6.01 Law of Sines

Standard notation for labeling a Triangle

Label the triangle below based off of the one in the lesson.

Solving an ________________triangle requires knowledge of one ___________ and any ________ other parts of the triangle: __________sides, _____________ angles, or one angle and one side.

The Four Cases

Solutions to oblique triangles are classified into four cases:

1.

2.

3.

4.

Law of Sines

If ABC is a triangle with sides a, b, and c, then ______=_______=_______

We solve triangles using the Law of Sines by noting that, if we are given two __________________ and ____________ side of the triangle, we are therefore given _______ parts of a proportion. We can therefore find the missing side of the triangle by solving for the missing part of the _______________________.

>>Test your skills<<<

In the triangle below, A = 35°, B = 50°, and a = 12 feet. Find the other values. 

Show work:

b= C=c=

The Ambiguous Case (SSA) Acute Angle Given

1.____________________________________________________________________2.____________________________________________________________________

3.____________________________________________________________________

When we are given _________, notice that we are always given________ sides and the non-included angle. That means, when we set up the _______________ using the Law of Sines, we will always be_____________ for the angle with the same letter as one of the given sides. For example, given side a, b, and angle A, we will be ___________ for angle B. Let's look at the _________ situations that can occur. Our chart will use letter h to represent the perpendicular of the triangle. Remember that h = _________ We will begin with the given angle, A, being acute.

a<________ No triangle The given side, a, is too_____________

to form a triangle

a=_____ one triangle The given side, a , is

_____________ equal to the perpendicular h.

a>________ one triangle The given side, a, is too

_____________ to form more than one triangle

h<______<b Two triangles The given side a, is

______________than the perpendicular and can swing along the base of the triangle to form __________ different triangles. 

To find out how many _____________ we have, we can use the chart above and examine the given side, a, or we can just proceed to solve the triangle.

By proceeding, if there is indeed no____________ possible when we try to find the value of angle B, our calculator will give us an______________ message.

However, if the calculator gives us a value for the measure of angle B, we must be cautious. Unless we have ____________the side opposite the given angle using the chart above, we will not know if there are ________ or ______ triangles.

This happens because the __________ function is positive in both quadrants I and II and our calculator can only give us the quadrant I angle. We can find the second corresponding quadrant II angle by subtracting the angle the calculator has given us from 180°. We should then make sure that the angle we have found

plus the angle that was given to us in the problem do not add up to more than 180°.

Solve the triangle: a = 6, b = 8, A = 35°

We set up the Law of Sines to solve for angle B.

_______=________

_______=_________

_________=________

B1= _________ and B2=180-______ =

For both choices of B, we have A + B < 180°. Hence there are two triangles, one containing the angle B1 = 49.9°, and the other containing the angle B2= 130.1°. The third angle C is eitherC1 = 180° - A -B1 = 95.1°     or      C2 = 180° - A - B2 = 14.9°

Remember that, although sometimes we can see there is no solution if we try to construct the triangle, we still need to verify it mathematically using the Law of Sines.

>>> Test your skills<<<

a. Show that there is no triangle for which a = 22 inches, c = 25 inches, and A = 68°. Part of the solution is to sketch the triangle.

Show work

b. Given A = 43.2°, a = 7.7 cm and b = 9.1 cm, how many triangles can satisfy these conditions? What is the value of B for each triangle? Part of the solution is to sketch the triangles.

Show work

The Ambiguous Case (SSA) Obtuse Angle Given

The ambiguous case, SSA, can also occur with the given angle being obtuse (> 90°). In this situation, there are only two cases:

1.

2.

a ____ b No triangle

a______b one triangle

>>>Test your skills<<<

In the triangle below, A = 123°, b = 23 cm, and a = 47 cm. Find the other values to the nearest tenth.

Show work:

Area of a Non-Right Triangle

The area of any triangle is given by one-half the product of the lengths of the two sides times the sine of their included angle. That is,

Area = ½bcsinA = ½absinC = ½acsinB

>> Test your skills<<

Find the area of the triangle to the right where C = 18.5°, a = 32.4 cm, b = 49.2 cm.

