Chapter 10 Trigonometric Applications
10.1 The Law of Cosines

We will now study oblique triangle – those without right angles
Standard Notation of Triangles:
1. each vertex is labeled with a capital letter
2. The length of the side opposite that vertex is denoted by the same letter in lower
case.
Law of Cosines – in any triangle ABC, with lengths a,b,c as in figure:
a² = b²+c² - 2bc cosA
b² = a² + c² - 2ac cos B
c² = a² + b² - 2ab cos C
Alternate Form for Law of Cosines:
In any triangle ABC, with sides of length a,b,c as in figure:
Cos A = b² + c² - a²
2bc
Cos B = a² + c² - b²
2ac
Cos C = a² + b² - c²
2ab
Use the Law of Cosines to solve triangles in the following cases:
#1. 2 sides and the angle between them is known (SAS)
#2. 3 sides are known (SSS)
Ex. Solve the triangle shown below
Ex. Find the angles of the triangle below
Ex. A ship R leaves port P and travels at the rate of 16 miles per hour. Another ship Q
leaves the same port at the same time and travels at 12 miles per hour. Both ships are
traveling in straight paths. The directions taken by the 2 ships make an angle of 55
degrees. How far apart are the ships after 3 hours?
10.2 The Law of Sines
*In any triangle ABC in standard notation:
a =
sin A
b
= c
sin B
sin C
Use Law of Sines in the following cases:
#1. Two angles and 1 side are known AAS
#2. 2 sides and the angle opposite one of them are known SSA
Ex.
A = 115º B = 27º a = 45
**When triangles follow the congruence theorems of AAS,SAS,SSS,ASA, there is
always only 1 possible unique solution.
However, with SSA triangles  that is not the case:
The ambiguous case  when 2 sides of a triangle and the angle opposite one of them are
known as in SSA, there may be 1, 2 or no triangles that satisfy the given data.
Possibilities: ambiguous case: SSA info. with A<90º
SinA = h/b
1) a<b and side a is too short to reach the 3rd side = No solution
2) a<b and side a just reaches the 3rd side and is perp. To it = 1 solution
3) a<b and an arc of radius a meets the 3rd side at 2 points to right of A = 2 solutions
4) a≥b so that an arc of radius a meets the 3rd side at just 1 point to the right of A =
1solution
Now if angle A > 90º then there are only 2 possibilities:
#1. if a≤b so that side a is too short to reach the 3rd side = No solution
#2. a>b so that an arc of radius a meets the 3rd side at just 1 point to the right of A =
1solution
Supplementary Angle Identity: if 0º≤Ө ≤90º then sinӨ = sin(180º- Ө)
Ex. Given a possible ΔABC with b=8, c=5, C = 54º find angle B
Ex. Lighthouse B is 3 miles east of lighthouse A. Boat C leaves lighthouse B and sails
in a straight line. At the moment that the boat is 5 miles form lighthouse B, an observer
at lighthouse A notes that the angle determined by the boat, lighthouse A (the vertex) and
lighthouse B is 65º. Approximately how far is the boat from lighthouse A at that
moment?
Solving Triangles with SSA information.
Ex. Solve ΔABC when a = 4.8, b=6, and A = 28º
Solving triangles with ASA information:
Ex. 2 surveyors standing at points A and B are measuring a building. The surveyor at
point A is 20 feet further away from the building than the surveyor at point B. The angle
of elevation of the top of the building form point A is 58º and the angle of elevation of
the top of the building from point B is 72º. How tall is the building?
