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11.2 the Law of Sines

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11.2 the Law of Sines 894 Applications of Trigonometry 11.2 The Law of Sines Trigonometry literally means `measuring triangles' and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to `solve' triangles { that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we've had some experience solving 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 length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For definitiveness, we label the triangle below. β = 7 c a α b = 4 2 2 2 To find the length of thep missing side a, we use the Pythagorean Theorem to get a + 4 = 7 which then yields a = 33 units. Now that all three sides of the triangle are known, there are several ways we can find α using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the 4 lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 7 . 4 ◦ Since α is an acute angle, α = arccos 7 radians. Converting to degrees, we find α ≈ 55:15 . Now that we have the measure of angle α, we could find the measure of angle β using the fact that α and β are complements so α + β = 90◦. Once again, we opt to use the data given to us in the 4 4 problem. According to Theorem 10.4, we have that sin(β) = 7 so β = arcsin 7 radians and we have β ≈ 34:85◦. A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α; a), (β; b) and (γ; c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it 1as well as the measure of said angle 2as 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 require solving triangles `in the wild' rely on degree measure, we shall adopt this convention for the time being.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn't 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 (α; a), (β; b) and (γ; c), the following ratios hold sin(α) sin(β) sin(γ) = = a b c or, equivalently, a b c = = sin(α) sin(β) sin(γ) The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle 4ABC below, all of whose angles are acute, with angle-side opposite pairs (α; a), (β; b) and (γ; 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 10.4 h h that sin(α) = c and sin(γ) = a so that h = c sin(α) = a sin(γ). After some rearrangement of the sin(α) sin(γ) last equation, we get a = c . If we drop an altitude from vertex A, we can proceed as above sin(β) sin(γ) using the triangles 4ABQ and 4ACQ to get b = c , completing the proof for this case. B B B β Q c β a c h a c h0 γ α γ α γ AC AC AC b Q b For our next case consider the triangle 4ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: 4ABQ and 4ACQ. Proceeding sin(β) sin(γ) as before, we get h = b sin(γ) and h = c sin(β) so that b = c . B B Q β a β a c α c h γ γ A b C A b C 3Your Science teachers should thank us for this. 4Don't 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. We 0 h0 0 0 0 ◦ 0 know that sin(α ) = c so that h = c sin(α ). Since α = 180 − α, sin(α ) = sin(α), so in fact, 0 h0 0 we have h = c sin(α). Proceeding to 4BCQ, we get sin(γ) = a so h = a sin(γ). Putting this sin(γ) sin(α) together with the previous equation, we get c = a , and we are finished with this case. B β a h0 c α0 α γ Q A b C The remaining case is when 4ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of 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. α = 120◦, a = 7 units, β = 45◦ 2. α = 85◦, β = 30◦, c = 5:25 units 3. α = 30◦, a = 1 units, c = 4 units 4. α = 30◦, a = 2 units, c = 4 units 5. α = 30◦, a = 3 units, c = 4 units 6. α = 30◦, a = 4 units, c = 4 units Solution. 1. Knowing an angle-side opposite pair, namely α and a, we may proceedp in using the Law of ◦ b 7 7 sin(45◦) 7 6 Sines. Since β = 45 , we use sin(45◦) = sin(120◦) so b = sin(120◦) = 3 ≈ 5:72 units. Now that we have two angle-side pairs, it is time to find the third. To find γ, we use the fact that the sum of the measures of the angles in a triangle is 180◦. Hence, γ = 180◦ − 120◦ − 45◦ = 15◦. To find c, we have no choice but to used the derived value γ = 15◦, yet we can minimize the propagation of error here by using the given angle-side opposite pair (α; a). The Law of Sines c 7 7 sin(15◦) 5 gives us sin(15◦) = sin(120◦) so that c = sin(120◦) ≈ 2:09 units. 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦. As in the previous 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 7 sin(15◦) becomes unnecessarily messy for the discussion at hand. Thus \exact" here means sin(120◦) . 11.2 The Law of Sines 897 a 5:25 angle-side pair available is (γ; c). The Law of Sines gives sin(85◦) = sin(65◦) . After the usual 5:25 sin(85◦) rearrangement, we get a = sin(65◦) ≈ 5:77 units. To find b we use the angle-side pair (γ; c) b 5:25 5:25 sin(30◦) which yields sin(30◦) = sin(65◦) hence b = sin(65◦) ≈ 2:90 units. β = 30◦ c = 5:25 a ≈ 5:77 β = 45◦ a = 7 ◦ c ≈ 2:09 α = 120 α = 85◦ γ = 65◦ γ = 15◦ b ≈ 5:72 b ≈ 2:90 Triangle for number1 Triangle for number2 3. Since we are given (α; a) and c, we use the Law of Sines to find the measure of γ. We start sin(γ) sin(30◦) ◦ with 4 = 1 and get sin(γ) = 4 sin (30 ) = 2. Since the range of the sine function is [−1; 1], there is no real number with sin(γ) = 2. Geometrically, we see that side a is just too short to make a triangle. The next three examples keep the same values for the measure of α and the length of c while varying the length of a. We will discuss this case in more detail after we see what happens in those examples. 4. In this case, we have the measure of α = 30◦, a = 2 and c = 4. Using the Law of Sines, sin(γ) sin(30◦) ◦ we get 4 = 2 so sin(γ) = 2 sin (30 ) = 1. Now γ is an angle in a triangle which also contains α = 30◦.
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