2. 7.1 & 7.2 Law of Sines Oblique triangle – A triangle that does not contain a right angle. C B A b a c A B C c a b sin A sin B sin C a b c == or a b c__ sin A sin B sin C ==

3. Solving SAA or ASA Triangles SAA and ASA Triangles – Oblique triangles in which you know 2 angles and 1 side. C B A b a 14cm sin A sin B sin C a b c = = or a b c__ sin A sin B sin C == 64° 82° Solving for <B There are 180 degrees in a triangle, so 64 + 82 + B = 180 146 + B = 180 B = 34° Solving for a Solving for b Sin (82) = Sin (64) Sin(82) = sin(34) 14 a 14 b aSin(82) = 14 Sin(64) a = 14 Sin(64) Sin (82) a = 12.7cm b = 7.9 cm

4. Try this one on your own A B C c a 20 sin A sin B sin C a b c = =or a b c__ sin A sin B sin C == 102°38° Check your answers: < C = 40° a = 31.8 c = 20.9

5. Application Problem: Bearings The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W. Draw a Picture to represent the situation Find the distance between the ship and the lighthouse at each location. sin A sin B sin C a b c or a b c__ sin A sin B sin C

6. 7-5 The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W. 7.1 Example 3 Using the Law of Sines in an Application (ASA) (page 305) Let x = the distance to the lighthouse at bearing N 52° W and y = the distance to the lighthouse at bearing N 23° W. Application Problem: Bearings

7. Jerry wishes to measure the distance across the Big Muddy River. He determines that C = 117.2°, A = 28.8°, and b = 75.6 ft. Find the distance a across the river. 7.1 Example 2 Using the Law of Sines in an Application (ASA) (page 305) Practice: Law of Sines

8. SSA Ambiguous Case Triangles SSA Triangles – Oblique triangles in which you know 2 of the sides and an angle opposite one of the known sides. You may have NO solutions, ONE solution, or TWO solutions Example (NO solutions) Solve the triangle ABC if A = 75°, a = 51, and b = 71 sin A sin B sin C a b c

9. SSA Ambiguous Case (1 & 2 Solution Triangles) Example (ONE solution) Solve the triangle ABC A = 61°, a = 55, and c = 35 sin A sin B sin C a b c Example (TWO solutions) Solve the triangle ABC A = 40°, a = 54, and b = 62

10. Solve triangle ABC if B = 68.7°, b = 25.4 in., and a = 19.6 in. Practice1:

11. Solve triangle ABC if A = 61.4°, a = 35.5 cm, and b = 39.2 cm. Practice2:

12. Area of Oblique Triangles C B A b a c sin A = h b h = b SinA h Since Area = ½ (Base)(Height) Area = ½ c (b SinA) Area = ½ b c Sin A You may use any one of 3 area formulas. Area = ½ b c Sin A Area = ½ a b Sin C Area = ½ a c Sin B Example: (Find the Area) A = 62° b = 10 c = 24 m

13. Copyright © 2008 Pearson Addison- Wesley. All rights reserved. 7-12 Find the area of triangle DEF in the figure.

14. Copyright © 2008 Pearson Addison- Wesley. All rights reserved. 7-13 Find the area of triangle ABC if B = 58°10′, a = 32.5 cm, and C = 73°30′. We must find AC (side b) or AB (side c) in order to find the area of the triangle.

15. 7.3 Law of Cosines C B A b a c a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C A B C c a b

16. Solving SAS Triangles C B A b=20 a c=30 60° Law of Cosines a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C Law of Sines sin A sin B sin C a b c = = Step 1: Use Law of Cosines to find side opposite given angle a 2 = b 2 + c 2 – 2bc cos A a 2 = (20) 2 + (30) 2 – 2(20)(30) cos (60) a = 26 Step 2: Use Law of Sines to find angle opposite the shorter of the 2 given sides. Step 3: Find the 3 rd angle in the triangle using concept of 180 degrees in a triangle. sin B = sin (60) 20 26 B = 41° C = 79°

17. Try this one on your own A B C 8 a 7 Law of Cosines a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C Law of Sines sin A sin B sin C a b c 120° Check your answers: < B = 28° a = 13 <C = 32°

