Name
Chapter 6 Notes
Day 17 Section 6-1 Law of Sines
C
Law of Sine
b
a
b
c


for any oblique triangle:
sin A sin B sin C
A
a
c
Remember that you are
setting up the ratio of the
side to the angle opposite
that side.
B
Depending on the information given in the problem states how many possible triangles you may have.
If you are given SAA you will only have one triangle.
Practice. Solve the triangle with A = 640, C = 820, and c = 14 centimeters. Round sides to tenths place.
If you are given ASA you will only have one triangle
Practice. Solve the triangle if A = 400, C = 22.50, and b = 12 inches. Round sides to tenths place.
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If you are given SSA this is when you may have No Triangle, One Triangle, or Two Triangles.
When you solve for the 1st angle in a problem given information about a side, side and an angle, always
check to see if the supplement of the answer you got also works to see if a 2nd triangle is possible.
You will have no triangle if when you solve for the 1st angle in the problem and you get an error when
you solve for the angle. Remember that sine is only defined between: −1 to 1.
Practice. Solve the triangle if A = 500, a = 10 feet, and b = 20 feet.
Practice. Solve the triangle if A = 570, a = 33 meters, and b = 26 meters.
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Practice. Solve the triangle if A = 350, a = 12, b = 16.
Area of An Oblique Triangle:
The area of a triangle equals one-half the product of the length of two sides times the sine of their
included angle. The wording can be expressed as the formulas:
1
1
1
Area  bc sin A  ab sin C  ac sin B
2
2
2
C
b
a
A
B
c
Practice: Find the area of ΔABC having A = 75○, a = 35 feet, and c = 25 feet. Round to the nearest square
unit.
Asn 17 p635#4-32 by 4’s, 34, 36, 38
Trig Review Ws#2-18(even)
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Notes 18: Section 6-2 Law of Cosines and Heron’s Area Formula
Objectives: To be able to
1. Use the Law of Cosines to solve oblique triangles.
2. Solve applied problems using Law of Cosines
3. Use Heron’s formula to find the area of a triangle.
Law of Cosines
If A, B, and C are the measures of the angles of a triangle, and a, b, and c are
the lengths of the sides opposite these angles, then
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
The square of a side of a triangle equals the sum of the squares of the other
two sides minus twice their product times the cosine of their included angle.
Only use Law of Cosines once
to find the largest angle then
change back to Law of Sines
and solve for the smallest
angle that is left.
Practice: Use the given information to solve the triangle. Round lengths of sides to two decimal places and
angle measures to nearest whole degree.
a  6, b  8, c  12
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Practice: Use the given information to solve the triangle. Round lengths of sides to two decimal places and
angle measures to nearest whole degree.
A  50o , b  15, c  30
Practice: Draw a triangle that describes the given problem. Then solve the problem.
A boat leaves port and travels a 100 miles due east, then adjusts its course 20o northward. After traveling
160 miles in the new direction, how far is the boat from port?
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Heron’s Area Formula
The area of a triangle with sides a, b, and c is:
A  s(s  a)(s  b)(s  c)
1
2
where s   a  b  c  .
Practice: Find the area of the triangle with sides 6 meters, 16 meters, and 18 meters.
Review Problems: Solving Trig Equations.
In 1-6, Solve each equation on the interval [0, 2π)
1. tan 3x  3
2. 2 cos2 x  2 cos x  3
3. 2sin 2 x  4sin x  6
4. 4cos2 2x  2  0
5. sin 4 x  1
6. 4sec2 x  2  0
Asn 18: P. 643#2-30(even)
Trig Review Ws (even)
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Notes 19: Section 6-6 Vectors
Objectives: To be able to
1. Use magnitude and direction to show vectors are equal.
2. Visualize scalar multiplication, vector addition, and vector subtraction as geometric vectors.
3. Represent vectors in the rectangular coordinate system.
4. Perform operations with vectors in terms of i and j.
5. Find the unit vector in the direction of v.
6. Write a vector in terms of its magnitude and direction.
7. Solve applied problems involving vectors.
A vector is a quantity that represents two quantities, magnitude and direction. Examples of quanities that are
expressed as vectors are Force, velocity, and tension in a wire.
How to represent vectors:
Vectors are represented as directed line segments:
•
Q
•
P = initial point
Q = terminal point
PQ =length of line segment (also called magnituded)
P
Component Form of a Vector: v1 , v2
Given a vector PQ on a plane shown:
•
Component Form of PQ is:
PQ = Q1  P1 , Q2  P2  v1 , v2  v
Q (Q1 , Q2 )
•
Another way to think of component
form is:
P ( P1 , P2 )
v  terminal  initial
Practice: Find component form a vector with an initial point of (─4, ─4) to a terminal point if (6, 2).
Magnitude of a vector: v
Magnitude of a vector is the distance between the initial point and terminal points.
Magnitude of vector PQ is given by the distance formula:
PQ 
 q1  p1 
2
  q2  p2 
2
where  p1 , p2  is the initial point and  q1 , q2  is the terminal point
If given component form v1 , v2
v  v12  v22
If v  1 , v is called a unit vector.
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Example 1: Find the component form and the magnitude of the vector shown in the graph.
(─2, 2)
(2, ─1)
Example 2: Find the component form and the magnitude of the vector given an initial point of (─3, ─5) and a
terminal point of (5, 1).
Vector Operations
Two basic vector operations:
1. Scalar multiplication
 A scalar is just a number (no direction or magnitude)
2. Vector addition
 To add vectors just add their “like” components.
Example 3: State the scalar multiplication for each (a-g) given vector v..
v
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Definition of Scalar Multiplication:
Let v = v1 , v2 then kv = kv1 , kv2 where k = scalar (number)
Example 4: Given v = 3, 2 evaluate
(a) ─2v
(b) 6v
(c)
1
v
3
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Definition of Vector Addition:
Let w = w1 , w2 and v = v1 , v2 then:
w + v = w1  v1 , w2  v2
Example 5: Let w = 3, 4 and v = 3, 2 evaluate and show the sketch each result.
(a) w + v
(b) 2w ─ 3v
(c) 2v ─ 3w
Unit Vectors (very important!)
Definition of a unit vector: u
A unit vector is a vector with a magnitude of 1 in the same direction as a given vector v.
To find a unit vector, u:
u
v
v
 Take the components of the vector and divide them by its magnitude.
Example 6: Find the unit vector in the direction of v = 5,12 .
Two Special Unit Vectors:
1. i : a unit vector in the direction of the positive x-axis
Component Form
i = 1, 0
i
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2. j: a unit vector in the direction of positive y-axis
Component Form
j = 1, 0
j
Finding vectors in the same direction as another vector.
In order to find a vector in the direction of another vector:
1. Find unit vector in the direction of the vector you want to go in:
u
v
v
2. Multiply unit vector by the magnitude of the vector.
Example 7: Find the vector v with the given magnitude and same direction as u.
v 7
u = 2, 4
Example 8: Find the vector v with the given magnitude and same direction as u.
v  10
u = 3i – 4j
Math Analysis Notes Prepared by Mr. Kevin Hopp
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Θ is called the direction angle and is measured counterclockwise
from the positive x-axis to v.
Direction Angles
y
To find the direction angle:
v
tan  
y  component
x  component
To find the direction angle if given
component form: v = ai + bj
θ
tan  
b
a
b
or   tan 1  
a
x
Example 9: Find the magnitude and direction angle of v. v = ─2i + 5j
Component Form of a vector given θ. (Called cis-form)
v  v cos i  v sin  j
where θ = direction angle and v = maginitude
Example 10: Find the component form of v given: v  7 and θ = 60○
Example 11: Find the component form of v given v =10 in the direction of w = 4i + 3j
Asn 19
P690#6-52(even)
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Notes 20: Section 6-7 Vectors and Dot Products
The dot product of u u1 , u2 and v v1 , v2 is u●v = u1v1  u2v2

