11.2 The Law of Sines - Sam Houston State kws006/Precalculus/5.3_Law_of_Sines_files/S&Z 11.2...

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  • 894 Applications of Trigonometry

    11.2 The Law of Sines

    Trigonometry literally means measuring triangles and with Chapter 10 under our belts, we aremore than prepared to do just that. The main goal of this section and the next is to developtheorems which allow us to solve triangles that is, find the length of each side of a triangleand the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, weve had some experiencesolving right triangles. The following example reviews what we know.

    Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length4 units, find the length of the remaining side and the measures of the remaining angles. Expressthe angles in decimal degrees, rounded to the nearest hundreth of a degree.

    Solution. For definitiveness, we label the triangle below.

    b = 4

    a

    c=

    7

    To find the length of the missing side a, we use the Pythagorean Theorem to get a2 + 42 = 72

    which then yields a =

    33 units. Now that all three sides of the triangle are known, there areseveral ways we can find using the inverse trigonometric functions. To decrease the chances ofpropagating error, however, we stick to using the data given to us in the problem. In this case, thelengths 4 and 7 were given, so we want to relate these to . According to Theorem 10.4, cos() = 47 .Since is an acute angle, = arccos

    (47

    )radians. Converting to degrees, we find 55.15. Now

    that we have the measure of angle , we could find the measure of angle using the fact that and are complements so + = 90. Once again, we opt to use the data given to us in theproblem. According to Theorem 10.4, we have that sin() = 47 so = arcsin

    (47

    )radians and we

    have 34.85.

    A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lowercase Greek letter denotes an angle1 and the corresponding lowercase English letter represents theside2 opposite that angle. Thus, a is the side opposite , b is the side opposite and c is the sideopposite . Taken together, the pairs (, a), (, b) and (, c) are called angle-side opposite pairs.Second, as mentioned earlier, we will strive to solve for quantities using the original data given inthe problem whenever possible. While this is not always the easiest or fastest way to proceed, it

    1as well as the measure of said angle2as well as the length of said side

  • 11.2 The Law of Sines 895

    minimizes the chances of propagated error.3 Third, since many of the applications which requiresolving triangles in the wild rely on degree measure, we shall adopt this convention for the timebeing.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handleany given right triangle problem, but what if the triangle isnt a right triangle? In certain cases,we can use the Law of Sines to help.

    Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (, a),(, b) and (, c), the following ratios hold

    sin()

    a=

    sin()

    b=

    sin()

    c

    or, equivalently,

    a

    sin()=

    b

    sin()=

    c

    sin()

    The proof of the Law of Sines can be broken into three cases. For our first case, consider thetriangle 4ABC below, all of whose angles are acute, with angle-side opposite pairs (, a), (, b)and (, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles:4ABQ and 4BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4that sin() = hc and sin() =

    ha so that h = c sin() = a sin(). After some rearrangement of the

    last equation, we get sin()a =sin()c . If we drop an altitude from vertex A, we can proceed as above

    using the triangles 4ABQ and 4ACQ to get sin()b =sin()c , completing the proof for this case.

    a

    b

    c

    A C

    B

    ac

    A C

    B

    Q

    h

    b

    c

    A C

    B

    Q

    h

    For our next case consider the triangle 4ABC below with obtuse angle . Extending an altitudefrom vertex A gives two right triangles, as in the previous case: 4ABQ and 4ACQ. Proceedingas before, we get h = b sin() and h = c sin() so that sin()b =

    sin()c .

    a

    b

    c

    A

    B

    C

    a

    b

    c

    A

    B

    C

    Q

    h

    3Your Science teachers should thank us for this.4Dont worry! Radians will be back before you know it!

  • 896 Applications of Trigonometry

    Dropping an altitude from vertex B also generates two right triangles, 4ABQ and 4BCQ. Weknow that sin() = h

    c so that h = c sin(). Since = 180 , sin() = sin(), so in fact,

    we have h = c sin(). Proceeding to 4BCQ, we get sin() = ha so h = a sin(). Putting this

    together with the previous equation, we get sin()c =sin()a , and we are finished with this case.

    a

    b

    c

    A

    B

    CQ

    h

    The remaining case is when 4ABC is a right triangle. In this case, the Law of Sines reduces tothe formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines tosolve a triangle, we need at least one angle-side opposite pair. The next example showcases someof the power, and the pitfalls, of the Law of Sines.

    Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations(rounded to hundredths) and sketch the triangle.

    1. = 120, a = 7 units, = 45 2. = 85, = 30, c = 5.25 units

    3. = 30, a = 1 units, c = 4 units 4. = 30, a = 2 units, c = 4 units

    5. = 30, a = 3 units, c = 4 units 6. = 30, a = 4 units, c = 4 units

    Solution.

    1. Knowing an angle-side opposite pair, namely and a, we may proceed in using the Law of

    Sines. Since = 45, we use bsin(45) =7

    sin(120) so b =7 sin(45)sin(120) =

    7

    63 5.72 units. Now that

    we have two angle-side pairs, it is time to find the third. To find , we use the fact that thesum of the measures of the angles in a triangle is 180. Hence, = 180 120 45 = 15.To find c, we have no choice but to used the derived value = 15, yet we can minimize thepropagation of error here by using the given angle-side opposite pair (, a). The Law of Sines

    gives us csin(15) =7

    sin(120) so that c =7 sin(15)sin(120) 2.09 units.

    5

    2. In this example, we are not immediately given an angle-side opposite pair, but as we havethe measures of and , we can solve for since = 180 85 30 = 65. As in theprevious example, we are forced to use a derived value in our computations since the only

    5The exact value of sin(15) could be found using the difference identity for sine or a half-angle formula, but that

    becomes unnecessarily messy for the discussion at hand. Thus exact here means7 sin(15)sin(120) .

  • 11.2 The Law of Sines 897

    angle-side pair available is (, c). The Law of Sines gives asin(85) =5.25

    sin(65) . After the usual

    rearrangement, we get a = 5.25 sin(85)

    sin(65) 5.77 units. To find b we use the angle-side pair (, c)which yields bsin(30) =

    5.25sin(65) hence b =

    5.25 sin(30)sin(65) 2.90 units.

    a = 7

    b 5.72

    c 2.09 = 120 = 15

    = 45

    a 5.77

    b 2.90

    c = 5.25

    = 85 = 65

    = 30

    Triangle for number 1 Triangle for number 2

    3. Since we are given (, a) and c, we use the Law of Sines to find the measure of . We start

    with sin()4 =sin(30)

    1 and get sin() = 4 sin (30) = 2. Since the range of the sine function is

    [1, 1], there is no real number with sin() = 2. Geometrically, we see that side a is just tooshort to make a triangle. The next three examples keep the same values for the measure of and the length of c while varying the length of a. We will discuss this case in more detailafter we see what happens in those examples.

    4. In this case, we have the measure of = 30, a = 2 and c = 4. Using the Law of Sines,we get sin()4 =

    sin(30)2 so sin() = 2 sin (30

    ) = 1. Now is an angle in a triangle whichalso contains = 30. This means that must measure between 0 and 150 in orderto fit inside the triangle with . The only angle that satisfies this requirement and hassin() = 1 is = 90. In other words, we have a right triangle. We find the measure of to be = 180 30 90 = 60 and then determine b using the Law of Sines. We findb = 2 sin(60

    )sin(30) = 2

    3 3.46 units. In this case, the side a is precisely long enough to form a

    unique right triangle.

    a = 1c = 4

    = 30

    a = 2c = 4

    b 3.46

    = 30

    = 60

    Diagram for number 3 Triangle for number 4

    5. Proceeding as we have in the previous two examples, we use the Law of Sines to find . In thiscase, we have sin()4 =

    sin(30)3 or sin() =

    4 sin(30)3 =

    23 . Since lies in a triangle with = 30

    ,

  • 898 Applications of Trigonometry

    we must have that 0 < < 150. There are two angles that fall in this range and havesin() = 23 : = arcsin

    (23

    )radians 41.81 and = arcsin

    (23

    )radians 138.19. At

    this point, we pause to see if it makes sense that we actually have two viable cases to consider.As we have discussed, both candidates for are compatible with the given angle-side pair(, a) = (30, 3) in that both choices for can fit in a triangle with and both have a sine of23 . The only other given piece of information is that c = 4 units. Since c > a, it must be truethat , which is opposite c, has greater measure than which is opposite a. In both cases, > , so both candidates for are compatible with this last piece of given information aswell. Thus have two triangles on our hands. In the case = arcsin

    (23

    )radians 41.81, we

    find6 180 30 41.81 = 108.19. Using the