18. Solving SSS Triangles Law of Cosines a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C Law of Sines sin A sin B sin C a b c = = Step 1: Use Law of Cosines to find angle opposite the longest side b 2 = a 2 + c 2 – 2ac cos B 2ac cos B = a 2 + c 2 - b 2 cos B = a 2 + c 2 - b 2 2ac cos B = 6 2 + 4 2 – 9 2 = -29/48 B = 127.2 2(6)(4) Step 2: Use Law of Sines to find either of the other angles. A = 32.1 Step 3: Find the 3 rd angle in the triangle using concept of 180 degrees in a triangle. C = 20.7 B A C c=4 b=9 a=6

19. Try this one on your own Law of Cosines a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C Law of Sines sin A sin B sin C a b c = = B A C c=8 b=16 a=10

20. Heron’s Formula for Triangle Area (Derived from Law of Cosines) If a triangle has sides of lengths a, b, c with semiperimeter s = (1/2)(a + b + c) Area = s(s - a)(s - b)(s – c) Example: Find the area of the triangular region: A C B c=2427 mi b=331 miles a=2451 miles S = (1/2)(2451+331+2427) = 2604.5 A = 2604.5 (2604.5 –2451)(2604.5 –331)(2604.5 –2427) A = 401,700 mi 2

21. 7-4 & 7.5 Vectors Scalar – A quantity that involves magnitude, but no direction. (Example: the temperature outside might be 75°) Vector – A quantity that involves both magnitude and direction. (Example: The jet is flying 120 mph 30° East of North) Magnitude – How big something is – the size Direction – The angle in degrees something is travelling Vectors can be represented as directed line segments Magnitude = length of segment (distance formula) ||v|| or | v| - the magnitude of vector v v u

22. Showing 2 Vectors are Equal Two vectors (u and v) are equal if they have the same magnitude and the same direction. Magnitude Use the distance formula to see if ||u|| = ||v|| Direction Find the slope of each vector. If the slopes are the same u and v have the same direction. Example Let u have initial point (-3, -3) and terminal point (0, 3) Let v have initial point (0, 0) and terminal point (3, 6) Does u = v? YES! - ||v|| = ||u|| =  45 = 3  5 and both slopes = 2

23. Scalar Multiplication (Geometric) v 2v2v ½ v-2v Multiplying a vector by a negative number reverses the direction Multiplying a vector by a positive number changes the magnitude but not the direction If k is a real number and v is a vector, then kv is a scalar multiple of vector v.

24. A geometric method for adding two vectors is shown below. The sum of u + v is called the resultant vector. Here is how we find this vector. 1. Position u and v so the terminal point of u extends from the initial point of v. 2. The resultant vector, u + v, extends from the initial point of u to the terminal point of v. Initial point of u u + v v u Resultant vector Terminal point of v The Geometric Method for Adding Two Vectors Opposite vectors: v and –v -- Same magnitude, opposite direction -- The sum of opposite vectors has magnitude 0 – called the zero vector v -v

25. 1 1 i j O x y Vectors in the X/Y Axes Vector i is the unit vector whose direction is along the positive x-axis. Vector j is the unit vector whose direction is along the positive y-axis. Position Vector: Vector with initial point at origin Position Vector, v, with endpoint at (a,b) is written i and j are unit vectors (they have magnitude of 1) i = and j = 1 1 i j O x y Vectors are also represented in terms of i and j. A vector, v, with initial point (0,0) and termianl point (a,b) is also represented Magnitude: a bj v=ai+bj (a, b) (Pythagorean Theorem)  a = ||u|| cos  b = ||u|| sin  v = =

26. Example 1: Sketch the vector, v = -3i + 4j and find its magnitude. -5-4-3-212345 5 4 3 2 1 -2 -3 -4 -5 Initial point Terminal point v = -3i + 4j Vector Representation & Examples General formula for representation of a vector v with initial point P 1 = (x 1, y 1 ) and terminal point P 2 = (x 2, y 2 ) is equal to the position vector and v = (x 2 – x 1 )i + (y 2 – y 1 )j. Example 2: Vector v has initial point (3, -1) and terminal point (-2, 5) a)Represent it in i, j format b) Represent it in formc) Find magnitude ||v|| v = (-2 – 3)i + (5 – (-1))j v = -5i + 6j  61

