Precalculus Notes: Unit 6 Law of Sines & Cosines, Vectors, & Complex Numbers

Precalculus Notes: Unit 6 – Law of Sines & Cosines, Vectors, & Complex Numbers

Date: 6.1 Law of Sines Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of Sines). Deriving the Law of Sines: Consider the two triangles. C C b a a h b h A B B c A c h h In the acute triangle, sin A = and sin B = . b a h In the obtuse triangle, sin(p -B) = sin B = . a Solve for h. h= bsin A and h= asin B sinB sin A Substitute. asin B= b sin A can be rewritten as = b a sinC sin B sin A The same type of argument can be used to show that = = . c b a Law of Sines: The ratio of the sine of an angle to the length of its opposite side is the same for all three angles of any triangle. C sinA sin B sin C a b c = =or = = b a a b csin A sin B sin C B A c

Solving a Triangle: finding all of the missing sides and angles

Note: The Law of Sines can be used to solve triangles given AAS and ASA.

Ex1: Solve the triangle DABC given that �B �15° ,= C 52 ° , and b 9 . We are given AAS, so we will use the Law of Sines. C 9 a B A c sinB sin C = Solve for c: b c Find m A using the triangle sum. m� A -180 +( 15 = 52) sinB sin A = Solve for a: b a DABC: m� A � _____, m � B = ____, m = C = ____, a ___, b ____, c _____

of 17 Precalculus – Graphical, Numerical, Algebraic: Larson Chapter 6 Precalculus Notes: Unit 6 – Law of Sines & Cosines, Vectors, & Complex Numbers

Ambiguous Case: When given SSA, there could be 2 triangles, 1 triangle, or no triangles that can be created with the given information.

Ex2: Solve the triangle DABC (if possible) when m� C=54° , a = 10, c 7 . sinC sin A Given SSA, use Law of Sines. = c a

Solve for A.

There is no possible triangle with the given information.

Ex3: Solve the triangle DABC (if possible) when m� C=31° , b = 46, c 29 . sinC sin B Given SSA, use Law of Sines. = c b Solve for B.

Note that the calculator only gives the acute angle measure for B. There does exist an obtuse angle B with the same sine. m� B -180 = 54.8 125.2° This is also an appropriate measure of an angle in a triangle, so there are 2 triangles that can be formed with the given information. Triangle 1 Triangle 2

Application Problems 1. Draw a picture! 2. Use the Law of Sines to solve for what is asked in the problem.

Ex4: The angle of elevation to a mountain is 3.5° . After driving 13 miles, the angle of elevation is 9° . Approximate the height of the mountain.

(not drawn to scale)

3.5° θ 9° 13

First, find θ:

You Try: Solve the triangle DABC (if possible) when m� B=98° , b = 10, c 3.

QOD: Explain why SSA is the ambiguous case when solving triangles.

Date: 6.2 Law of Cosines Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of Cosines).

Law of Cosines: For any triangle, ABC

c2= a 2 + b 2 - 2 ab cos C C b2= a 2 + c 2 - 2 ac cos B b a 2 2 2 a= b + c - 2 bc cos A B A c

Note: In a right triangle, c2= a 2 + b 2 -2 ab cos90° � c 2 + a 2 - b 2 � 2 ab( 0) + c 2 a 2 b 2 (Pythagorean Theorem)

The Law of Cosines can be used to solve triangles when given SAS or SSS.

Ex1: Solve the triangle ABC when m� A=49° , b = 42, & c 15 . Note: The given information is SAS. Use a2= b 2 + c 2 - 2 bc cos A .

Now that we have a matching pair of a side and angle, we can use the Law of Sines. a b = or B =180 - 68.4 = sinA sin B Now find the two possibilities for m C using the triangle sum:

Since c is the shortest side, it must be opposite the smallest angle. So m� C .

Ex2: Solve the triangle ABC when a=31, b = 52, & c = 28 . Note: The given information is SSS. Use a2= b 2 + c 2 - 2 bc cos A . (Or any of them!)

