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  • Elementary Functions Part 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 identities and 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 the origin.

    In the drawing sinC = yb and cosC = x b . We may relabel the x and y

    coordinates 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 and so

    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 and its two adjacent sides. This is the Side-Angle-Side (SAS) case, in which we may label the angle C and its two sides a and b and so we can solve for the side c.

    Or, if we have the Side-Side-Side (SSS) situation, in which we know all three sides, we can label one angle C and solve for that angle in terms of the 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 of cosines (c2 = a2 + b2 − 2ab cosC),

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

    c ≈ √ 7059 ≈ 84.018 feet.

    Now we apply the law of sines:

    sin 32◦

    84.018 =

    sinA

    100 =

    sinB

    150 This 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 angles of 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 cosines tells 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 ,

    so sinA = 2 sin 26.38◦.

    and so A = 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

    3 sin 26.38◦

    so B = 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 of cosines 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 in which 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 of sines to find two possible values of the angle B (one in which B is acute and 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 Sines to 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 would eventually take the arccosine of a number larger than 1, which is impossible.)

    Smith (SHSU) Elementary Functions 2013 12 / 22

  • The Law of Cosines

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

    (End)

    Smith (SHSU) Elementary Functions 2013 13 / 22

    Elementary Functions Part 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 the product of its base and height. Let K represent the area of a triangle. Looking at the drawing below, we see that

    K = 12ch.

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

    K = 12cb sinA and K = 1 2ac sinB

    Or we could call the known angle C and just write

    K = 1

    2 ab sinC (2)

    Smith (SHSU) Elementary Functions 2013 15 / 22

    The area of an oblique triangle

    K = 1

    2 ab sinC (3)

    We can summarize this by saying that the area of a triangle is one-half of the 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 is fixed and we can compute the three angles, we should be able to compute the area of the triangle. A first step is the formula we found when we proved the Law of Sines:

    K = 1

    2 ab sinC

    A succinct formula for the area of a triangle, given the three sides, was worked 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 = a 2+b2−c2

    2ab

    to replace cos2C:

    K2 = 14a 2b2 − (14a

    2b2)(a 2+b2−c2

    2ab ) 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 for the 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 can get this into the form

    K2 = (a+b+c2 )( −a+b+c

    2 )( a−b+c

    2 )( a+b−c

    2 )

    The perimeter of the triangle is a+ b+ c. Half of the perimeter, written s = a+b+c2 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 =

    sinC c .

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

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

    Smith (SHSU) Elementary Functions 2013 20 / 22

    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