lesson #64 first degree trigonometric equations · lesson #52 - inverse trigonometric functions...

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Lesson #64 – First Degree Trigonometric Equations A2.A.68 Solve trigonometric equations for all values of the variable from 0° to 360°

How is the acronym ASTC used in trigonometry?

If I wanted to put the reference angle, 75° into the 2nd, 3rd,

and 4th quadrants, how would I do so?

Find sin30°. Find sin150°.

Find sin390°. Find sin510°.

Find sin(-330°). Find sin(-210°).

Why do all of these angles have the same sine value?

Solve the equation, 1

sin( )2

x . Is your initial answer the only solution?


cos( )2

x . Find all solutions between 0° and 360°.


sin( )2



cos( )2


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Even though there are really an infinite number of solutions to most trigonometric equations, we

will only consider the solutions between 0° and 360° for this course. Here is the method for

doing so.

1. Isolate the trigonometric part of the equation

using SADMEP.

2. Determine what quadrants your answer will be in

based upon the sign of the trig. value (ASTC).

TIP: Write ASTC next to the trig. value. Circle

the quadrants where the answers will be.

3. Use the inverse trig function to solve for the

angle.

4. If necessary, find the reference angle.

5. Put the reference angle in the quadrants you

chose.

a. QII: subtract reference angle from 180°.

b. QIII: add reference angle to 180°.

c. QIV: subtract reference angle from 360°.

Ex) Solve for x. Round to the nearest degree. 3cos 6 8x

Example: Solve for in the interval 0 360 to the nearest degree.

1) Solve for x on the interval, 0 360x .

xx sin2sin

2) Solve for x on the interval, 0 360x to the

nearest degree. 2tan 1 0x

8cos 2 5 15cos

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nearest degree. 3(sin 5) 4x


nearest degree. tan 2 2x

In this unit we will be working with a couple of formulas that are used with triangles. Since we

are working with triangles we will only have to consider angles between 0° and 180°.

The Law of Sines a

A

b

Bsin sin

The Law of Cosines

a b c bc A2 2 2 2 cos

Solve the following equations for x on the interval, 0°<x<180°. Round to the nearest degree.

5 9

sin 30 sin x

2 2 210 4 7 2(4)(7)cos x 2

2 25 3.2 (6) 2(3.2)(6)cos x

Notice, with the law of sines you can get 2 answers, but with the law of cosines you can get only

one answer for your angle. Why is this true?

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Lesson #52 - Inverse Trigonometric Functions A2.A.63 Restrict the domain of the sine, cosine, and tangent functions to

ensure the existence of an inverse function

A2.A.64 Use inverse functions to find the measure of an angle, given its sine,

cosine, or tangent

A2.A.65 Sketch the graph of the inverses of the sine, cosine, and tangent

functions

At this point, many students ask the following questions:

1. If there are two answers between 0° and 360°, why does my calculator only give

me one of them?

2. Why do I sometimes get a negative answer for my angle?

These questions have loaded answers. We will have to use a lot of our knowledge about

one-to-one functions, inverses, trig. graphs, and domains to answer them.

1sin ( )y x

Graph y=sin(x) on your calculator, in

radian mode, with a zoom trig

window. You should see the

following graph. Let’s consider the

equation, sin(x)=0. Where is the y-

value of the sine curve equal to 0?

Convert these values to degrees.

Are these the only places where the sine curve is equal to 0? How many answers are

there?

Your calculator cannot give you an infinite number of answers. It works with functions,

which only give one output for each input, and expects you to find any other answers

you want using your knowledge of reference angles and ASTC. Since functions are

predictable, your calculator is predictable in what answer it will give you.

Circle a portion of the sine curve that is one-to-one (passes the Horizontal Line Test) and

is also closest to the origin.

The domain of this piece of the graph is: __________.

Converted to degrees this would be: _________.

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This is called a restricted domain. When you use the the inverse sine function, 1sin ( )x ,

the calculator will always give you an answer between -90° and 90°, inclusive. In radians

this would be:

Graph 1sin ( )y x , and sketch it on the graph provided.

1cos ( )y x

Graph y=cos(x) on your calculator,

in radian mode, with a zoom trig

window. You should see the

following graph. Let’s consider

the equation, cos(x)=0. Where is

the y-value of the cosine curve

equal to 0?


Are these the only places where the cosine curve is equal to 0? How many answers are

there?





Restricted Domain for Sine so that

the inverse will be a function.

Note: You must be in

radian mode when

graphing inverse trig.

functions.