Now that we have a matching pair of a side and angle, we can use the Law of Sines. a b = or B =180 - 56.5 = sinA sin B Now find the two possibilities for m C using the triangle sum: Since c is the shortest side, it must be opposite the smallest angle. So m� C . Application Problems

1. Draw a picture! 2. Use the Law of Cosines to solve for what is asked in the problem.

Ex3: A plane takes off and travels 60 miles, then turns 15° and travels for 80 miles. How far is the plane from the airport? (not drawn to scale)

15° 80 165° 60 c Using the picture, we can find the angle in the triangle to give us SAS. Use the Law of Cosines: c2= a 2 + b 2 - 2 ab cos C

Area of a Triangle 1 h 1 Recall: A= bh sinq= � h c sin q ; \A = bcsinq 2 c 2

c h h θ b b 1 Formula for the Area of a Triangle Given SAS: A= absin C 2

Ex4: Find the area of triangle ABC shown. We will use . B 120° 1 32 A= acsin B 50 2

Heron’s Area Formula a+ b + c Semi-Perimeter: s = 2 Area of a Triangle Given SSS: A= s( s - a)( s - b)( s - c)

Ex5: Find the area of the triangle with side lengths 5 m, 6 m, and 9 m. a+ b + c 5 + 6 + 9 Semiperimeter: s= � s � s 10 2 2 A= s( s - a)( s - b)( s - c)

You Try: Two ships leave port with a 19° angle between their planned routes. If they are traveling at 23 mph and 31 mph, how far apart are they in 3 hours?

QOD: Can there be an ambiguous case when using the Law of Cosines? Explain why or why not.

Date: 6.3 Vectors in the Plane Syllabus Objectives: 5.1 – The student will explore methods of vector addition and subtraction. 5.2 – The student will develop strategies for computing a vector’s direction angle and magnitude given its coordinates. 5.4 – The student will resolve vectors into unit vectors. 5.7 – The student will solve real-world application problems using vectors in two and three dimensions.

Directed Line Segment: a segment with direction and distance A: Initial Point (start); B: Terminal Point (end) B v AB

Coordinates of A: ( Ax, A y ) Coordinates of B: (Bx, B y )

 v v 2 2 Magnitude (length) of a Directed Line Segment AB : AB=( Bx - A x) +( B y - A y ) Note: This is the distance formula!

Vector (v): the set of all directed line segments that are equivalent to a given directed line segment Note: Equivalent means same magnitude and direction.

Component Form of a Vector: Bx- A x, B y - A y

v Ex: Graph the vector AB =3, - 2 and find the magnitude. One possible graph:

y Note: could be placed anywhere on the coordinate grid. We have 6

4 2 v AB =3, - 2 -6 -4 -2 2 4 6 x -2

-6 placed it in standard position, which is with the initial point at the origin. v Magnitude: AB = Note: If a vector u is written in component form, , then the magnitude of u is u = u1, u 2

2 2 0,0 u =(u1) + ( u 2 ) . This is because the initial point is the origin, ( ) .

Vector Addition: Let u = u1, u 2 and v = v1, v 2 . Then u+ v =u1 + v 1, u 2 + v 2 .

Scalar Multiplication: Let u = u1, u 2 and k be any constant. Then ku = ku1, ku 2 . Note: If k < 0 , then ku is in the opposite direction. Ex: Use the graph of the vectors to complete each example below.

1. Show that u= v . Show that u= v . u = v = Show that the direction of u is the same as the direction of v. Use slope: Direction of u = ; direction of v = The direction and magnitude are the same, so u= v .

2. Find the component form and the magnitude of uand w . Component form of u: u = u = (see above) Component form of w: w = w =

3. Find the component form of 2u- 3 w . 2u- 3 w =

Unit Vector: a vector with a magnitude of 1 A unit vector in the direction of a vector v can be found by dividing v by the magnitude of v. v Unit Vector in the Direction of v: v

Standard Unit Vectors: unit vectors i and j in standard position along the positive x- and y-axes i=1,0 & j = 0,1 Any vector can be written in terms of the standard unit vectors. Ex: Write the vector v = -2,5 in terms of the standard unit vectors. v =

Ex: Find a unit vector in the direction of the given vector. Verify your answer is a unit vector and give your answer in component form and standard unit vector form. 2i- 4 j

Find the magnitude: 2i- 4 j = 2i- 4 j Divide the original vector by its magnitude: = 2 5

Component Form:

5 2 5 Verify magnitude of unit vector: ,- = ☺ 5 5

Recall: In the unit circle, x=cosq , y = sin q . This leads into another way of expressing a vector, in terms of its direction angle, θ.