27. Vector Operations: Addition, Subtraction, and Scalar Multiplication v = a 1 i + b 1 j and w = a 2 i + b 2 j v + w = (a 1 + a 2 )i + (b 1 + b 2 )j v – w = (a 1 – a 2 )i + (b 1 – b 2 )j Example: v = 5i + 4j and w = 6i – 9j v + w = __________________ v – w = ___________________ Let k be a constant and v = a 1 i + b 1 j then kv = (ka)i + (kb)j Example: v = 5i + 4j 6v = ___________________-3v = __________________ 11i – 5j -i +13j 30i + 24j -15i - 12j

28. Vector Operations Revisited v = and w = v + w = v – w = Example: v = and w = v + w = __________________ v – w = ___________________ Let k be a constant and v = then kv = Example: v = 6v = ___________________-3v = __________________ If v = then –v =  Example: v =, so –v =

29. Zero Vector and Unit Vector The vector whose magnitude is 0 is called the zero vector. The zero vector has no direction: 0 = 0i + 0j = The vector whose magnitude is 1 is called the unit vector. To find the unit vector in the same direction as a given vector v, divide v by its magnitude Unit Vector = _v_ ||v|| Example: Find the unit vector in the same direction as v = 5i – 12j First find ||v|| = || 5i –12j|| =  5 2 + 12 2 =  25 + 144 =  169 = 13 Unit Vector = _v_ = 5i – 12j = _5 i - 12 j ||v|| 13 13 13

30. Using Magnitude and Direction to write a Vector (a,b) ||v||  v = ai + bj v = ||v||cos  i + ||v||sin  j = Example: Wind is blowing at an angle  = 120  Magnitude ||v|| = 20mph Express the wind’s velocity as a vector v = ||v||cos  i + ||v||sin  j v = 20cos(120)i + 20sin(120)j v = 20(- ½ )i + 20 (  3/2)j v = -10i + 10  3 j =

31. Dot Product v = w = The dot product v w is defined as follows: Examples: v = w = v · w = 5(-3) + (-2)(4) = -15 – 8 = -23 w · v = (-3)(5) + (4)(-2) = -15 – 8 = -23 v · v. = (5)(5) + (-2)(-2) = 25 + 4 = 29 ac + bd

32. Another Dot Product example u = v = w = Find: u (v + w) (4)(2) + (6)(2) 8 + 12 20 v = w = ac + bd

33. Dot Product Properties u v = v u u (v + w) = u v + u w (u + v) w = u w + v w(ku) v = k(u v) = u kv 0 u = 0u u = |u| 2

34. Finding the angle between vectors with an alternative dot product formula If v and w are two nonzero vectors and  is the smallest nonnegative angle between them, then Example: Find the angle  between v = and w = Dot Product Angle Positive Acute 0 Right Negative Obtuse v w = 6 + 4 = 10 ||v|| = (9 + 16) = 25 =5 ||w||=(4 + 1) = 5  = cos –1 10/[(5)(  5) ]  = cos –1.894427191  = 26.57 degrees You try one: Find the angle between u = and v =

35. Parallel and Orthogonal Vectors Two vectors are parallel if the angle between them is 0 or 180° Two vectors are orthogonal if the angle between them is 90°  If the dot product of two vectors is 0 then the vectors are orthogonal. Example: v = and w = Are v and w orthogonal? Check the dot product: v w = (3)(3) + (-2)(2) 9 + -4 5 Since the dot product is NOT zero, these vectors are NOT orthogonal.

36. Find the magnitude of the equilibrant of forces of 54 newtons and 42 newtons acting on a point A, if the angle between the forces is 98°. Then find the angle between the equilibrant and the 42-newton force. Equilibrant is –v. & ||-v|| = ||v|| Application: Equilibriant Law of cosines In parallelograms Opposite angles : Equal Adjacent angles : Supplementary Angle α = 180 - <CAB. Use Law of Sines to find < CAB.

37. Find the force required to keep a 2500-lb car parked on a hill that makes a 12° angle with the horizontal. BA : force of gravity. BC: force of weight pushing against hill BF: force that would pull car up hill BA = BC + (–AC) Application: Forces BF & AC are equal, so, |AC| gives the required force. BF and AC are parallel, BA acts as transversal; < EBD = < A. <C and <EDB are right angles Thus, Triangles CBA & DEB have 2 corresponding angles =>are similar triangles. Therefore, the measure of angle ABC equals the measure of angle E, which is 12°. A 520 lb force will keep the car parked on the hill.