Section 7.1 Oblique Triangles & Law of Sines

Section 7.1 Oblique Triangles & Law of Sines

Section 7.2 Ambiguous Case & Law of Sines

Section 7.3 The Law of Cosines

Section 7.4 Vectors and the Dot Product

Section 7.5 Applications of Vectors

Chapter 7Applications of Trig and Vectors

Section 7.1 Oblique Triangles & Law of Sines

• Congruency and Oblique Triangles

• Law of Sines

• Solving using AAS or ASA Triangles

Congruency and Oblique Triangles

• If we use A for angles and S for sides what are all of the three letter combinations you could create?

• Which of these can we use to prove the triangles are congruent?

ASA

SAA

AAA

YES

YES

NO

Congruence Shortcuts

SSS

SAS

SSA

YES

YES

NO

Congruence Shortcuts

Data required for SolvingOblique Triangles

1. One side and two angles (ASA or AAS)2. Two sides and one angle not included

between the two sides (ASS). Yep this one can create more than one triangle

3. Two sides and the angle between them (SAS)

4. Three sides (SSS)5. Three angles (AAA) Yep this one only

creates similar triangles.

E

C

A

R

Given:

AR = ER

EC = AC

Show /_E = /_A

~

~

~

1. AR = ER

2. EC = AC

3. RC = RC

ΔRCE = ΔRCA /_E = /_A

~

~

~

~ ~

Given

Given SSS CPCTC

Reflexive

E

C

A

R

Given:

/_E = /_A

/_ECR = /_ACR

Show AR = ER

~

~

~

2. / ECR = / ACR

3. RC = RC

ΔRCE = ΔRCA AR = ER

~

~

~

~ ~

Given

Given AAS CPCTC

Reflexive

1. /_E = /_A

E

C

A

R

Given:

/_E = /_A

/_ERC = /_ARC

Show AR = ER

~

~

~

2. / ERC = / ARC

3. RC = RC

ΔRCE = ΔRCA AR = ER

~

~

~

~ ~

Given

Given ASA CPCTC

Reflexive

1. /_E = /_A

Law of Sines

In any triangle ABC, with sides a, b, and c,

= , = , =

This can be written in compact form as

= =

asin A

bsin B

asin A

csin C

bsin B

csin C

asin A

csin C

bsin B

Area of a Triangle

In any triangle ABC, the area A is given by any of the following formulas:

A = ½bc sin A

A = ½ab sin C

A = ½ac sin B

Section 7.2 Ambiguous Case & Law of Sines

• Description of the Ambiguous Case

• Solving SSA Triangles (Case 2)

• Analyzing Data for Possible Number

Ambiguous Case AcuteNumber of

Possible Triangles SketchCondition Necessary for Case to Hold

0 a<h

1 a=h

1 a>b

2 b>a>h

Ambiguous Case

Number of Possible Triangles

SketchCondition Necessary

for Case to Hold

0 a<b

1 a>b

SSA Cases

• Remember since SSA results in two possible triangles we must check the angles supplement as well. So if we find the angle is 73 then we also have to check 180 – 73 = 107.

Section 7.3 The Law of Cosines

• Derivation of the Law of Cosines

• Solving SAS Triangles Case 3

• Solving SSS Triangles Case 4

• Heron’s Formula for the Area of a Triangle

Triangle Side Length Restriction

• In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.

Law of Cosines

Law of Cosines

In any triangle ABC, with sides a, b, and c,

a2 = b2 + c2 – 2bc cos A

b2 = a2 + c2 – 2ac cos B

c2 = a2 + b2 – 2ab cos C

Oblique Triangle Case 1

• One side and two angles AAS or ASA1. Find the remaining angle using the angle

sum formula (A+B+C)=1802. Find the remaining sides using the Law of

Sines

Oblique Triangle Case 2

• Two sides and a non-included angle SSA1. Find an angle using the Law of Sines

2. Find the remaining angle using the Angle Sum Formula

3. Find the remaining side using the Law of Sines

There may be no triangle or two triangles

Oblique Triangle Case 3

• Two sides and an included angle SAS1. Find the third side using the Law of Cosines

2. Find the smaller of the two remaining angles using the Law of Sines

3. Find the remaining angle using the angle sum formula

Oblique Triangle Case 4

• Three sides SSS1. Find the largest angle using the Law of

Cosines

2. Find either remaining angle using the Law of Sines

3. Find the remaining angle using the angle sum formula

Heron’s Area Formula

If a triangle has sides of lengths a, b, and c, and if the semi-perimeter is

s= ½(a+b+c)

then the area of the triangle is

A = s(s-a)(s-b)(s-c)

Section 7.4 Vectors andthe Dot Product

• Basic Vector Terminology

• Finding Components and Magnitudes

• Algebraic Interpretation of Vectors

• Operations with Vectors

• Dot Product and the Angle between Vectors

Basic Terminology

• scalars – quantities involving only magnitudes

• vector quantities – quantities having both magnitude and direction

• vector – a directed line segment• magnitude – length of a vector• initial point – vector starting point• terminal point – second point through which

the vector passes

Sum of vectors

To find the sum of two vectors A and B:

A+B

resultant vector

or

Difference of vectors

To find the difference of 2 vectors A and B:

A+(-B)

resultant vector

or

To find the product of a real number k and a vector A : kA=A+A+…+A (k times)

Example: 3A

Scalar Product

Magnitude and DirectionAngle of a Vector <a,b>

The magnitude of vector u=<a,b> is

given by

|u| = a2 + b2

The direction angle satisfies

tan =b/a,

where a ≠ 0.

Horizontal and Vertical Components

The horizontal and vertical components, respectively, of a vector u having magnitude |u| and direction angle are given by

= |u| cos and

=|u| sin

Vector Operations

For any real numbers a, b, c, d, and k,

<a, b> + <c, d> = <a+c, b+d>

k ·<a, b> = <ka, kb>

If a = <a1, a2>, then -a = <-a1, -a2>

<a, b> - <c, d> = <a, b> + -<c, d>

Unit Vectorsi = <1, 0> j = <0, 1>

i, j Form for VectorsIf v = <a, b>, then v = ai + bj

Dot Product

The dot product of the two vectors

u = <a, b> and v = <c, d> is denoted by

u·v, read “u dot v,” and given by

u·v = ac + bd

Properties of the Dot Product

For all vectors u, v, w and real numbers k

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

Geometric Interpretation ofDot Product

If is the angle between the two nonzero vectors u and v, where 0< <180, then

u·v = |u||v| cos

Orthogonal Vectors

Two nonzero u and v vectors are orthogonal vectors if and only if u · v = 0

Section 7.5 Applications of Vectors

• The Equilibrant

• Incline Applications

• Navigation Applications

Equilibrant

A vector that counterbalances the resultant is called the equilibrant. If u is a vector then –u is the equilibrant.

u + -u = 0

Equilibrant Force

• Use the law of Cosines

48

60

60

48

v-v

B

A

130à

|v|2 = 482 + 602 – 2(48)(60)cos(130à)|v|2 ≈ 9606.5|v| ≈ 98 newtons

The required angle can be found by subtracting angle CAB from 180à.

C

98 60

sin 130à sin CAB=

CAB ≈ 28à so £ = 180à - 28à =152à

Inclined Application

20à50

x

sin 20à = |AC|/50|AC| ≈ 17 pounds of force

AC