Show work:

Bearings

Directional bearings are given in three parts. 

1. ______________________ direction is given2. ______________________ degree value 3. _______________________ direction. 

Let’s begin with the basic compass.

A direction of S 55° E is read South 55 degrees East.  Imagine standing at the origin facing south and then merely turning 55 degrees toward the east. 

If we were to measure this on the general axis, we would see that the degree measure to this bearing is 325 degrees. 

These three values are all you need to determine a bearing. 

>>> Test your skills<<

Draw the following bearings on a compass:

1. N 41° W

2. S 22° W

Applications of the Law of Sines

Most application problems using the Law of Sines use bearings to represent degree values.  Let’s practice using both the Law of Sines and bearings to solve real-world problems!

>>> Test your skills<<<

Sarah and Casey are standing on the beach. The top of the 200 ft. tall light house is located on a bearing of N 33° W from Sarah and on a bearing of N 9° W from Casey. How far apart are Sarah and Casey?

Show your work:

6.02a Law of CosinesIf you are given three ______________or two _____________and its included angle, the ratios would be incomplete. To solve those types of problems, we need the Law of _________________.

Law of Cosines

Standard Form Alternate Form1. 1.2. 2.3. 3.

Three Sides (SSS)

Law of Cosines when we are given the three sides of a triangle, a = 8, b = 3, c = 9. This means that we must find the values of angles A, B, and C. We will start by finding angle A, so we use the version of the formula that says:

Cos A=

Plugging in the values for a, b, and c, we get:

Cos A=

We now repeat the process to find angle C:

Cos C=

We could of course repeat again to find angle B, but it is faster to use the fact that the three angles add up to 180°. We therefore find:

B=

>>> Test your skills<<<

Find the three angles (to the nearest tenth) of the triangle whose sides have lengths a = 23.5 cm, b = 44.2 cm, and c = 30.1 cm. Part of the solution is to sketch the triangle.

Show work here:

Find the value of a if A = 60°, b = 25 m, and c = 42 m in triangle ABC. Part of the solution is to sketch the triangle.

Show work here:

Herons’ Area Formula (SSS)

Find the value of a if A = 60°, b = 25 m, and c = 42 m in triangle ABC. Part of the solution is to sketch the triangle.

Area=

Where s=

We use the Law of Cosines to work through the proof of the formula, and you should read through that proof in your text. Let's look at an example that uses Heron's formula. We will find the area of triangle ABC given that a = 75.4, b = 52, and c = 52:

We begin by finding the semi-perimeter, s.

S=

Now, we simply have to replace s with 89.7 in the formula and evaluate:

Area=

>>> Test you skills<<<

Find the area of the triangular region having sides of lengths a = 6 meters, b = 3 meters, and c = 4 meters.

Show work here:

A boat leaves Port A and travels east for 230 miles and then 130 miles in the direction S 25° E.  How far is the boat from Port A (round to the nearest mile)?

Show work here:

6.03 Vectors in the Plane

Definition of a Vector

A vector is a___________________ _________________ ________________. The directed line segment PQ has ______________ point P and ______________ point Q. Its length is shown by ||PQ||.

Equivalence

Equivalent if they have the same __________________and the same _____________________.

To show that two vectors are equivalent, you show that the _________________and the __________________are the same.

Vocab Words Examples/definitions/equationsStandard Position

Component Form

Slope

Length or (magnitude)

Unit Vector

>>>Test your skills<<<

Find the component form and length of the vector v that has initial point (3,4) and terminal point (7,-1).

Show work here:

Vector Operations

Scalar Multiplication

Vector Addition

Properties of Vector Operations1.

2.

3.

4.

5.

6.

7.

8.

9.

Unit Vector:

>>> Test your skills<<<

Let u be a vector with initial point (3,8) and terminal point (1,1).

Let v be a vector with initial point (3,-4) and terminal point (0,0).

Find the component form of the vectors and calculate the value of 3u - 2v.

Show work

Find a unit vector in the direction of v = <1, -2 >, and verify that the result has length 1.