A way to think about dot product is that you multiply the “x-component” plus the
multiplication of the “y-component”
Example 1: Find the dot product of:
(a) u = 4,5 , v= 2,3
(b) u = ─i + 3j, v = 2i – 4j
(c) u = 2i – j , v = i + 2j
Example 2: Given u = 8, 2 , v = ─6i + 5j, and w = 2, 6 to find the indicated quantity. Sate whether the
result is a vector or a scalar.
(a) u●u
(b) 2v●3w
Math Analysis Notes Prepared by Mr. Kevin Hopp
(c) (u●u)w
12
Another way to find magnitude of a vector is by taking the dot product:
New Formula for magnitude: v  v  v
Example 3: Use the dot product to find the magnitude of vector u given u = 7i – 2j
Angle (θ) Between
To find the angle between to vectors u and v:
cos  
uv
u v
or
 uv
 u v
  cos1 



Example 4: Find the angle between the given vectors:
(a) u = 1,0 , v = 0, 2
(b) u = 3i + 4j, v = ─2i + 3j
Example 5: Find u●v, where θ is the angle between u and v, given: u  5, v  9,  
Math Analysis Notes Prepared by Mr. Kevin Hopp
3
4
13
Trigonometric Form of a vector (cis-form)
v = v1 , v2
or v = v1i  v2 j
(Standard Form)
v = v cos i  v sin  j
(Trig Form)
where v = magnitude and θ = direction angle (not the angle between two vectors)


3
3
Example 6: Find the angle between two vectors given: u = cos   i  sin   j and v = 4cos   i  4sin   j
3
3
 4 
 4 
Orthogonal and Parallel Vectors
Two vectors are orthogonal if the angle between the two vectors is 90○
Orthogonal means Perpendicular
If two vectors f and u are orthogonal then the dot product of f and u equals zero.
f●u = 0 the vectors are orthogonal
Example 7: Determine if the given vectors are orthogonal.
(a) f = 2i + 3j and u = 6i + 2j
(b) f = 3i – 5j and u = 6i +
18
j
5
Math Analysis Notes Prepared by Mr. Kevin Hopp
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Two vectors are parallel if the angle between the vectors is 0○ (going in same direction) or 180○ (going in
opposite direction).
Vectors that are parallel have scalar products of the same vector or the vector in the opposite direction.
Vectors f = f1 , f 2 and u = u1 , u2 are parallel if:
u = k x, y and v = c x, y
Example 8: Determine if the vectors are parallel:
(a) f = 4, 2
and u = 2, 1
(b) f = 36, 12 and v = 1, 
1
3
Example 9: Determine if the vectors are orthogonal, parallel or neither. Show work to justify answers.
(a) v = 4i + 2j and u = 8i + 16j
(b) v = 5,17 and u = 10, 34
(c) v = 3,7 and u = 7,3
Asn 20
P700#2-32(even)
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