Laws of Sines and Cosines Sections 8.1 and 8.2

Objectives Apply the law of sines to determine the lengths of side and measures of angle of a triangle. Solve word problems requiring the law of sines. Apply the law of cosines to determine the lengths of side and measures of angle of a triangle. Solve word problems requiring the law of cosines. Solve a word problem requiring Heron's formula.

Law of sines Law of cosines Heron’s formula The formulas listed below will allow us to more easily deal with triangles that are not right triangles. Law of sines Law of cosines Heron’s formula

Formulas Law of sines Law of cosines Heron’s formula or or a, b, and c are the lengths of the sides of the triangle P is the perimeter of the triangle A is the area of a triangle

Law of Sines

Law of Cosines

Use the Law of Sines to find the value of the side x. We are told to use the law of sines to find x. In order to use the law of sines, we need to have the lengths of two sides and the measures of the angle opposite those sides. In this case we have one side and the side we are looking for. We have the measure of the angle opposite the side we are looking for, but are missing the measure of the angle opposite the side we have. continued on next slide

Use the Law of Sines to find the value of the side x. Since we have the measures of two of the three angles, we can use the fact that the sum of the measures of the angles of a triangle add up to 180 degrees. This will give us: Now that we have the measure of the angle opposite the side AB, we can apply the law of sines to find the value of x. continued on next slide

Use the Law of Sines to find the value of the side x.

Use the Law of Cosines to find the value of the side x. In order to use the law of cosines, we need the lengths of two sides and the measure of the angle between them. We have that here. We can let side a be x and angle α be the 39 degree angle. Sides b and c are the lengths 21 and 42. continued on next slide

Laws of Sines and Cosines Sections 8.1 and 8.2

Objectives Apply the law of sines to determine the lengths of side and measures of angle of a triangle. Solve word problems requiring the law of sines. Apply the law of cosines to determine the lengths of side and measures of angle of a triangle. Solve word problems requiring the law of cosines. Solve a word problem requiring Heron's formula.

Formulas Law of sines Law of cosines Heron’s formula or or a, b, and c are the lengths of the sides of the triangle P is the perimeter of the triangle A is the area of a triangle

Use the Law of Cosines to find the value of the side x. Now we plug into the law of cosine formula to find x. Since length is positive, x is approximately 28.88104097

Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them.  If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours? For this problem, the first thing that we should do is draw a picture. Once we have the picture, we may be able to see which formula we can use to solve the problem. continued on next slide

Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them.  If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours? 26mph*3hr = 78 miles harbor ship 1 ship 2 140° 34mph*3hr = 102 miles x 26mph*3hr = 78 miles harbor ship 1 ship 2 140° 34mph*3hr = 102 miles x harbor harbor 140° 34mph*3hr = 102 miles x ship 1 ship 2 Looking at the labeled picture above, we can see that the have the lengths of two sides and the measure of the angle between them. We are looking for the length of the third side of the triangle. In order to find this, we will need the law of cosines. x will be side a. Sides b and c will be 78 and 102. Angle α will be 140°. continued on next slide

Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them.  If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours? 26mph*3hr = 78 miles harbor ship 1 ship 2 140° 34mph*3hr = 102 miles x 26mph*3hr = 78 miles harbor ship 1 ship 2 140° 34mph*3hr = 102 miles x harbor harbor 140° 34mph*3hr = 102 miles x ship 1 ship 2 Since distance is positive, the ships are approximately 169.3437309 miles apart after 3 hours.

Approximating the area of a triangle Heron’s Formula where P is the perimeter of the triangle and a,b, and c are the lengths of the sides of the triangle. OR The area of a triangle equals one-half the product of the lengths of any two sides and the sine of the angle between them. i.e. 1/2ab sinγ = A

48°20′ 37.0 cm Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-19

8.3 Vectors, Operations, and the Dot Product Basic Terminology ▪ Algebraic Interpretation of Vectors ▪ Operations with Vectors ▪ Dot Product and the Angle Between Vectors Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-20