Vocab Words Examples/definitions/equationsHorizontal and vertical components of V

Linear Combinations

>>> Test your skills<<<

Let u be the vector with initial point (4,1) and terminal point (2,4). Write u as a linear combination of the standard unit vectors i and j.

Show work

Let u = i -2j and v = 5i +2j. Find 2u + 3v.

 

Resolving the Vector

If you are given the ____________________and ______________ angle of a vector, you may find its horizontal and vertical components through the use of trigonometry. This is called ______________ the vector.

If v = <v1, v2> = v1i + v2j, then v1 = |v| cos Θ and v2 = |v| sin Θ. In other words,

v = <|v| cos Θ, |v| sin Θ>

Direction Angles

The _________________angle of a vector is defined to be the angle______________ in ________________form (from the positive x-axis counter clockwise to the vector). The __________ direction angle is found by evaluating tan-1 |b/a|. We then find the actual reference angle by placing the reference angle into the correct quadrant.

Let's find the reference angle for -10i + 10j. We start by evaluating the reference angle, tan-1 |10/-10| or tan-1(1). This is

45°. Now, we place that reference angle into quadrant II to get the direction

angle (measured in standard position) of 135°.

>>> Test your skills<<<

Find the direction angle of the vector u = 8i + 5j.

Show work:

Applications of Vectors

Navigation is one example of an application of vectors.  There are several other applications of vectors found in your text.

An airplane is flying at a speed of 520 miles per hourat an angle of 30°.  At one point during flight, the plane comes across a wind with a velocity of 50 miles per hour at an angle of 15°.  Find the resulting direction and speed of the airplane.

Sketch these vectors on a coordinate system.

Convert each vector to component form and add them together to get the ground speed.

(**fill in each of the part below from your lesson)

Velocity of airplane=u=

Velocity of the wind = v=

||w||=

Direction of the airplane:

6.04 Dot Product of Vectors

Dot Product

Definition: ___________________________________________________

>>> Test your skills<<<

Find each dot product. Show work

a. <-2,-7> ▪ <-3,4>

b. <9,5> ▪ <0,8>

Properties of the Dot Product

Let u, v, and w be vectors in the plane or in space and let c be a scalar.

1.

2.

3.

4.

5.

>>> Test your skills<<<

a. Let u = <-4, -2> and v =<3, 7>. Find (u ▪ v)u

Show work

b. Use the dot product to find the length of v given v = <-6, 8>.

Angle Between Two Vectors

If  is the angle between two nonzero vectors u and v, then

cos= ----------------------

>>> Test your skills<<<

Find the angle between u = <-3, 2> and v =<2, 3>

Show work

Orthogonal Vectors

Definition of Orthogonal Vectors :________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

>>> Test your skills<<<

Are the vectors u = (6,-3) and v = (4,8) orthogonal, parallel, or neither?

Show work

Rectangular Coordinates in Space

vector v = < ____,_____,______>

The component form of vector v is found by subtracting the coordinates of the initial point from the terminal point:

vector v = < ____,_____,______> = <_____-____,_____-______,______-_____>

The zero vector is written by: 0 = <_____,_____,_____>

The standard vector notation for vector v is:

vector v=<v1i, v2j ,v3k>

Vectors in Space

1. ________________________________________________________________________________________________________________________________________________________________________________________________

2. ________________________________________________________________________________________________________________________________________________________________________________________________

3. ________________________________________________________________________________________________________________________________________________________________________________________________

4. ________________________________________________________________________________________________________________________________________________________________________________________________

5. ________________________________________________________________________________________________________________________________________________________________________________________________

>>> Test your skills<<<

Find the component form and length of vector v having initial point (0, -1, 2) and terminal point (-3, 4, 5).

Show work

Find the dot product of u = 5i +4j – 3k and v = i – j +2k

Angle Between Two Vectors

Let u and v be two nonzero vectors and , 0 ≤   ≤ π be the angle between them.  Then, angle   can be determined using the following:

cos= ----------------------

Please note that if the dot product of u and v is zero, the angle between the 2 vectors must be 90°. Therefore the vectors are orthogonal.

