Trigonometry (Chapter 6) Sample Test #1 - · PDF fileTrigonometry (Chapter 6) – Sample...

Trigonometry(Chapter6)–SampleTest#1

First, a couple of things to help out:

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Trigonometry (Chapter 6) – Sample Test #1

More Formulas (memorize these):

Law of Sines:

sin sin sin

Law of Cosines:

2 cos

2 cos

2 cos

Area of a Triangle:

12 sin

12 sin

12 sin

12

Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest

degree.

1) A = 38° B = 32° a = 42.1

∠ 180° 38° 32° °

42.1sin 38° sin 32°

⇒

42.1 ∙ sin 32°sin 38°

.

42.1sin 38° sin 110°

⇒

42.1 ∙ sin 110°sin 38°

.

To solve: find the third angle, and then use the Law of Sines:

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Two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle, two triangles, or no triangle at all. Solve each triangle that results. Round lengths to the nearest tenth and angle measures to the nearest degree.

2) C = 35° a = 18.7

c = 16.1

3) B = 88° b = 2

a = 23

sin sin

⇒

18.7sin

16.1sin 35°

⇒sin 0.6662

First, calculate the measure of ∠ using the Law of Sines:

So, ∠ sin 0.6662 42° or ∠ 180° 42° 138°

Note that 16.1 18.7 , so we must compare to the height of the triangle:

sin 18.7 sin 35° 10.7 so that 10.7 16.1

Then, , so we will have two triangles and must solve each.

18.7, 16.1 ∠ 35°, ∠ °

∠ 180° 35° 42° °

sin sin

⇒

sin 103°16.1sin 35°

.

Triangle 1 – Start with:

Then,

And,

18.7, 16.1 ∠ 35°, ∠ °

sin sin

⇒

sin 7°16.1sin 35°

.

Triangle 2 – Start with:

Then,

∠ 180° 35° 138° °

And,

sin sin

⇒

23sin

2sin 88°

⇒ sin 11.5

First, calculate the value of using the Law of Sines:11.5 is not in the range of

the function sin .

Therefore, the given values

do not define a triangle.

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Find the area of the triangle having the given measurements. Round to the nearest square unit.

4) A = 37°, b = 10 inches, c = 9 inches

sin ∙ 10 ∙ 9 ∙ sin 37° inches2

Use Heron's formula to find the area of the triangle. Round to the nearest square unit.

5) a = 10 yards, b = 11 yards, c = 15 yards

12

First calculate: 10 11 15 18

Then, 18 18 10 18 11 18 15

√18 ∙ 8 ∙ 7 ∙ 3 √ ~ yards2

Solve the problem.

6) Two tracking stations are on the equator 127 miles apart. A weather balloon is located on a bearing of N36°E from the western station and on a bearing of N13°W from the eastern station. How far is the balloon from the western station? Round to the nearest mile.

90° 36° 54°

90° 13° 77°

180° 54° 77° 49°

127sin 49° sin 77°

⇒

The angles given are those shown in orange in the diagram

at right. The first step is to calculate the angles shown in

magenta in the diagram.

Then, use the Law of Sines, as follows:

is called the semi‐

perimeter because it is

half of the perimeter.

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Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree.

When you are given the lengths of two sides and the measure of the angle between them:

Use the Law of Cosines to determine the third side length,

Use the Law of Sines to determine the measure of one of the two unknown angles, and

Subtract the two known angle measures from 180° to get the measure of the last unknown angle.

2 cos

2 cos Law of Cosines

2 cos

Note: in problems with a lot of calculations, it is a good idea to keep more accuracy than you

are required to provide in the final answer, and round your answer at the end. This avoids the

compounding of rounding errors throughout your calculations.

7) a = 6, c = 11, B = 109°

6 11 2 6 11 cos 109° ~14.14125~ .

sin sin

⇒

6sin

14.14125sin 109°

⇒sin 0.4012

⇒ ∠ °

∠ 180° 109° 24° °

8) a = 7, b = 7, c = 5

2 cos

7 7 5 2 7 5 cos

Simplifying, we get49 84 70 cos

Then, we solve to get: cos , and so ∠ °

Since this is an isosceles triangle (note that ), ∠ °

∠ 180° 69° 69° °

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Solve the problem.

9) Two sailboats leave a harbor in the Bahamas at the same time. The first sails at 25 mph in a direction 320°. The second sails at 30 mph in a direction 200°. Assuming that both boats maintain speed and heading, after 4 hours, how far apart are the boats?

Boat 1 travels: 25mph ∙ 4hours 100mi.

Boat 2 travels: 30mph ∙ 4hours 120mi.