8.3 Vectors A vector is an object that has a magnitude and a direction. Given two points P1: and P2: on the plane, a vector v that connects the points from P1 to P2 is v = i + j. Unit vectors are vectors of length 1. i is the unit vector in the x direction. j is the unit vector in the y direction. A unit vector in the direction of v is v/||v|| A vector v can be represented in component form by v = ai + bj. The magnitude of v is ||v|| = Using the angle that the vector makes with x-axis in standard position and the vector’s magnitude, component form can be written as v = ||v||cos(θ)i + ||v||sin(θ)j 21

Magnitude and Direction Angle Find the magnitude and direction angle for Magnitude: Direction angle: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-22

Horizontal and Vertical Components Vector v has magnitude 14.5 and direction angle 220°. Find the horizontal and vertical components. Horizontal component: –11.1 Vertical component: –9.3 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-23

Example 4 Finding the Magnitude of a Resultant Two forces of 32 and 48 newtons act on a point in the plane. If the angle between the forces is 76°, find the magnitude of the resultant vector. because the adjacent angles of a parallelogram are supplementary. Law of cosines Find square root. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-24

Let u = 6, –3 and v = –14, 8. Find the following. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-25

Unit Vector has the same direction as a given vector, but is 1 unit long Unit vector = (original vector)/length of vector Simply involves scalar multiplication once the length of the vector is determined (recall the length = length of hypotenuse if legs have lengths = a & b) Given vector, v = -2i + 7j, find the unit vector:

8.1-8.3 Review Answers  

8.4 Dot Product Objectives Find dot product of 2 vectors Find angle between 2 vectors Use dot product to determine if 2 vectors are orthogonal Find projection of a vector onto another vector Express a vector as the sum of 2 orthogonal vectors Compute work.

8.4 Vector Operations Scalar multiplication: A vector can be multiplied by any scalar (or number). Example: Let v = 5i + 4j, k = 7. Then kv = 7(5i + 4j) = 35i + 28j. Alternate Dot Product formula v · w = ||v||||w||cos(θ). The angle θ is the angle between the two vectors. v θ w Addition/subtraction of vectors: Add/subtract same components. Example Let v = 5i + 4j, w = –2i + 3j. v + w = (5i + 4j) + (–2i + 3j) = (5 – 2)i + (4 + 3)j = 3i + 7j. 3v – 2w = 3(5i + 4j) – 2(–2i + 3j) = (15i + 12j) + (4i – 6j) = 19i + 6j. 29

Definition of Dot Product The dot product of 2 vectors is the sum of the products of their horizontal components and their vertical components Example: Let v = 5i + 4j, w = –2i + 3j. v · w = (5)(–2) + (4)(3) = –10 + 12 = 2.

Find the dot product of v&w if v=3i+j and w= -2i - j 7 -5 -7 -4

Properties of Dot Product If u,v, & w are vectors and c is scalar, then

Angle between vectors, v and w

Find the angle between a=<4,-3> and b=<1,2>.

Parallel Vectors Parallel: the angle between the vectors is either 0 (the vectors on top of each other) or 180 (vectors are in opposite directions). a and b are parallel if θ = 0 or θ = ∏.

Perpendicular Vectors Two vectors v and w are orthogonal (perpendicular) iff v · w = 0 Two vectors v and w are orthogonal (perpendicular) if the angle between them is θ=∏/2.

Let a = 1/2i – 3j and b = -2i+12j. Show that a and b are parallel.

Show that the pair of vectors is orthogonal. 2i+3j; 6i-4j

Formula for compba: If a and b are nonzero vectors, then compba = a • b

If c = 10i+4j and d = 3i-2j, find compdc and compcd and illustrate.

Work done by a force F moving an object from A to B The work W done by a constant force a as its point of application moves along a vector b is W = a • b.