Compare this

graph with the

piece of y=sin(x)

you circled on the

previous page.

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Circle a portion of the cosine curve that is one-to-one (passes the Horizontal Line Test)

and is also closest to the origin.

The domain of this piece of the graph is: _________.

Converted to degrees this would be: _______.

This is the restricted domain for cosine. When you use the inverse cosine function, 1cos ( )x , the calculator will give you an answer between 0° and 180°, inclusive. In radians

this would be:

Graph 1cos ( )y x , and sketch it on the graph provided.

1tan ( )y x

Graph y=tan(x) on your calculator, in radian mode, with a zoom trig window. You should

see the following graph. Let’s

consider the equation, tan(x)=0.

Where is the y-value of the tangent

curve equal to 0?


Are these the only places where the

tangent curve is equal to 0? How many answers are there?





Restricted Domain for Cosine so

that the inverse will be a function.

Compare this

graph with the

piece of y=cos(x)

you circled on the

previous page.

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Circle a portion of the tangent curve that is one-to-one (passes the Horizontal Line Test)

and is also closest to the origin.

The domain of this piece of the graph is: _________.

Converted to degrees this would be: _______.

This is the restricted domain for tangent. When you use the inverse cosine function, 1cos ( )x , the calculator will give you an answer between -90° and 90°, exclusive. In

radians this would be:

Graph 1tan ( )y x , and sketch it on the graph provided.

This whole explanation is important for your math understanding which ultimately leads to

better retention and better grades, but the information you will be directly tested on is in the

thickly outlined textboxes.

Other important information about trigonometric inverses

1. The trigonometric functions can have alternate names, Arc ___.

a. 1siny x (also known as siny Arc x )

b. 1cosy x (also known as rccosxy A )

c. 1tany x (also known as tany Arc x )

2. The value for the angle that your calculator gives you is called the principal value.

3. Unless the problem says to solve for x between 0° and 360°, you can assume that you are

looking for the principle value.

4. For the following problems we will not be graphing. Just as in Unit #6, if you are asked to

find the answer in radians, complete the problem in degree mode and convert at the end.

Restricted Domain for Tangent so

that the inverse will be a function.

Compare this

graph with the

piece of y=tan(x)

you circled on the

previous page.

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1. What is the principal value of ?

1)

3)

2)

4)

2 The value of is

1) 0 3) 2)

4)

3 The value of is

1) 120° 3) 90° 2) 105° 4) 75°

4 If , then x is equal to

1) 3) 2) 4)

5 If and , the measure of angle x

is

1) 45º 3) 225º 2) 135º 4) 315º

6 What is the value of x in the equation ?

1)

3)

2)

4)

7 If , what is the measure of angle , in

degrees?

8 What is the principal value of ?

1) 3) 2) 4)

9 What is the principal value of , in

degrees and radians.

10 What is the smallest positive value of x, in radians,

that satisfies ?

11 Find the value of , in

degrees.

12 If , what is the value of angle A to

the nearest minute?

1) 3) 2) 4)

13 If , find the value of positive acute

angle A to the nearest minute.


angle x to the nearest minute.


angle to the nearest minute.

16 If , find the measure of positive acute

angle to the nearest minute.

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c

a

b

m BAC = 36

m BCA = 51

m ABC = 93

A

B

C

x

22

70

43

B

A

C

Lesson #65- Trigonometric Application Formulas A2.A.73 Solve for an unknown side or angle, using the Law of Sines or the Law of Cosines

For all problems in this lesson, round to the nearest tenth.

Finding the sides and angles in triangles that are not right triangles requires the use of the

“trig laws.” You are given these formulas on the A2&T reference sheet. Therefore the main

focus of this unit is learning how and when to use them to solve different types of problems.

ALWAYS DRAW A PICTURE!!!!

The Law of Sines:

sin sin sin

a b c

A B C

1. Given: a=12, 25m A , 58m C . Find side c.

2. Example: Solve for x.

Triangle Review Sum of the Degrees in a Triangle:

Labeling a Triangle:

lowercase letters for sides.

UPPERCASE letters for angles.

The same letter for a side and the opposite angle.

The smallest angle is across from the smallest side, ___.

The largest angle is across from the largest side, ____.

In what types of triangles can you use the Pythagorean

Theorem and SOH-CAH-TOA?

“Trick”

CIRCLE The PAIRS You must have 1 Angle/Side

Pair where you know the

values.