Using the Law of Cosines and the diagram at right,

we can calculate:

100 120 2 100 120 cos 120° 36,400⇒ .

Find another representation, , , for the point under the given conditions.

10) 5, , r < 0 and 2 4

5, can also be represented as 5, 5,

The resulting point is still in the interval 0, 2 . To get to the interval 2 , 4 , add 2 .

5, 2 ,

Note: given the parameters of this question, another solution would be:

5, 2 5,

Polar coordinates of a point are given. Find the rectangular coordinates of the point. Give exact answer.

11) (‐3, 120°)

First, note that 3, 120° 3, 120° 180° 3, 300°

Then, cos 3 cos 300° 3 ∙

And, sin 3 sin 300° 3 ∙ √ √

So, the rectangular coordinates are: , √

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The rectangular coordinates of a point are given. Find polar coordinates of the point. Express in radians.

12) (4, ‐4√3)

4 4√3 8

tan4√34

tan √3 in 4 53

So, the polar coordinates are: ,

Convert the rectangular equation to a polar equation that expresses r in terms of .

13) 4

Since cos , we make that substitution and solve for .

cos 4

14) 16

Substitute cos and sin .

cos sin 16

cos sin 16 (recall that cos sin 1)

16

note that we take the positive root of only

15) 8 3 10 0

Substitute cos and sin .

8 ∙ cos 3 ∙ sin 10 0

8 cos 3 sin 10

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Convert the polar equation to a rectangular equation.

16) r=5

Substitute

5

17) r=‐3cos

Substitute cos and

3

3

3

3 0

394

94

18) r=8cos +9sin

Substitute cos , sin and

8 9

8 9

8 9

8 9 0

8 16 9814

16814

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Graph the polar equation.

19) r 2 2 sin

This cardioid is also a limaçon of form sin with . The use of the sine function indicates that the large loop will be symmetric about the ‐axis. The sign indicates that the large loop will be above the ‐axis. Let’s create a table of values and graph the equation:

The four Cardioid forms:

0 2

/6 3 7π/6 1

/3 3.732 4π/3 0.268

/2 4 3π/2 0

2 /3 3.732 5 /3 0.268

5 /6 3 11 /6 1

2 2 2

Once symmetry is

established, these values

are easily determined.

Generally, you want to look at

values of in 0, 2 . However,

some functions require larger

intervals. The size of the interval

depends largely on the nature of the

function and the coefficient of .

Orange points on the

graph correspond to

orange values in the table.

Blue points on the graph

correspond to blue values

in the table.

The portion of the graph

above the ‐axis results

from in Q1 and Q2,

where the sine function is

positive.

Similarly, the portion of

the graph below the ‐axis

results from in Q3 and

Q4, where the sine

function is negative.

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20) r 4 sin 2

This function is a rose. Consider the forms sin θ and cos θ. The number of

petals on the rose depends on the value of .

If is an even integer, the rose will have 2 petals.

If is an odd integer, it will have petals.

The rose and other polar functions can be investigated in more detail using the Algebra

App, available at www.mathguy.us.

Let’s create a table of values and graph the equation:

The four Rose forms:

0 0

/12 2 7π/12 2

/6 3.464 2π/3 3.464

/4 4 3π/4 ‐4

/3 3.464 5 /6 3.464

5 /12 2 11 /12 2

/2 0 0

Once symmetry is

established, these values

are easily determined.

Because this function involves an

argument of 2θ, we want to start by looking at values of θ in 0, 22 0, . You could plot more

points, but this interval is sufficient

to establish the nature of the curve;

so you can graph the rest easily.

Orange points on the

graph correspond to

orange values in the table.

Blue points on the graph

correspond to blue values

in the table.

The values in the table

generate the points in the

two petals right of the ‐axis.

Knowing that the curve is a

rose allows us to graph the

other two petals without

calculating more points.

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Write the complex number in polar form. Express the argument in radians.

21) 5√3 5

The process of putting a complex number in polar form is very similar to converting a set

of rectangular coordinates to polar coordinates. So, if this process seems familiar, that’s

because it is.

5√3 5 10

tan5

5√3tan

√33

in 3 76

So, the polar form is: 10 cos sin

Note: 10cis 10 ∙ because cos sin . However, the student may

not be required to know this at this point in the course.

Write the complex number in polar form. Express the argument in degrees.

22) 6 8

6 8 10

tan86

tan43in 2 126.9°

So, the polar form is: . °

Note: To convert a polar form to exponential form, degrees

must be converted to radians. So, 10cis129.6°~10 . .

However, the student may not be required to know this at this point in the course.

Write the complex number in rectangular form. Give exact answer.