The Area of a Triangle - the area of a triangle containing an angle C with adjacent sides
of length a and b is:
= ½absinC
--or if contains angle B with adjacent sides of length a and c
=½acsinB
--or if contains angle A with adjacent sides of length b and c
=½bcsinA
Ex. find the area of the triangle in the figure shown below:
Heron’s Formula: the area of a triangle with sides, a, b, c is:
√s(s-a)(s-b)(s-c)
where s = ½(a+b+c)
Ex. find the area of the triangle whose sides are a =8, b=10, c=14
10.3 The Complex Plane and Polar Form for complex #’s
Complex Number system is represented by the coordinate plane: Complex Plane
a + bi  corresponds to the point (a,b) in the plane
Ex. (2,3) in complex system is 2 + 3i
**each number a = a + 0i and corresponds to the point (a,0) on the horizontal axis
= REAL AXIS
** the vertical axis = IMAGINARY AXIS b/c every imaginary #bi = 0 + bi and
corresponds to (0,b) on vertical axis
Absolute Value (or Modulus) of a Complex #, (represents distance from a + bi to the
Origin
|a + bi| = √a²+ b²
Ex. find the absolute value of each complex #
a) -6 + 3i
b) 1-5i
This above all leads to other ways to represent complex #’s
1) denote |a + bi| by r
a) r = length of the line segment joining (a,b) and (0,0) in the plane
b) let θ = be the angle in standard position with the line segment at the
terminal side. Then in terms of trig:
cosθ=a/r
sinθ=b/r
a=rcosθ
b=rsinθ
So: a + bi = rcosθ + (rsinθ)i = r(cosθ + isinθ)
** When complex #’s are written this way it is called POLAR FORM/TRIG. FORM
** Θ angle  represents the argument and is expressed in radians
** r = |a + bi| = modulus
Polar Form of a Complex Number: every complex # a+bi can be written in polar
form r(cosθ + isinθ)
where r = |a + bi| = √a²+ b² and a = rcosθ and b = rsinθ
Ex. express 1 – i in polar form
a = 1 b = -1
r = √a²+ b² = √1²+ (-1)² = √2
cosθ=a/r = 1/√2
sinθ=b/r = -1/√2
(1 – i) is represented by (1,-1) which lies in Q4 in the complex plane
(√2/2,-√2/2) = 7π/4  1 – i =√2(cos 7π/4 + isin7π/4)
Ex. express -1-4i in polar form
a= -1 b = -4
r = √-1² + -4² = √17
cosθ=a/r = -1/√17
sinθ=b/r = -4/√17
tan = -4/√17 = 4
tan-14 = 1.3258
-1/√17
So θ = 1.3258 + π = 4.4674
And -1-4i = √17(cos4.4674 + isin4.4674)
Q1 – stays same
Q2/Q3 – add π
Q4 – add 2π
Multiplication and Division of Complex Numbers:
If z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2) are any 2 complex numbers
Then: z1 z2 = r1r2[(cos(θ1 + θ2) + isin(θ1 + θ2))]
And


z1 = r1[(cos(θ1 - θ2) + isin(θ1 - θ2))]
z2 r2
to multiply 2 #’s  multiply the moduli and add the arguments
to divide the 2 #’s  divide the moduli and subtract the arguments
Ex. find z1 z2 when z1 = 5(cos π/4 + isinπ/4) and z2= 4(cos 2π/3 + isin 2π/3)
z1 z2 = (5)(4)[cos (π/4 + 2π/3) + isin (π/4 + 2π/3)]
= 20[cos 11π/12 + isin 11π/12]
Ex. find: z1
z2
when z1 = 12(cos 4π/3 + isin 4π/3) and z2 = 6(cos 3π/4 + isin 3π/4)
= 12 [cos (4π/3 - 3π/4) + isin (4π/3 - 3π/4)
6
= 2cos 7π/12 + isin 7π/12
10-4 DeMoivre’s Theorem and nth roots of Complex #’s
**For any complex # z= r(cosθ + isinθ) and any positive integer n,
zn = rn (cosnθ + isinnθ)
Ex. z³ = r³(cos3θ + isin3θ)
Ex. evaluate (1+i√3)8  must convert from standard to polar and back to standard
Step 1: express in polar form: 1 + i√3
a = 1 b = √3 r = √(√3² + 1² = 2
sinθ=b/r = √3/2
cosθ=a/r = ½
= 2(cosπ/3 + isinπ/3)
