one angle and the law of cosines - sam houston state...
TRANSCRIPT
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.
Smith (SHSU) Elementary Functions 2013 2 / 22
Solving Triangles and the Law of Cosines
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
Smith (SHSU) Elementary Functions 2013 3 / 22
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.
Smith (SHSU) Elementary Functions 2013 4 / 22
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.
Smith (SHSU) Elementary Functions 2013 6 / 22
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◦.
Smith (SHSU) Elementary Functions 2013 8 / 22
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◦.
Smith (SHSU) Elementary Functions 2013 9 / 22
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◦.
Smith (SHSU) Elementary Functions 2013 10 / 22
Solving Triangles and the Law of Cosines
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◦.
Smith (SHSU) Elementary Functions 2013 11 / 22
Solving Triangles and the Law of Cosines
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.)
Smith (SHSU) Elementary Functions 2013 12 / 22
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.)
(End)
Smith (SHSU) Elementary Functions 2013 13 / 22
Elementary FunctionsPart 5, Trigonometry
Lecture 5.4b, Heron’s Formula, Five Guys
Dr. Ken W. Smith
Sam Houston State University
2013
Smith (SHSU) Elementary Functions 2013 14 / 22
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)
Smith (SHSU) Elementary Functions 2013 15 / 22
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.
Smith (SHSU) Elementary Functions 2013 16 / 22
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.
Smith (SHSU) Elementary Functions 2013 17 / 22
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.
Smith (SHSU) Elementary Functions 2013 20 / 22
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.
Smith (SHSU) Elementary Functions 2013 21 / 22
Five Guys
In the next presentation, we will look at polar coordinates.
(End)
Smith (SHSU) Elementary Functions 2013 22 / 22