Arkansas Tech University MATH 1203: Trigonometry Dr. Marcel B. Finan 24 The Law of Sines One important use of trigonometry is to solve problems that can be mod- eled by a triangle. Determining the measures of all the sides and angles of a triangle is referred to as solving the triangle. In Section 8, we used trigonometric functions to solve right triangles. These functions can also be used to solve oblique triangles, that is, triangles with no right angles. This can be done by using the Law of Sines to be discussed in this section and the Law of Cosines to be discussed in the next section. To simplify our discussion, we will agree that in the triangle ∆ABC, the vertices are A; B; and C, and the sides opposite these vertices are a; b; and c respectively. See Figure 24.1. Also, when writing A = 42◦ we will mean that the measure of the angle at vertex A is 42◦: Figure 24.1 The Law of Sines The Law of Sines is a relationship between the angles and the sides of a triangle. This law requires that either two angles and a side are given (AAS) or two sides and an angle are given (SSA)(note that the angle must not be the angle between the two given sides). We derive the Law of Sines as follows. Consider a triangle with sides a; b; c and angles A; B; C. Let CD be the altitude drawn from C. See Figure 24.2. 1 Figure 24.2 From Figure 24.2, we see that h = b sin A and h = a sin B: Thus, b sin A = sin A sin B 0 a sin B or a = b : Next, let AE be the altitude drawn from A: Then h = h0 0 c sin B and sin C = sin (π − \ECA) = sin \ECA = b so that h = b sin C: sin B sin C Thus, c sin B = b sin C or b = c : We conclude that a b c = = : sin A sin B sin C This relationship is known as the Law of Sines. Example 24.1 Solve the triangle ABC if a = 74:1;A = 52:1◦; and C = 35:9◦: Solution. Using the Law of Sines we can write the equality sin A sin C = : a c Solving this for c we find a sin C (74:1) sin (35:9◦) c = = ≈ 55:1: sin A sin (52:1◦) To find the angle B we use the fact that the sum of the interior angles is 180◦: Thus, B = 180◦ − (A + C) = 180◦ − (52:1◦ + 35:9◦) = 92:0◦: To find b; we use the Law of sines again sin A sin B = a b and solving for b we find a sin B 74:1 sin (92:0◦) b = = ≈ 93:8 sin A sin (52:1◦) 2 Example 24.2 Solve the triangle in Figure 24.3. Figure 24.3 Solution. Note first that B = 180◦ − (20◦ + 25◦) = 135◦: By the Law of Sines we have a c = : sin A sin C Solving this equation for a we find c sin A 80:4 sin 20◦ a = = ≈ 65:1: sin C sin 25◦ b c Similarly, to find b we use sin B = sin C : Solving for b c sin B 80:4 sin 135◦ b = = ≈ 134:5: sin C sin 25◦ The Ambiguous Case (SSA) Given two sides and the angle opposite one of them. One must compute the angle opposite the other side. Recall that if the sine function is positive, there are two possible answers, a first quadrant angle and a second quadrant angle. Since the sum of all angles of a triangle is 180 degrees, there may be two possible answers. Recall further that the sine function must be less than or equal to one. If the sine exceeds one, there is no solution, and thus no triangle. The three examples below show the three possible scenarios for this case: no solution, one solution, and two solutions. Example 24.3 (No Solution) If a = 5:7; b = 8:8 and A = 68:7◦; find c, B and C for any possible triangles. 3 Solution. This is SSA so we use the Law of Sines where we have ambiguity. We first sin B sin A b sin A have b = a which gives sin B = a : From that we get sin B = 8:8 sin 68:7◦ 5:7 or sin B ≈ 1:438: This is impossible since the sine of an angle can not exceed 1, so there are no possible triangles. See Figure 24.4. Figure 24.4 Example 24.4 (One Solution) If a = 2:0; b = 1:4; and A = 44:5◦, find c; B; and C for any possible triangles. Solution. This is SSA so we use the Law of Sines where we have ambiguity. We first sin B sin A b sin A have b = a which gives sin B = a : From that we get sin B = 1:4 sin 44:5◦ ◦ ◦ 2:0 ≈ 0:4906: Hence, either B = arcsin (0:4906) ≈ 29:4 or B = 180 − 29:4◦ ≈ 150:6◦ ( reject since 44:5◦ +150:6◦ > 180◦). Now, C = 180◦ −44:5◦ − 29:4◦ = 106:1◦: sin A sin C a sin C To find c; we use the Law of Sines again, a = c or c = sin A = (2:0) sin (106:1◦) sin 44:5◦ ≈ 2:7: So there is one possible triangle as shown in Figure 24.5. Figure 24.5 Example 24.5 (Two Solutions) If a = 32:2; b = 20:3andB = 20◦; find c, B and C for any possible triangles. 4 Solution. This is an SSA problem so we will use the Law of Sines. Keep in mind sin B sin A that this is the ambiguous case. We first have b = a which gives a sin B 32:2 sin 20◦ sin A = b : From that we get sin A = 20:3 ≈ 0:5425: Now, A = arcsin 0:5425 ≈ 32:9◦: However, the angle 180◦ − 32:9◦ = 147:1◦ gives the same sine value. We therefore can construct two triangles. ◦ ◦ ◦ ◦ ◦ Triangle # 1 - A1 = 32:9 : Then C1 = 180 − 20 − 32:9 = 127:1 : By sin C1 sin A1 a sin C1 the Law of Sines, = , giving c1 = ≈ 47:3: This does leave c1 a sin A the largest side opposite the largest angle and the smallest side opposite the smallest angle so we suspect our work was probably right. ◦ ◦ ◦ ◦ ◦ Triangle # 2 - A2 = 147:1 : Then C2 = 180 − 147:1 − 20 = 12:9 : By sin C2 sin A2 a sin C2 the Law of sines, = giving c2 = ≈ 13:2: This also leaves c2 a sin A2 the largest side opposite the largest angle and the smallest side opposite the smallest angle so we suspect our work was probably right. See Figure 24.6. Figure 24.6 Example 24.6 A forest fire is spotted by observers in two fire towers 12 miles apart. Tower B is on a bearing of S12◦100E from Tower A. If the bearing of the fire from Tower A is S45◦400W and from Tower B is N75◦200W , how far is the fire from Tower B? Solution. Using Figure 24.7 we see that the triangle ABC has interior angles of measure A = 57◦500;B = 63◦100 and C = 180◦ − 57◦500 − 63◦100 = 59◦: The problem sin A sin C c sin A is to find a: By the Law of Sines, a = c : This implies that a = sin C = 12 sin 57◦500 sin 59◦ ≈ 11:9 miles. 5 Figure 24.7 6.