Connection to Proofs:

Use when given:

ASA or AAS

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x

7

41

103

Remember, with the Law of Sines you always need a known side angle pair and one other piece

of information.

3. Example: Solve for x.

Law of Cosines: 2 2 2 2 cosa b c bc A

4. Given: c=12, b=15, 84m A . Find a.

5. Given: a=10, b=15, and c=20, find m A .

The letters are less important in the formula than the actual placement of the sides.

Law of Cosines

WORKING WITH 3 SIDES An Angle/Side pair must start

and finish the equation. One

of them will be unknown since

it is what you are finding.

Connection to Proofs:

Use when given:

SSS or SAS

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The gist of the Law of Cosines is:

(1st side)Ü= (2nd side)Ü+ (3rd side)Ü– 2(2nd side)(3rd side)COS(angle opposite the 1st side)

You can choose which side you want for the 1st side based upon what you want to find.

6. Solve for x.

7. Find the measure of angle B.

x

7

536

12

15

11

B

A

C

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Directions: Round sides to the nearest tenth of a unit. Round angles to the nearest degree.

1) In triangle ABC, a=12, b=15, and 60m C . Find c.

2) If 82m C , 55m A , and a=8, find c.

3) In triangle ABC, a=20, b=16, and c=32. Find m B .

4) In triangle ABC, if 110m B , 30m A , and a=15, find b.

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5) In triangle ABC, a=20, c=25, and 98m B . Find b.

6) In triangle ABC, if 16m A , b=92, and 120m B find c.

7) In triangle ABC, b=20, c=23, and a=30, find m A.

8) If 75m A , 55m B , and c=5, and find a.

9) In triangle ABC, a=19, b=14, and c=12. Find m C .

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Lesson #66 –The Ambiguous Case & the Donkey Theorem (SSA) A2.A.75 Determine the solution(s) from the SSA situation (ambiguous case)

Abiguous/Ambiguity (from Webster dictionary)– 1 a : doubtful or uncertain especially from obscurity or indistinctness

<eyes of an ambiguous color>

**2 : capable of being understood in two or more possible senses or ways

<an ambiguous smile> <an ambiguous term> <a deliberately ambiguous reply>

In lesson #72, we looked at solving triangles when given AAS, ASA, SAS, and SSS. You will

remember from last year that these are all ways to prove triangles congruent. When two

triangles are congruent it means we could find all of their sides and angles, so we know that

they are EXACTLY THE SAME.

What about the donkey theorem, SSA? This is not one of our ways to prove triangles

congruent, which means that given this pattern, we do not really know what the remaining parts

of the triangle will be. It is AMBIGUOUS.

You will want to look for this SSA pattern, but the fact that this situation is unclear arises

naturally when we use the law of sines.

Why: When given SSA we have a known SIDE-ANGLE pair, so we would use the law of sines to

find the other Angle.

There are two possible answers for the angle, one between 0 and 90 degrees as well as an

answer between 90 and 180 degrees.

You just have to figure out if one, both, or none of the angles will fit in your triangle with the

angle you are given.

Solve for x in each equation on the interval 0°<x<180° because these are the only angles

that could be in a triangle. Round your answers to the nearest degree.

1. sin .5678x

2. cos .5678x 3. cos .5678x

When working with triangles, what is the only trigonometric function that can give us

two answers for the angle?

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Before we start solving these problems, the following three pictures show how there could be 0, 1, or 2 different possible triangles when we are given the SSA pattern. In each triangle the lengths of sides a, b, as well as ∠B are given.

a) Determine the number of possible triangles.

b) Find the measures of the three angles of each possible triangle. Express approximate values

to the nearest degree.

Steps: a=4, b=6, and m A=30

1. Once you recognize the SSA pattern,

draw 2 triangles.

2. Set up proportions to perform

law of sines to find a missing angle.

3. If sin(x) 1, find the missing angle.

(If sin(x) > 1, you know there are ______ triangles)

4. Since sine is positive in QI and Q II,

find the 2 possibilities for the angle.

5. Put each answer into one of the triangles you drew.

See if neither (0), one (1), or both (2) of them fit

with your given angle.

6. For each possible triangle, find the remaining angle measure.

No triangles

Two triangles

One triangle

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Practice: c) Determine the number of possible triangles.

d) Find the measures of the three angles of each possible triangle. Express approximate values

to the nearest degree.