Step 2: Apply DeMoivre’s Theorem (1+i√3)8
=28(cos 8·π/3 + isin 8·π/3)
= 256(cos 8·π/3 + isin 8·π/3)
Step 3: Convert back to standard form
8π/3 = 2π/3 + 6π/3  2π/3 + 2π  so 2π/3 can be substituted for 8π/3
( we substitute b/c we want to get to a radian value on our unit circle)
(1+i√3)8 = 256(cos 2π/3 + isin 2π/3)
= 256 (-1/2 + √3/2i)
= -128 + 128i√3
Ex. Finding Powers of Complex #’s
Evaluate (1-i)6
Step1: Express in polar form
a=1 b = -1
r=√2
cosθ=a/r = 1/√2 = √2/2
sinθ=b/r = -1/√2 = -√2/2
= √2(cos7π/4 + isin7π/4)
Step 2: Apply DeMoivre’s Theorem: (1-i)6 = √26(cos6·7π/4 + isin6·7π/4)
= 8(cos21π/2 + isin21π/2)
21π/2 = π/2 + 20π/2 = π/2 + 10π or π/2 + 5(2π)
Step 3: Convert back to standard form:
(1-i)6 = 8(cosπ/2 + isinπ/2) = 8(0 + i) = 8i
Nth Roots  if a +bi is a complex # and n is a positive integer, the equation zn = a+bi
May have n different solutions in the complex #’s system
 Any solution of the equation zn = a+bi is called an nth root of a+bi
 Every real # is a complex #  when the definition of nth root of a
complex # is applied to a real # -- the terminology for real #’s no longer
applies
Ex. complex # 16 has 4 fourth roots – 2, -2, 2i, -2i are all solutions of
z4 = 16  whereas in the real #’s, 2 is the fourth root of 16
 The radical symbol – will be used only for nonnegative real #’s
-- if r is a nonnegative real #  then n√r denotes the unique nonnegative
real # whose nth power is r
Ex. find the 4th roots of -81/2 - 81√3/2i
Step 1: Express in polar form: a = -81/2 b= -81√3/2 r = 81
Cos θ = -81/2 - 1
sinθ = √3/2
in Q3
81
2
81(cos4π/3 + isin4π/3)
Step 2: z4 = 81(cos4π/3 + isin4π/3)  rewrite in terms of s (for r) and β for the
radian
4
s(cos β + isinβ) = 81(cos 4π/3 + isin4π/3)
Step 3: Use DeMoivre’s Theorem to rewrite the left side:
s4 = 81 (take the 4√ )  s = 3
4β = 4π/3 + 2kπ  β = π/3 + kπ/2
So z = 3[cos (π/3 + kπ/2) + isin (π/3 + kπ/2)] where k is any integer
Letting k = 0, 1, 2, and 3 produces 4 distinct solutions:
k=0 z= 3(cosπ/3 + isinπ/3) = 3(½ + i√3/2) = 3/2 + 3i√3/2
k = 1 z = 3(cos(π/3+1π/2) + isin(π/3+1π/2)) = 3(cos5π/6+ isin5π/6)= -3√3/2 + 3/2i
k=2 z=3(cos(π/3+2/2π) + isin(π/3+2/2π) = 3(cos4π/3 + isin4π/3) = -3/2 -3√3/2i
k=3 z=3(cos(π/3+3π/2) + isin(π/3+3π/2)) = 3(cos11π/6 + isin11π/6) = 3√3/2-3/2i
Formula for nth roots = for each positive integer n, the nonzero complex #
r(cosθ + isinθ) has exactly n distinct nth roots. They are given by
n√r[cos(θ + 2kπ) + isin(θ + 2kπ)] where k=0,1,2,3…n-1
n
n
Ex. find the fifth root of -16√2 – (16√2)i
Step 1: find r, sinθ, cosθ
Step 2: Express in polar form r(cosθ + isinθ)
Step 3: Apply the root formula with n=5 ---
So have the following form: 2[cos 5π/4 + 2kπ + isin 5π/4 + 2kπ ]for k = 0,1,2,3,4
5
5
K=0 2[cos 5π/4 + 2π + isin 5π/4 + 2π ] = 2[cos π/4 + isin π/4]
5
5
K=1 2[cos 5π/4 + 2π + isin 5π/4 + 2π ] = 2[cos 13π/20 + isin 13π/20]
5
5
K=2 2[cos 5π/4 + 4π + isin 5π/4 + 4π ] = 2[cos 21π/20 + isin 21π/20]
5
5
K=3 2[cos 5π/4 + 6π + isin 5π/4 + 6π ] = 2[cos 29π/20 + isin 29π/20]
5
5
K=4 2[cos 5π/4 + 8π + isin 5π/4 + 8π ] = 2[cos 37π/20 + isin 37π/20]
5
5
Roots of Unity – the n distinct nth roots of 1 (the solutions of 2n=1) are called the nth
Roots of unity  polar form of 1 = cos 0 + isin 0
Formula for Roots of Unity – For each positive integer n, these are n distinct nth roots