Trigonometric Form of Complex Numbers Lesson 8.5

Graphical Representation of a Complex Number Graph in coordinate plane Called the complex plane Horizontal axis is the real axis Vertical axis is the imaginary axis 3 + 4i • -2 + 3i • • -5i 43

Absolute Value of a Complex Number Defined as the length of the line segment From the origin To the point Calculated by using Pythagorean Theorem 3 + 4i • 44

Find That Value, Absolutely Try these Graph the complex number Find the absolute value 45

Trig Form of Complex Number Consider the graphical representation We note that a right triangle is formed a + bi • r b θ a How do we determine θ? 46

Trig Form of Complex Number Now we use and substitute into z = a + bi Result is Abbreviation is often 47

Try It Out Given the complex number -5 + 6i Given z = 3 cis 315° Write in trigonometric form r = ? θ = ? Given z = 3 cis 315° Write in standanrd form a = ? b = ? 48

Product of Complex Numbers in Trig Form Given It can be shown that the product is Multiply the absolute values Add the θ's 49

Quotient of Complex Numbers in Trig Form Given It can be shown that the quotient is 50

Try It Out Try the following operations using trig form Convert answers to standard form 51

8.6 Complex Numbers in Polar Form: DeMoivre’s Theorem Objectives Plot complex numbers in the complex plane Fine absolute value of a complex # Write complex # in polar form Convert a complex # from polar to rectangular form Find products & quotients of complex numbers in polar form Find powers of complex # in polar form Find roots of complex # in polar form

Complex number = z = a + bi a is a real number bi is an imaginary number Together, the sum, a+bi is a COMPLEX # Complex plane has a real axis (horizontal) and an imaginary axis (vertical) 2 – 5i is found in the 4th quadrant of the complex plane (horiz = 2, vert = -5) Absolute value of 2 – 5i refers to the distance this pt. is from the origin (continued)

Find the absolute value Since the horizontal component = 2 and vertical = -5, we can consider the distance to that point as the same as the length of the hypotenuse of a right triangle with those respective legs

Expressing complex numbers in polar form z = a + bi

Express z = -5 + 3i in complex form

DeMoivre’s Theorem [r (cos θ + i sin θ)]n = rn (cos n•θ + i sin n•θ) Ex: Change (1 + i)20 to the form a + bi

Taking a root (DeMoivre’s Theorem) Taking the nth root can be considered as raising to the (1/n)th power Now finding the nth root of a complex # can be expressed easily in polar form HOWEVER, there are n nth roots for any complex number & they are spaced evenly around the circle. Once you find the 1st root, to find the others, add 2pi/n to theta until you complete the circle

DeMoivre’s Theorem and nth roots: Wk = n√r [cos (θ+2∏k ) + i sin (θ + 2∏k) n n where k = 0,1,2,…n-1.

Example: Find the fourth roots of -8-8√3i.

Example: Find the 6th roots of z= -2 + 2i If you’re working with degrees add 360/n to the angle measure to complete the circle. Example: Find the 6th roots of z= -2 + 2i Express in polar form, find the 1st root, then add 60 degrees successively to find the other 5 roots.

Example: Find the 6th roots of z= -2 + 2i If you’re working with degrees add 360/n to the angle measure to complete the circle. Example: Find the 6th roots of z= -2 + 2i Express in polar form, find the 1st root, then add 60 degrees successively to find the other 5 roots.

Example: Find the 6th roots of z= -2 + 2i If you’re working with degrees add 360/n to the angle measure to complete the circle. Example: Find the 6th roots of z= -2 + 2i Express in polar form, find the 1st root, then add 60 degrees successively to find the other 5 roots.

Formulas for index cards: Law of Sines: Sin A = Sin B = Sin C a b c Law of Cosines: a2 = b2 + c2 – 2bc Cos A a2-b2-c2 = Cos A -2bc Area of a Triangle A = ½ ab Sin C A = √(s(s-a)(s-b)(s-c)); s = ½ (a+b+c)

a1 = ||a||cos θ a2 = ||a||sin θ Unit Vector • a ||a|| = √(a12 + a22) a1 = ||a||cos θ a2 = ||a||sin θ Unit Vector • a cos θ = a•b angle between vectors ||a||||b|| compba = a•b ||b||

Products and Quotients of Complex #’s z1z2 = r1r2(cos(θ1+ θ2)+isin(θ1+ θ2)) z1 = r1 (cos(θ1- θ2)+isin(θ1- θ2)) z2 r2 DeMoivre’s Theorem [r(cos θ + isin θ)]n = rn(cosnθ + isin nθ) nth roots Wk = n√r [cos (θ +2пk) + isin (θ +2пk)] n n where k = 0,1,…n-1

Dot Product <a1,a2> • <b1,b2> = a1b1 + a2b2 Work Done = a • b (force * distance)

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