7 1 7 2 Law of Sines Oblique

7. 1 & 7. 2 Law of Sines Oblique triangle – A triangle that does not contain a right angle. C b C a a b A c B B c A sin A a = sin B b or a sin A = = sin C c b = sin B c__ sin C

Solving SAA or ASA Triangles SAA and ASA Triangles – Oblique triangles in which you know 2 angles and 1 side. sin A = a sin B = b C 82° b 64° A 14 cm sin C c or a sin A = b sin B = c__ sin C Solving for <B There are 180 degrees in a triangle, so 64 + 82 + B = 180 146 + B = 180 a B = 34° Solving for a Solving for b Sin (82) = Sin (64) Sin(82) = sin(34) B 14 a 14 b a. Sin(82) = 14 Sin(64) a = 14 Sin(64) Sin (82) a = 12. 7 cm b = 7. 9 cm

Try this one on your own sin A a = sin B = b sin C c a sin A or C a 20 102° A 38° c Check your answers: < C = 40° a = 31. 8 c = 20. 9 B = b sin B = c__ sin C

Application Problem: Bearings sin A a sin B b sin C c or a sin A b sin B • 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. c__ sin C

• 7. 1 Example 3 Using the Law of Application Problem: Bearings infrom ana Application The bearing of a Sines lighthouse ship was found to be N 52° W. After (ASA) (page 305) the ship sailed 5. 8 km due south, the new bearing was N 23° W. Let x = the distance to the lighthouse at bearing N 52° W and y = the distance to the lighthouse at bearing N 23° W. 7 -5

Example Using Law of Practice: 2 Law ofthe Sines Application Jerry wishes to. Sines measureinthean distance across the Big Muddy River. He(ASA) determines that C = 117. 2°, (page 305) 7. 1 • A = 28. 8°, and b = 75. 6 ft. Find the distance a across the river.

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 sin A a sin B b sin C c Example (NO solutions) Solve the triangle ABC if A = 75°, a = 51, and b = 71

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

Practice 1: Solve triangle ABC if B = 68. 7°, b = 25. 4 in. , and a = 19. 6 in.

Practice 2: • Solve triangle ABC if A = 61. 4°, a = 35. 5 cm, and b = 39. 2 cm.

Area of Oblique Triangles C b h A c sin A = h b a Since Area = ½ (Base)(Height) Area = ½ c (b Sin. A) Area = ½ b c Sin A h = b Sin. A B Example: (Find the Area) A = 62° b = 10 c = 24 m 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

• Find the area of triangle DEF in the figure. 7 -12

• 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. 7 -13

7. 3 Law of Cosines C C b a a b A c B B A a 2 = b 2 + c 2 – 2 bc cos A b 2 = a 2 + c 2 – 2 ac cos B c 2 = a 2 + b 2 – 2 ab cos C c

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

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

Solving SSS Triangles C Law of Cosines a 2 = b 2 + c 2 – 2 bc cos A b=9 a=6 b 2 = a 2 + c 2 – 2 ac cos B A B c=4 Step 1: Use Law of Cosines to find angle opposite the longest side b 2 = a 2 + c 2 – 2 ac cos B = a 2 + c 2 - b 2 2 ac cos B = 62 + 42 – 92 = -29/48 2(6)(4) c 2 = a 2 + b 2 – 2 ab cos C Law of Sines sin A sin B= = a b sin C c B = 127. 2 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

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

Heron’s Formula for Triangle Area (Derived from Law of Cosines) If a triangle has sides of lengths a, b, c with semiperimeter Area = s = (1/2)(a + b + c) s(s - a)(s - b)(s – c) Example: Find the area of the triangular region: S = (1/2)(2451+331+2427) = 2604. 5 A A= c=2427 mi B b=331 miles C a=2451 miles 2604. 5 (2604. 5 – 2451)(2604. 5 – 331)(2604. 5 – 2427) A = 401, 700 mi 2