23) 5 cos sin

We want this in the form

cos 5cos 5 ∙

sin 5cos 5 ∙ √ √

So, the rectangular form is: √

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Find the product of the complex numbers. Leave answer in polar form.

24) 8 cos sin

3 cos sin

To multiply two numbers in polar form, multiply the ‐values and add the angles.

∙ 8 ∙ 3 ∙ cis6 2

Note: this may be easier to understand in exponential form, since exponents are added when values with the same base are multiplied:

8 ∙ ∙ 3 ∙ 24 24

25) 2 2

√3

This problem is probably best approached by converting each number to polar form and then multiplying.

Relating to :

2 2 2√2

tan22

tan 1 in 1 4

So, the polar form is: 2√2 cos sin 2√2

Relating to :

√3 1 2

tan1

√3tan

√33

in 4 6

So, the polar form is: 2 cos

sin

2

Then, multiply:

∙ 2√2 4

∙ 2 6

4√2 4 6

√

This angle could be or .

Thinking ahead, we know

we will be adding it to . By

using , we will not need

to simplify after adding.

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Find the quotient of the complex numbers. Leave answer in polar form.

26) 8 cos sin

3 cos sin

To divide two numbers in polar form, divide the ‐values and subtract the angles.

∙83∙ cis

2 6

Note: this may be easier to understand in exponential form, since exponents are subtracted when values with the same base are divided:

8 ∙

3 ∙

83

∙ 83

Use DeMoivre's Theorem to find the indicated power of the complex number. Write the answer in rectangular form.

27) 3 cos15° sin15°

To take a power of two numbers in polar form, take the power of the ‐value and multiply the angle by the exponent. (This is the essence of DeMoivre’s Theorem.)

3 cos15° sin15° 3 ∙ cis 4 ∙ 15° 81cis 60°

81cis 60° 81 cos60° sin60° 8112

√32

√

Note: Consider this in exponential form. Exponents are multiplied when a value with an exponent is taken to a power:

3cis12

3 ∙ 3 ∙ ∙ 81 ∙

28) 3 3 √3

3 3√3 6

tan3√33

tan √3 in 2 23

Then, 3 3 √3 6cis 6 cis 3 ∙

216cis2 216 ∙ cos 2 sin 2

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⟨ 3, 6⟩ ⟨ 6, 8⟩ ⟨ 3 6, 6 8⟩

⟨ , ⟩

⟨ 4, 21⟩ ⟨ 2,7⟩

⟨2, 14⟩

‖ ‖ 2 14

√200 √100 ∙ √2 √

Subtracting is the same as adding .

To get – , simply change the sign of each

element of . If you find it easier to add

than to subtract, you may want to adopt

this approach to subtracting vectors.

Find the specified vector or scalar.

29) u = ‐3i ‐ 6j, v = 6i + 8j; Find u + v.

An alternative notation for a vector in the form is ⟨ , ⟩. Using this alternative notation makes many vector operations much easier to work with.

To add vectors, simply line them up

vertically and add:

30) u = ‐2i ‐ 7j and v = ‐4i ‐ 21j; Find ‖ ‖.

Find the unit vector that has the same direction as the vector v.

A unit vector has magnitude 1. To get a unit vector in the same direction as the original vector, divide the vector by its magnitude.

31) v = 5i ‐ 12j

The unit vector is: ‖ ‖

√

32) v = ‐8j

The unit vector is: ‖ ‖

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Write the vector v in terms of i and j whose magnitude v and direction angle are given. Give exact answer.

33) ‖ ‖ = 10, θ = 120°

The unit vector in the direction θ 120° is:

⟨cos 120°, sin 120°⟩ ⟨12,√32⟩

12

√32

Multiply this by ‖ ‖ to get :

1012

√32

√

34)‖ ‖ = 7, θ = 225°

The unit vector in the direction θ 225° is:

⟨cos 225°, sin 225°⟩ ⟨√22,

√22⟩

√22

√22

Multiply this by ‖ ‖ to get :

7√22

√22

√ √

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⟨56.2917,32.5⟩ ⟨24.7487, 24.7487⟩

⟨81.0404,7.7513⟩

Solve the problem.

35) The magnitude and direction of two forces acting on an object are 65 pounds, N60°E, and 35 pounds, S45°E, respectively. Find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.

This problem requires the addition of two vectors. The approach I prefer is:

1) Convert each vector into itsi and j components, call them and ,

2) Add the resulting and values for the two vectors, and

3) Convert the sum to its polar form.

Keep additional accuracy throughout and round at the end. This will prevent error

compounding and will preserve the required accuracy of your final solutions.