>>> Test your skills<<<

Find the angle between u = <1, 0, 3> and v = <2, 2, 1>

Show work

Parallel Vectors

Definition of Parallel Vectors: ________________________________________________________________________________________________________________________________________________________

>>> Test your skills<<<

Determine whether u and v are parallel.

u = <1, 2, -3>  and  v = < -4/3,-8/3,-4)

Show work

Cross Product

Definition of Cross Product:

__________________________________________________________________________________________________________________________________________________________________________________________________________________

>>> Test your skills<<<

Find each cross product, u x v

a. u = <-1, 5, 0> and v = <1, -3, 1>

Show work

b. u = 7i – k and v = i – 3j – 4k

Algebraic Properties of the Cross Product

1. ________________________________________________________

2. ________________________________________________________

3. ________________________________________________________

4. ________________________________________________________

5. ________________________________________________________

6. _____________________________________________________

Geometric Properties of the Cross Product

Let u, v, and w be vectors in space and let   be the angle between u and v.

1. ________________________________________________________

2. ________________________________________________________

3. ________________________________________________________

4. ________________________________________________________

Finding a Vector Orthogonal To Two Vectors

Given u = 2i and v = i – 2j + 3k, find a vector orthogonal to u and v.

According to the geometric properties of cross product, property #1, u x v is the vector orthogonal to u and v.

Answer: ____________________________________________________________

Finding the Area of a Parallelogram

A parallelogram has adjacent sides u = 2j and v= -2i + 3j

Since u and v are adjacent sides, || u x v || is the area of the parallelogram based on geometric property #4

Answer:____________________________________________________________________________________________________________________________________

6.05 DeMoivre’s Theorem and nth Roots

Name Definitions/Examples

Absolute value of a complex Number

Trigonometric Form of a Complex Number

>>> Test your skills<<<

Problem Show your work Convert each of the following to trigonometric form.

a. z = 2 + 2i√3 b. z = 4i 

Now, convert each of the following complex numbers into standard form (a + bi):

a. z1 = 2(cos  + isin ) 

b. z2 = 6(cos180° + isin180°)

See how similar this is to writing a vector in component form, <a, b> and in trigonometric form, where we write the vector with a magnitude ||v|| and a direction angle,  . It is also a similar

operation to writing ordered pairs of the rectangular form (x, y) and polar form (r,  ). Look at the

table below to make this comparison.

Vectors Numbers Ordered PairsRec./Comp. Trigonometric/Polar Rec./Comp. Trigonometric/Polar Rec./Comp. Trigonometric/Polar

Multiplying and Dividing complex numbers

Definition:

Example:

Look at the following example. Given:

z1 = 2(cos π/3 + isin π/3 )

z2 = 6(cos180° + isin180°)

Find the product z1z2 by multiplying the 2 and 6 and vy adding the 60° and 180°.

We get:

For the quotient, z1/z2, we divide the 2 and 6 and subtract the 60° and 180°.

We get :

Raising a Complex Number to a Power

Definition/example

Test your skills

Find the 5th power of the following complex number, and write your answer in standard (a + bi) form.

z = 2(cosπ/3 + isin π/3)

Show work

DeMoivre's Theorem for Roots

For a positive integer n, the complex number z = r(costheta + isintheta) has exactly n distinct nth roots given by:

By looking at the diagram on the right, we can see that the n roots of the complex number are equally spaced around a circle of radius nth root of r. This means that, once we find the first root, we simply add 2pi/n (or 360°/n, if we are working in degrees) to each successive root. Let's look at an example.

Find the 6th roots of z = 1(cos0° + isin0°)

Test your skills

Find the fourth roots of 81(cos 300° + i sin 300°).

Show work :

Review for Module 6 DBA

Make sure not to only know the terms below but also how to apply them .

Concepts Definitions examplesLaw of Sines

Law of Cosines

Heron's Formula

Component Form

Length (Magnitude)

Unit Vector

Vector Direction

Dot Product

Angle between two vectors

Cross Product

DeMoivre's Theorem for Roots