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

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

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

The Geometric Method for Adding Two Vectors 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. Resultant vector u+v u Initial point of u v Terminal point of v Opposite vectors: v and –v -- Same magnitude, opposite direction -- The sum of opposite vectors has magnitude 0 – called the zero vector v -v

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 <a, b> y 1 j i and j are unit vectors (they have magnitude of 1) i = <0, 1> and j = <1, 0> i O 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 <a, b> Magnitude: (Pythagorean Theorem) y v=ai+bj (a, b) 1 j O a = ||u|| cos x 1 bj ai 1 x b = ||u|| sin v = <a, b> = <||u|| cos , ||u|| sin >

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 <(x 2 – x 1) , (y 2 – y 1) > and v = (x 2 – x 1)i + (y 2 – y 1)j. Example 1: Sketch the vector <-3. 4> , v = -3 i + 4 j and find its magnitude. Terminal point 5 4 v = -3 i + 4 j 3 2 1 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 1 2 3 4 5 Initial point Example 2: Vector v has initial point (3, -1) and terminal point (-2, 5) a) Represent it in i, j format v = (-2 – 3)i + (5 – (-1))j v = -5 i + 6 j b) Represent it in <x , y> form <-5, 6> c) Find magnitude ||v|| 61

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 = 5 i + 4 j and w = 6 i – 9 j 11 i – 5 j v + w = _________ -i +13 j v – w = __________ Let k be a constant and v = a 1 i + b 1 j then kv = (ka)i + (kb)j Example: v = 5 i + 4 j 30 i + 24 j 6 v = __________ -15 i - 12 j -3 v = _________

Vector Operations Revisited v = <a, b> and w = <c, d> v + w = <(a + c), (b + d)> v – w = <(a – c), (b – d)> Example: v = <5, 4> and w = <6, -9> <11 – 5> v + w = _________ Let k be a constant and v =< a , b> <-1, 13> v – w = __________ then kv = <(ka), (kb)> Example: v = <5, 4> <30, 24> 6 v = __________ <-15, – 12> -3 v = _________ If v = <a, b> then –v = <-a, -b> Example: v = <6, -2>, so –v = <-6, 2>

Zero Vector and Unit Vector The vector whose magnitude is 0 is called the zero vector. The zero vector has no direction: 0 = 0 i + 0 j = <0, 0> 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 = 5 i – 12 j First find ||v|| = || 5 i – 12 j|| = 52 + 122 = 25 + 144 = 169 = 13 Unit Vector = _v_ ||v|| = 5 i – 12 j 13 = _5 i 13 - 12 j 13

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

Dot Product v = <a, b> The dot product w = <c, d> v • w is defined as follows: ac + bd Examples: v = <5, – 2> w = <-3, 4> v · w = 5(-3) + (-2)(4) = -15 – 8 = -23 w · v = (-3)(5) + (4)(-2) = -15 – 8 = -23 v · v. = (5)(5) + (-2) = 25 + 4 = 29

Another Dot Product example v = <a, b> w = <c, d> ac + bd u = <4, 6> v = <5, – 2> Find: u • (v + w) <4, 6> • <5– 3, -2 + 4> <4, 6> • <2, 2> (4)(2) + (6)(2) 8 + 12 20 w = <-3, 4>

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=0 u • u = |u|2

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 Dot Product Angle Positive 0 Negative Acute Right Obtuse Example: Find the angle between v = <3, 4> and w = <2, 1> = cos – 1 10/[(5)( 5) ] v • w = 6 + 4 = 10 ||v|| = (9 + 16) = 25 =5 = cos – 1. 894427191 = 26. 57 degrees ||w||=(4 + 1) = 5 You try one: Find the angle between u = <2, -6> and v = <6, 2>

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 = <3, -2> and w = <3, 2> 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.

Application: Equilibriant • 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|| Law of cosines Angle α = 180 - <CAB. Use Law of Sines to find < CAB. In parallelograms • Opposite angles : Equal • Adjacent angles : Supplementary

Application: Forces 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) 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.