos 20◦ ≈ 124.1

So c ≈√124.1 ≈ 11.14.

Use the law of sines to find two possible values of the angle A (one inwhich A is acute and one in which A is obtuse.)

Solution. The sine of A is 0.307 so A = 17.88◦ or 162.12◦.


Some Worked Problems.

A triangle has angle C = 20◦ and sides a = 10, b = 20. Use the law ofsines to find two possible values of the angle B (one in which B is acuteand one in which B is obtuse.)

Solution. The sine of B is 0.614 so B ≈ 37.88◦ or 142.12◦.

Solve the triangle.Solution. c ≈ 11.14, A ≈ 17.88◦ and B ≈ 142.12◦.



Solve the triangle with sides a = 15, b = 12, c = 10.

Solution. By the Law of Cosines, C = 41.65◦. Now use the Law of Sinesto get

a = 15, b = 12, c = 10, A = 85.46◦, B = 52.89◦, C = 41.65◦.



Solve the triangles with sides a = 15, b = 12, c = 30.

Solution. Since one side is longer than the sum of the other two sides,no such triangle is possible.

(No calculations are necessary here. If we did a calculation, we wouldeventually take the arccosine of a number larger than 1, which isimpossible.)


The Law of Cosines

In the next presentation, we will look at Heron’s formula and the “FiveGuys”, (five trig identities I give to my classes.)

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Elementary FunctionsPart 5, Trigonometry

Lecture 5.4b, Heron’s Formula, Five Guys

Dr. Ken W. Smith

Sam Houston State University

2013


The area of an oblique triangle

Since a triangle is half of a parallelogram, its area is one-half of theproduct of its base and height.Let K represent the area of a triangle. Looking at the drawing below, wesee that

K = 12ch.

But earlier, in our proof of the Law of Sines, we solved for h and we wroteh = b sinA and h = a sinB. So we can substitute for h and write the areaas

K = 12cb sinA and K = 1

2ac sinB

Or we could call the known angle C and just write

K =1

2ab sinC (2)


The area of an oblique triangle

K =1

2ab sinC (3)

We can summarize this by saying that the area of a triangle is one-half ofthe product of the sine of an angle and its neighboring sides.


Heron’s formula

If we know the three sides a, b and c then in theory, since the triangle isfixed and we can compute the three angles, we should be able to computethe area of the triangle.A first step is the formula we found when we proved the Law of Sines:

K =1

2ab sinC

A succinct formula for the area of a triangle, given the three sides, wasworked out long ago by Heron of Alexandria.


Heron’s formula

Let’s take the area equation

K = 12ab sinC

and square both sides.

K2 = 14a

2b2 sin2C.

By the Pythagorean identity replace sin2C by 1− cos2C

K2 = 14a

2b2(1− cos2C).

and distribute

K2 = 14a

2b2 − 14a

2b2 cos2C.

Now use the Law of Cosines in the form

cosC = a2+b2−c22ab

to replace cos2C:

K2 = 14a

2b2 − (14a2b2)(a

2+b2−c22ab )2

and simplify

K2 = 14a

2b2 − ( 116)(a

2 + b2 − c2)2Smith (SHSU) Elementary Functions 2013 18 / 22

Heron’s formula

Using the law of sine and the law of cosines, we worked out a formula forthe area:

K2 = 14a

2b2 − ( 116)(a

2 + b2 − c2)2

K2 = 116(4a

2b2 − (a2 + b2 − c2)2)

Now, with a little bit of algebra ... (skipping a bunch of steps!) ... we canget this into the form

K2 = (a+b+c2 )(−a+b+c2 )(a−b+c2 )(a+b−c2 )

The perimeter of the triangle is a+ b+ c. Half of the perimeter, writtens = a+b+c

2 is called the semiperimeter.

We can use the semiperimeter s to shorten this equation

K2 = s(s− a)(s− b)(s− c).So the area of a triangle is

K =√s(s− a)(s− b)(s− c).

This is Heron’s formulaSmith (SHSU) Elementary Functions 2013 19 / 22

Summary & Five Guys

Summary of our identities

We have come across two very useful identities that are easy to remember:

1 The Pythagorean identity, cos2 θ + sin2 θ = 1 (and its two siblings),and

2 The Law of Sines, sinAa = sinB

b = sinCc .

There are five more identities that are very useful and if we know them (orhave them handy) then the other trig identities follow easily from them.

Our goal here is understanding, not memorization! But if one were tomemorize trig identities, in addition to the easy two above, I recommendthe following five.


http://en.wikipedia.org/wiki/Heron_of_alexandria

Five Guys

1 The formulas for the sine and cosine of the sum of two angles:

cos(α+ β) = cosα cosβ − sinα sinβ

sin(α+ β) = sinα cosβ + cosα sinβ

2 The power reduction formulas:

cos2 α = 1+cos(2α)2

sin2 α = 1−cos(2α)2

3 The law of cosines: c2 = a2 + b2 − 2ab cosC

In my precalculus and trigonometry classes these five identities, boxedabove, will be provided on quizzes and exams. Since I have nevermemorized these – and I’ve done well in mathematics, emphasizingunderstanding over memorization – then I won’t ask my students tomemorize them.Students should focus on understanding how to wield these five guys in avariety of environments.


Five Guys

In the next presentation, we will look at polar coordinates.

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