of unity, which have the following form: cos 2kπ + isin 2kπ for k = 0,1,2…,n-1
n
n
Ex. find the eighth root of unity
Apply formula for roots of unity with n=8 k=0,1,2,3,4,5,6,7
K=0 cos0 + isin0 = 1
K=1 cos2π/8 + isin2π/8 = cosπ/4 + isinπ/4 = √2/2 + √2/2i
K=2 cos4π/8 + isin4π/8 = cosπ/2 + isinπ/2 = 0 + √1i
K=3 cos6π/8 + isin6π/8 = cos3π/4 + isin3π/4 = -√2/2 + √2/2i
K=4 cos8π/8 + isin8π/8 = cosπ + isinπ = -1
K=5 cos10π/8 + isin10π/8 = cos5π/4 + isin5π/4 = -√2/2 + -√2/2i
K=6 cos12π/8 + isin12π/8 = cos3π/2 + isin3π/2 = -1i
K=7 cos14π/8 + isin14π/8 = cos7π/4 + isin7π/4 = √2/2 - √2/2i
All Roots of Unity – let n be a positive integer with n>1. Then the number
cos 2kπ + isin 2kπ is an nth root of unity and all the nth roots of unity are
n
n
z,z²,z³,zn-1,zn = 1
ex. Find the ninth roots of unity
10.5 Vectors in the Plane
Vectors – quantities that require 2 #’s – magnitude and direction to describe them
a) represented by: a directed line segment or arrow
P→Q

P = initial point Q= terminal point and written as PQ
Length is denoted by: ║PQ║
**when endpoints are not specified  Vectors are denoted by boldface letters such as:
u, v, w
** length of vector u is ║u║ and is called the magnitude of u
** if vectors have the same magnitude and direction, the vectors u and v are said to be
equivalent
Ex.
**Need to use distance formula to show if vectors have the same length, then check for
same directions


Let P = (1,2) Q = (5,4) O = (0,0) R = (4,2) Show PQ = OR
Step 1: Find Magnitude of each:
║PQ║= √(5-1)²+ (4-2)² = √20
║OR║=√(4-0)²+ (2-0)² = √20
Step 2: Make sure direction is the same – if can’t tell use the slope formula
PQ = 4-2 = 1
OR = 2-0 = 2 = 1
5-1 2
4-0 4 2

same slope = same direction, and same length, so they are equivalent
** If 2 vectors are equivalent – then 1 vector may be moved to another location, provided
that the direction and magnitude don’t change, and still be equivalent
Equivalent Vectors – Every vector PQ is equivalent to vector OR with initial point at its
Origin if P=( x1y1) and Q=(x2y2) then PQ = OR where
R = (x2 - x1, y2 - y1)
A vector with initial point (0,0) and terminal point (a,b) is denoted by:
<a,b> where a and b are components of the vector
Magnitude (Norm) of a vector <a,b> is ║v║ = √a²+b²
Ex. find the components and magnitude of the vector with initial point P=(-2,6) and
terminal point Q=(4,-3)
Step 1: Find R by (x2 - x1, y2 - y1) = ( 4- -2, -3-6) = (6,-9)
Step 2: Find magnitude of r ║PQ║=║OR║= √6²+ -9² = √117
Scalar Multiplication
a) ordinary real numbers = Scalars
b) Scalar Multiplication = an operation in which a scalar k is multiplied by vector v to
produce another vector denoted by kv
**if k is a real # and v=<a,b> is a vector then kv = <ka,kb>
Ex. let v =<3,1> Find the components of 3v and -2v
a) 3v = 3<3,1> = <9,3>
b) -2v = -2<3,1> = <-6,-2>
Geometric Interpretation of Scalar Multiplicationj
a) The magnitude of the vector kv is |k| times the length of v:
║kv║=|k|·║v║
 the direction of kv is the same as that of v when k is positive and opposite
that of v when k is negative
Ex. ║-2v║=║<6,2>║= √6²+2²= √40 = 2√10
2√10 = 2║v║= |-2|·║v║
Vector Addition – is an operation in which 2 vectors u and v are added resulting in a new
vector u+v
**if u = <a,b> and v= <c,d> then u+v = <a+c, b+d>
Ex. let u=<-5,2> and v =<3,1> find the components of u + v
