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Recall In a triangle, the sum of the interior angles is 180º. No triangle can have two obtuse angles. The height of a triangle is less than or. equal to the length of two of the sides. Copyright 2007 Pearson Education, Inc. Slide 10-3

Recall The sine function has a range of 1 sin 1 If the θ is a positive decimal < 1, the θ can lie in the first quadrant (acute <) or in the second quadrant (obtuse <).. sin sin(180 ) Copyright 2007 Pearson Education, Inc. Slide 10-4

Case 1 Case 2 Case 3 Case 4 9.3 Data Required for Solving Oblique Triangles One side and two angles known:» SAA or ASA Two sides and one angle not included between the sides known:» SSA» This case may lead to more than one solution. Two sides and one angle included between the sides known:» SAS Three sides are known:» SSS Copyright 2007 Pearson Education, Inc. Slide 10-5

9.3 Derivation of the Law of Sines For our purposes we will use the triangles labelled as : The orientation of the triangle will not make a difference. Copyright 2007 Pearson Education, Inc. Slide 10-6

9.3 Derivation of the Law of Sines Start with an acute or obtuse triangle and construct the perpendicular from B to side AC. Let h be the height of this perpendicular. Then c and a are the hypotenuses of right triangle ADB and BDC, respectively. h sin A c h sin C a or or asin C csin A h csin A, h asin C a c sin A sin C Copyright 2007 Pearson Education, Inc. Slide 10-7

9.3 The Law of Sines In a similar way, by constructing perpendiculars from other vertices, the following theorem can be proven. The Law of Sine Use to find ANGLES sin A sin B sin C = = a b c Use to find SIDES a sin A = b sin B = c sin C Copyright 2007 Pearson Education, Inc. Slide 10-8

9.3 Using the Law of Sines to Solve a Triangle Example Solve triangle ABC if A = 32.0, B = 81.8, and a = 42.9 centimeters. Solution Draw the triangle and label the known values. Because A, B, and a are known, we can apply the law of sines involving these variables. a sin A b sin B 42.9 b sin 32.0 sin 81.8 b 80.1cm Copyright 2007 Pearson Education, Inc. Slide 10-10

9.3 Using the Law of Sines to Solve a Triangle To find C, use the fact that there are 180 in a triangle. A B C Now we can find c a sin A C c sin C 180 180 AC 180 32.0 81.8 66.2 42.9 sin 32.0 c sin 66.2 c 74.1cm Copyright 2007 Pearson Education, Inc. Slide 10-11

Law of Sines Day 2 For each given question 1-9: Draw a triangle with proper labels and place the given values on the diagram. Identify the combination given i.e. SAS, etc. We will be using these diagrams to solve each triangle and to record notes so allow room around each shape for further writing. Copyright 2007 Pearson Education, Inc. Slide 10-12

For each diagram: Identify whether L.O.S. can be used. If LOS can be used: How many solutions:0, 1 or 2? Can you make some generalization; or write a program? As you make any generalization, update your triangle table. Copyright 2007 Pearson Education, Inc. Slide 10-14

Repeat the process with the following 1)A = 75 ; a = 8, b = 7 2)A = 135 ; a = 9, b = 12 3)A = 25 ; a = 4, b = 12 4)A = 95 ; B = 35, b = 13 5)A = 43 ; a = 15, b = 17 6)C = 82 ; a = 14, b = 10 7)A = 56 ; C = 84, b = 6 8)A = 150 ; a = 9, b = 6 9)A = 140 ; C = 150, b = 8 Copyright 2007 Pearson Education, Inc. Slide 10-16

9.3 Using the Law of Sines in an Application (ASA) Example Two stations are on an east-west line 110 miles apart. A forest fire is located on a bearing of N 42 E from the western station at A and a bearing of N 15 E from the eastern station at B. How far is the fire from the western station? Solution Angle BAC = 90 42 = 48 Angle B = 90 + 15 = 105 Angle C = 180 105 48 = 27 Using the law of sines to find b gives b 110 sin 105 sin 27 b 234 miles. Copyright 2007 Pearson Education, Inc. Slide 10-17

9.3 Ambiguous Case If given the lengths of two sides and the angle opposite one of them, it is possible that 0, 1, or 2 such triangles exist. Some basic facts that should be kept in mind: For any angle, 1 sin 1, if sin = 1, then = 90 and the triangle is a right triangle. sin = sin(180 ). The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming unequal sides). Copyright 2007 Pearson Education, Inc. Slide 10-18

9.3 Number of Triangles Satisfying the Ambiguous Case Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate sin B.) 1. If sin B > 1, then no triangle satisfies the given conditions. 2. If sin B = 1, then one triangle satisfies the given conditions and B = 90. 3. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions (a) If sin B = k, then let B 1 = sin -1 k and use B 1 for B in the first triangle. b) Let B 2 = 180 B 1. If A + B 2 < 180, then a second triangle exists. In this case, use B 2 for B in the second triangle. Copyright 2007 Pearson Education, Inc. Slide 10-19

9.3 Solving the Ambiguous Case: One Triangle Example Solve the triangle ABC, given A = 43.5, a = 10.7 inches, and c = 7.2 inches. Solution sin C 7.2 sin 43.5 10.7 sin C.46319186 The other possible value for C: C = 180 27.6 = 152.4. C 27.6 Add this to A: 152.4 + 43.5 = 195.9 > 180 Therefore, there can be only one triangle. Copyright 2007 Pearson Education, Inc. Slide 10-23

9.3 Solving the Ambiguous Case: No Such Triangle Example Solve the triangle ABC if B = 55 40, b = 8.94 meters, and a = 25.1 meters. Solution Use the law of sines to find A. sin A sin B a b sin A 25.1 sin 55 40' 8.94 sin A 2.3184379 Since sin A cannot be greater than 1, the triangle does not exist. Copyright 2007 Pearson Education, Inc. Slide 10-25

Example Solve the triangle ABC if A = 55.3, a = 22.8 feet, and b = 24.9 feet. Solution 9.3 Solving the Ambiguous Case: Two Triangles sin A sin B a b sin 55.3 sin B 22.8 24.9 sin B.8978678 B 2 180 63.9 116.1 B 63.9 1 Copyright 2007 Pearson Education, Inc. Slide 10-26

9.3 Solving the Ambiguous Case: Two Triangles To see if B 2 = 116.1 is a valid possibility, add 116.1 to the measure of A: 116.1 + 55.3 = 171.4. Since this sum is less than 180, it is a valid triangle. Now separate the triangles into two: AB 1 C 1 and AB 2 C 2. C 1 180 A B1 180 55.3 63.9 a sin A c1 sin C 22.8 c1 sin 55.3 sin 60.8 1 c 1 24.2 feet 60.8 Copyright 2007 Pearson Education, Inc. Slide 10-27

9.3 Solving the Ambiguous Case: Two Triangles Now solve for triangle AB 2 C 2. C 2 180 A B2 180 55.3 116.1 8.6 a sin A c2 sin C 22.8 c2 sin 55.3 sin 8.6 c 2 4.15 feet 2 Copyright 2007 Pearson Education, Inc. Slide 10-28

Practice: Answer in pairs. Find m B, m C, and c, if they exist. 1) a = 9.1, b = 12, ma = 35 o 2) a = 25, b = 46, ma = 37 o 3) a = 15, b = 10, ma = 66 o Copyright 2007 Pearson Education, Inc. Slide 10-29

Practice Problems on your own: The problems below is a sampling of what you should try; feel free to try other problems also: Sec 9-3; Written Exercises, page 347 #8,14, 17-22 all Copyright 2007 Pearson Education, Inc. Slide 10-31