Step 1: Convert each vector into itsi and j components

Let u be a force of 65 lbs. at bearing: N60°E

From the diagram at right,

θ 90° 60° 30°

65 cos 30° 56.2917

65 sin 30° 32.5

Let v be a force of 35 lbs. at bearing: S45°E

From the diagram at right,

φ 90° 45° 45°

35 cos 45° 24.7487

35 sin 45° 24.7487

Step 2: Add the results for the two vectors

Step 3: Convert the sum to its polar form

Direction Angle tan .

.. °

Distance √81.0404 7.7513 . lbs.

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⟨6, 9⟩ ⟨7, 3⟩

⟨13, 12⟩ ∙ ⟨6,5⟩

13 ∙ 6 12 ∙ 5 78 60

cos ∙

‖ ‖‖ ‖ √ ∙ √ √

⟨ 1, 6⟩ ∙ ⟨ 4, 4⟩

1 ∙ 4 6 ∙ 4 28

‖ ‖ 1 6 √37

‖ ‖ 4 4 4√2

cos7

√74. °

Use the given vectors to find the specified scalar.

36) u = 6i + 5j, v = 6i ‐ 9j,w = 7i ‐ 3j; Find u ∙ (v + w).

The alternate notation for vectors comes in especially handy in doing these types of

problems:

Find the angle between the given vectors. Round to the nearest tenth of a degree.

37) u = ‐i + 6j, v = 4i ‐ 4j

cos ∙

‖ ‖‖ ‖

0° 180°

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Neither vector is a multiple of the other.

Therefore, the vectors are neither

parallel nor orthogonal.

⟨1, √3⟩ ∙ ⟨1, 4⟩

1 ∙ 1 √3 ∙ 4 0

Clearly, 2

Therefore, the vectors are parallel.

⟨3, 1⟩ ⟨6, 2⟩

Using the dot product, we check to see if:

∙ ‖ ‖ ‖ ‖

Therefore, the vectors are parallel.

⟨3, 1⟩ ∙ ⟨6, 2⟩

2 20 18

‖ ‖ 1 √10 3

‖ ‖ 2 √40 6

‖ ‖‖ ‖ √10 ∙ √40 20

It is clearly easier to check whether one

vector is a multiple of the other than to

use the dot product method. The

student may use either, unless

instructed to use a particular method.

To determine if two vectors are parallel using the

dot product, we check to see if:

∙ ‖ ‖‖ ‖

Therefore, the vectors are not parallel.

⟨1, √3⟩ ∙ ⟨1, 4⟩

1 ∙ 1 √3 ∙ 4 1 4√3

‖ ‖ √3 2 1

‖ ‖ 4 √17 1

‖ ‖‖ ‖ 2 ∙ √17 2√17

Use the dot product to determine whether the vectors are parallel, orthogonal, or neither.

If the vectors are parallel, one is a multiple of the other; also ∙ ‖ ‖‖ ‖ .

If the vectors are perpendicular, their dot product is zero.

38) v = i + √3j, w = i ‐ 4j

39) v = 3i ‐ j, w = 6i ‐ 2j

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Let’s do the calculations.

Then,

proj ∙

‖ ‖

⟨2,1⟩ ⟨ , ⟩

And,

⟨45,25⟩

⟨95 ,185 ⟩

⟨1, 4⟩ ∙ ⟨2,1⟩

1 ∙ 2 4 ∙ 1 2

‖ ‖ 2 1 5

⟨1, 4⟩

Decompose v into two vectors and , where is parallel to w and is orthogonal to w.

40) v = i ‐ 4j,w = 2i + j

The formulas for this are:

proj ∙

‖ ‖ and

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Solve the problem.

41) A force is given by the vector F = 5i + 2j. The force moves an object along a straight line from the point (5, 7) to the point (18, 13). Find the work done if the distance is measured in feet and the force is measured in pounds.

Work is a scalar quantity in physics that measures the force exerted on an object over a

particular distance. It is defined using vectors, as shown below. Let:

F be the force vector acting on an object, moving it from point to point .

be the vector from to .

be the angle between F and .

Then, we define work as:

∙

‖ ‖ cos

For this problem it is sufficient to use the first work formula, ∙

We are given ⟨5, 2⟩.

We can calculate as the difference between the two given points.

18, 13

5,7

⟨13,6⟩ Note that the difference between two points is a vector.

Then, calculate ∙

⟨5, 2⟩

∙ ⟨13, 6⟩

∙ 5 ∙ 13 2 ∙ 6 foot pounds

Both of these formulas are useful.

Which one you use in a particular

situation depends on what

information is available.

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