Chapter 10: Applications of Trigonometry; Vectors

2. Chapter 10: Applications of Trigonometry; Vectors 10.1 The Law of Sines 10.2 The Law of Cosines and Area Formulas 10.3 Vectors and Their Applications 10.4 Trigonometric (Polar) Form of Complex Numbers 10.5 Powers and Roots of Complex Numbers 10.6 Polar Equations and Graphs 10.7 More Parametric Equations

3. 10.1 The Law of Sines Congruence Axioms Side-Angle-Side (SAS) If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent. Angle-Side-Angle (ASA) If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent. Side-Side-Side (SSS) If three sides of one triangle are equal to three sides of a second triangle, the triangles are congruent.

4. 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. • The sine function has a range of • If the θ is a positive decimal < 1, the θ can lie in the first quadrant (acute <) or in the second quadrant (obtuse <). • .

5. 10.1 Data Required for Solving Oblique Triangles Case 1 One side and two angles known: • SAA or ASA • Case 2 Two sides and one angle not included • between the sides known: • SSA • This case may lead to more than one solution. • Case 3 Two sides and one angle included between • the sides known: • SAS • Case 4 Three sides are known: • SSS

6. 10.1 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.

7. 10.1 The Law of Sines In a similar way, by constructing perpendiculars from other vertices, the following theorem can be proven. Law of Sines In any triangle ABC,with sides a, b, and c, Alternative forms are sometimes convenient to use:

8. 10.1 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.

9. 10.1 Using the Law of Sines to Solve a Triangle To find C, use the fact that there are 180° in a triangle. Now we can find c

10. 10.1 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

11. 10.1 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).

12. 10.1 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.) • If sin B > 1, then no triangle satisfies the given conditions. If sin B = 1, then one triangle satisfies the given conditions and B = 90°. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions • If sin B = k, then let B1 = sin-1 k and use B1 for B in the first triangle. • Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists. In this case, use B2 for B in the second triangle.

13. 10.1 Ambiguous Case a < b sinA a < b sinA a = b sinA a = b sinA a > b sinA

14. How does it work? • Web • demo

15. 10.1 Ambiguous Case for Obtuse Angle A

16. 10.1 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 The other possible value for C: C = 180° – 27.6° = 152.4°. Add this to A: 152.4° + 43.5° = 195.9° > 180° Therefore, there can be only one triangle.

17. 10.1 Solving the Ambiguous Case: One Triangle

18. 10.1 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. Since sin A cannot be greater than 1, the triangle does not exist.

19. 10.1 Solving the Ambiguous Case: Two Triangles Example Solve the triangle ABC if A = 55.3°, a = 22.8 feet, and b = 24.9 feet. Solution

20. 10.1 Solving the Ambiguous Case: Two Triangles To see if B2 = 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: AB1C1 and AB2C2.

21. 10.1 Solving the Ambiguous Case: Two Triangles Now solve for triangle AB2C2.

22. Practice: Answer in pairs. Find mB, mC, and c, if they exist. 1) a = 9.1, b = 12, mA = 35o  2) a = 25, b = 46, mA = 37o 3) a = 15, b = 10, mA = 66o

23. Answers: 1)Case 1: mB=49o,mC=96o,c=15.78 Case 2: mB=131o,mC=14o,c=3.84 2)No possible solution. 3)mB=38o,mC=76o,c=15.93

24. Homework • P. 436 • #1-5 all • #8-18, evens • DUE FRIDAY – we will grade it 5 minutes after the tardy bell rings