Direction Angle: in standard position, the angle the vector makes with the positive x-axis (counterclockwise)

Resolving a Vector: in terms of its direction angle, θ, a vector can be written as u�cosq ,sin q+ u cos q i u sin q j

Ex: Find the magnitude and direction angle of v= -2 i + 6 j. Magnitude: v = Direction angle: vcosq i+ v sin q j

but since we know v= -2 i + 6 j is in Quadrant II, q =180 - 71.57 =

Ex: Find the component form of v given its magnitude and its direction angle. v =5,q = 30° v= vcosq i + v sin q j

Application: Resultant Force

° ° Ex: Two forces act on an object: u =3,qu = 45 and v =4,qu = - 30 . Find the direction and magnitude of the resultant force.

Write each vector in component form: u =

v= vcosqv i + v sin q v j The resultant force is the sum u+ v :

Application: Bearing

Ex: A plane flies due east at 500 km/h and there is a 60 km/h with a bearing of 45°. Find the ground speed and the actual bearing of the plane. 60 km/hr 45° θ 500 km/hr

Sketch a diagram: w v

Find the vectors p and w: p = w = Note: The 45° is the direction angle, not the bearing. Vector v is the sum p + w: v = The second component of vector v must equal zero, because the plane is headed due east. 60sin 45° + 500sinq = 0 Bearing of the plane: 90° + q Ground speed of the plane: v =

You Try: Find the component form of v given its magnitude and the angle it makes with the positive x- axis. v = 2 , direction: 2i+ 3 j OD: In the examples in your notes, we used sine or cosine to find the direction angle of a vector. Explain how you could use tangent to find the direction angle.

Date: 6.4 Vectors and Dot Product Syllabus Objective: 5.3 – The student will explore methods of vector multiplication. 5.5 – The student will determine if two vectors are parallel or perpendicular (orthogonal). 5.6 – The student will derive an equation of a line or plane by using vector operations. 5.7 – The student will solve real-world application problems using vectors in two and three dimensions.

Dot Product: Let u = u1, u 2 and v = v1, v 2 . The dot product is ug v =u1 v 1 + u 2 v 2 . Note: The dot product is a scalar.

Ex: Evaluate 5,- 2g 3,4 .

Properties of the Dot Product: 1. ug v= v g u 2 2. ug u = u 3. 0g u= 0 4. ug( v + w) = u g v+ u g w 5. (cu)g v= u g( c v) = c ( u g v)

Ex: Evaluate the following given u= -3,6 ; v = 1,0 ; w = 5, - 2 (a) wg w

(c) (v+ w)g u

(d) vg u+ w g u

ug v-1 骣 u g v Angle Between Two Vectors: cosq= � q cos 琪 u v桫 u v Proof: Use the triangle.

v u − v θ u

Law of Cosines: v- u2 = u 2 + v 2 - 2 u v cosq 2 2 Property of Dot Product: (v - u)g( v - u) = u + v - 2 u v cosq 2 2 Expand: v� v- vg u + u g v = u g u + u - v2 u v cosq 2 2 2 2 Property of Dot Product: v-2 ug v + u = u + v - 2 u v cosq

ug v -2ug v = - 2 u v cosq� cos q Property of Equality: u v 5p Ex: Find ug v , where θ is the angle between u and v. u=6, v = 8, q = 6 u v cosq = g u v

Application Ex: Find the interior angles of the triangle with vertices (3,0) ,( 4,2) ,( 5,1) . y 4

3 B 2 1 C

-2 -1 1 2A3 4 5 6 x -1

-2 v AB = cos A =

v BA = cos B =

Angles:

Orthogonal Vectors: two vectors whose dot product is equal to 0

What is the angle between two non-zero orthogonal vectors? u v 0 cosq= g � cos q � � cos q 0 q 90° u v u v Note: If the angle between the vectors is 90°, we may also say they are perpendicular. The word orthogonal is used instead for vectors because the zero vector is orthogonal to any other vector, but is not perpendicular.

What is the dot product of two vectors that are parallel? The angle “between” them would have to be either 180° or 360°. ug v° u g v ug v° u g v cosq =� cos180 � = 1 ug v cosq = � cos360 � 1 ug v u v u v or u v u v

Parallel Vectors: two vectors whose dot product is equal to −1 or 1

Ex: Are the vectors orthogonal, parallel, or neither? v=3 i - 2 j , w = 3 i + 4 j Find vg w : The vectors are______.