u+v = <-5+3, 2+1> = <-2,3>
Geometric Interpretation of Vector Addition

1. If u and v are vectors with the same initial point, P, then u+v is the vector PQ,
where PQ is the diagonal of the parallelogram with adjacent sides u and v
2. If the vector v is moved without changing its magnitude or direction so that its initial
Point lies on the endpoint of the vector u, then u+v is the vector with the same
Initial point P as u and the same terminal point Q, as v
Vector Subtraction = using the negative with a vector
* If u=<a,b> and v = <c,d> then u – v is the vector:
u + (-v) = <a,b> + <-c,-d> = <a-c,b-d>
Ex. let u = <2,5> and v=<6,1>. Find the components of u – v
Zero Vector <0,0> and denoted 0
Ex. Perform the combined vector operations:
u = <-1,6>
v=<⅔,-4>
w=<2,5/2>
a) 2u + 3v=
b) 4w – 2u =
Vector Properties:
1. u + (v+ w) = (u+v) + w
2. u+v=v+u
3. v+0=v
4. v+(-v) = 0
5. r(u+v) = ru + rv
6. (rs)v= r(sv) = s(rv)
7. 1v = v
8. 0v=0
Associative for addition
Commutative
Additive Identity
Additive Inverse
Distributive
Associative for Multiplication
Multiplicative Identity
Multiplication by 0
10.6 Applications of Vectors in the Plane
Unit Vector = a vector with length 1
 If v is a nonzero vector, then 1/║v║·v is a unit vector with the same
direction as v
Ex. <3/5, 4/5> is a unit vector b/c √(3/5)²+(4/5)² = 1
Ex. Find a unit vector u, with the same direction as the vector v=<20,21>
║v║=║<20.21>║= √20²+21² = √841 = 29
Let u = 1/29v = 1/29<20,21> = <20/29, 21/29>
Can check to make sure 1/║v║·v  1/29 · 29 = 1
Unit Vectors:
i = <1,0> and j=<0,1>
Alternate Vector Notation – written in terms of i and j
Ex. u=<5,-7> then u= <5,0> + <0,-7> = 5<1,0> + -7<0,1> = 5i -7j
i and j are linear combinations  when written in terms of i and j the properties of vector
addition and scalar multiplication can be used.
Ex. if u = 5i-4j and v=8i+2j find 4u-5v
4(5i-4j) – 5(8i+2j) = 20i – 16j – 40i – 10j  -20i-26j
Direction Angles – if v=<a,b> = ai+bj is a vector, then the direction of v is completely
determined by the standard position angle between 0° and 360°
whose terminal side is v
θ=direction angle
cosθ = a
║v║
sinθ = b
║v║
** In other words—If v=<a,b>= ai+bj then:
a=║v║cosθ and b= ║v║sinθ
Ex. find the component form of the vector that represents the initial velocity of a
baseball thrown at a speed of 80 feet per second, at an angle of 32° with the horizontal
Step 1: v= (80cos32)i + (80sin32)j
= 67.84i + 42.39j
=<67.84,42.39>
***If v=ai+bj is a nonzero vector with direction angle θ, the
Tan θ = sinθ =b
cosθ a
Ex. find the direction angle of each vector
a) v=6i+10j (in Q1 – so angle must be between 0 and 90)
tanθ=b/a = 10/6
θ = tan-1(10/6) = 59.04°
b) v=-9i-5j
(in Q3, angle must be between 180 and 270)
Vector Application
Resultant Force = the sum of all forces acting on an object
**every force has direction and magnitude – therefore each can be represented by
a vector.
Ex. An object at the origin is acted upon by 2 forces. A 300 lb. Force makes an angle of
35° with the positive x-axis, and the other force of 125 lbs makes an angle with 85° with
the positive x-axis. Find the direction and magnitude of the resultant force.
OP = (300cos35°)i + (300sin35°)j
OQ = (125cos85°)i + (125sin85°)j
Resultant (OR) is the sum of OP + OQ
OR =
Magnitude of OR =
Direction Angle =