骣u v proj u= 琪 g v Vector Projection: the projection of u onto v is denoted by: v 琪 2 桫v u u = u – u 2 1

u = proj u 1 v v

Ex: Find the projection of v onto w. Then write v as the sum of two orthogonal vectors, with

one the projwv . v=1,3 ; w = 1,1

骣v w proj v= 琪 g w w 琪 2 桫w

v-projw v = v =

Application: Force Ex: Find the force required to keep a 200-lb cart from rolling down a 30° incline. Draw a diagram and label:

f 200 30°

The force due to gravity: g= -200 j (gravity acts vertically downward) 3 1 Incline vector: v=cos30° i + sin30 ° j = i + j ( ) ( ) 2 2 骣g v f=proj g = 琪 g v Force vector required to keep the cart from rolling: v 琪 2 桫v

骣g v f=琪 g v = 琪 2 桫v Magnitude of Force: f =

Application: Work W = cosq ( force)( distance) Ex: A person pulls a wagon with a constant force of 15 lbs at a constant angle of 40° for 500 ft. What is the person’s work?

w = cos40° ( 15 lbs)( 500 ft)

You Try: Find the projection of v onto u. Then write v as the sum of two orthogonal vectors, with one the projuv . v=2 i - 3 j ; u = i - j

QOD: If u is a unit vector, what is ug u ? Explain why.

Syllabus Objectives: 7.1 – The student will graph a complex number on the complex/Argand plane. 7.2 – The student will represent a complex number in trigonometric (polar) form. 7.3 – The student will simplify expressions involving complex numbers in trigonometric (polar) form. 7.4 – The student will compute the powers of complex numbers using DeMoivre’s Theorem and find the nth roots of a complex number.

Complex Number Plane (Argand Plane): horizontal axis – real axis; vertical axis – imaginary axis a+bi Plotting Points in the Complex Plane Ex: Plot the points A(3+ 4 i) , B( - 1 + 3 i) , &( 2 - i) in the complex plane. Imaginary 6 A 4 B 2

-6 -4 -2 2 4 6 Real -2 C

Absolute Value (Modulus) of a Complex Number: the distance a complex number is from the origin on the complex plane a+ bi = a2 + b 2 (This can be shown using the Pythagorean Theorem.)

Ex: Evaluate 3- i .

Recall: Trigonometric form of a vector: u cosq ,sin q

Trigonometric Form of a Complex Number z = a + bi: z= r(cosq + i sin q ) Note: This can also be written as . z= rcisq

b a= rcosq , b = r sin q , r = a2 + b 2 tanq = a r = modulus; θ = argument

Writing a Complex Number in Trig Form Ex: Find the trigonometric form of -2 - 2 3i . 1. Find r: r= a2 + b 2 = b 2. Find θ: tanq= � tan q a 3. Find Quadrant 4. z= r(cosq + i sin q )

Writing a Complex Number in Standard Form (a + bi) Ex: Write 9cisp in standard form. Expand: 9cisp= 9( cos p +i sin p ) =

Multiplying and Dividing Complex Numbers

Let z1= r 1(cosq 1 + i sin q 1 ) and z2= r 2(cosq 2 + i sin q 2 ) .

Multiplication: z1� z 2 � r 1+ r 2臌轾cos(q 1 + q 2) i sin( q 1 q 2 )

z1 r 1 Division: =臌轾cos(q1 - q 2) +i sin( q 1 - q 2 ) z2 r 2

Ex: Express the product of z1 and z2 in standard form.

nn n Powers of a Complex Number: De Moivre’s Theorem z=臌轾 r(cosq + i sin q) = r( cos n q + i sin n q ) 5 Ex: Evaluate (-2 + i 2 ) .

Rewrite in trig form: n th Roots of a Complex Number:

n n轾骣q2 pk 骣 q 2 p k n 骣 q+ 2 p k z= r犏cos琪 + + i sin 琪 + = r cis 琪 , k=0,1, 2,...( n - 1) 臌桫n n 桫 n n 桫 n Note: Every complex number has a total of “n” nth roots.

Ex: Find the cube roots of 8i. Write in trig form:

Evaluate the roots:

Roots of Unity: the nth roots of 1 Ex: Express the fifth roots of unity in standard form and graph them in the complex plane. 5th Roots of Unity: 5 1+ 0i r =1,q = 0 1cis0

5 5 骣0+ 2pk 骣 2 p k 1cis0= 1cis琪 = cis 琪桫5 桫 5

You Try: 1. Write each complex number in trigonometric form. Then find the product and the quotient. -1 + 3i , - 2 - 2 3 i 2. Solve the equation x4 +1 = 0 . (You should have 4 solutions!)

QOD: Is the trigonometric form of a complex number unique? Explain.

of 17 Precalculus – Graphical, Numerical, Algebraic: Larson Chapter 6