1. a=7, b=6, and m B=150

2. a=6, b=4, and m A=150

3. a=6, b=8, and m A=40

4. In triangle ABC, if A=30 , a=6, and b=8, the number of distinct (different) triangles that

can be constructed is:

a. 1 b. 2 c. 3 d. 0

5. In triangle ABC, if A=30 , a=5, and b=10, the number of distinct (different) triangles

that can be constructed is:

a. 1 b. 2 c. 3 d. 0

Note: Some of these

problems will make

intuitive sense.

Look at your

answers to #1 and

#2. How could you

figure out those

answers without

using the law of

sines?

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Lesson #67 – Area of Triangles and Parallelograms A2.A.74 Determine the area of a triangle or a parallelogram, given the measure of two sides

and the included angle

Degrees-Minutes-Seconds We typically use decimals and the base ten system to express parts of a number. There is

another way to represent a part of an angle. A full rotation is split into 360°. We can also

split a degree up into smaller measurements based on multiples of 60 using words that will be

familiar to you. One minute of a degree is 1/60th of a degree. One second of a degree is

1/60th of a minute (1/360th of a degree). You can easily convert between decimal degrees

and degrees-minutes-seconds on your calculator. Follow the directions below.

Convert 57° 45' 17'' to decimal degrees:

In either Radian or Degree Mode: Type 57° 45' 17'' and hit Enter.

° is under Angle (above APPS) #1

' is under Angle (above APPS) #2

'' use ALPHA (green) key with the quote symbol above the + sign.

Answer: 57.75472222

Convert 48.555° to degrees, minutes, seconds:

Type 48.555 ►DMS Answer: 48° 33' 18''

The ►DMS is #4 on the Angle menu (2nd APPS). This function works even if Mode is set

to Radian.

A. Convert the following measures to decimal degrees. Round to the nearest hundredth.

a) 120°40’ 34’’ b) 18° 23’ c) 45° 50’ 10’’

B. Convert the following measures to degrees-minutes-seconds. Then round them to the

nearest minute.

d) 85.784° e) 26.33333° f) 98.760°

For this unit, we will always work in decimal degrees. Therefore, if you are given an angle in

DMS, convert it to decimal degrees. If you are asked to find an angle in DMS, convert it at

the end.

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Area of a Triangle There is one more trigonometric formula that you will be given to you on the regents. You

already know that the area of a triangle can be calculated with the formula . This

formula is limited because you must know the base and the height of the triangle. If you know

one side of the triangle, you can make that side the base, but you do not always know the

height.

We can use trigonometry to substitute known information for the height. Observe below:

A similar proof can be used to show that this formula works an obtuse triangle like the second

triangle ABC above.

To summarize, the area of a triangle, K, is given by the following formula:

1sin

2K ab C

You will notice that the formula looks slightly different than the one in the proof. You should

be comfortable with the fact that the letters do not matter; it is their relative position on the

triangle. Therefore, the information we need is SAS or 2 sides and the included angle.

1. Find the area of a triangle where: a=18, b=12, and 100m C . Round to the nearest square unit.

2. Find the area of a triangle where: b=20, c=30, and 34m A . Round to the nearest square unit.

Area of a Triangle

You need a known “corner”

(SAS).

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3. Find the area of the triangle below. Round to the nearest tenth.

4. Challenge: A triangular plot of land has sides that measure 5 meters, 7 meters, and 10

meters. What is the area of this plot of land, to the nearest tenth of a square meter?

Area of Parallelograms A parallelogram can be divided into two equal triangles. Therefore, what formula could we use

to find the area of the parallelogram below?

5. To the nearest tenth, find the area of a parallelogram with sides of 16 and 18 and an

angle of 60°.

6. Find the area of the parallelogram below to the nearest unit.

45

38

52

9

7

104

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Lesson #68 – Using Trig. Apps. in Word Problems A2.A.74 Determine the area of a triangle or a parallelogram, given the measure of two sides


A2.A.73 Solve for an unknown side or angle, using the Law of Sines or the Law of Cosines

This lesson we will be looking at different situations where we can use the trig. laws and the

area of a triangle formula. Below are some common shapes that arise in these problems and

their most important properties. Fill in everything else that you know on each shape.

Isosceles Triangles

2 sides congruent (called the legs)

Base angles congruent

Parallelograms

Opposite Sides Parallel

Opposite Sides Congruent

Opposite Angles Equal

Adjacent Angles Supplementary

Can be cut into two congruent triangles

Rhombuses

A parallelogram with congruent sides

Isosceles Trapezoids

Legs Congruent

Base Angles Congruent

Diagonals Congruent

1) If the area of an isosceles triangle is 25 square feet, and the leg length is 9 feet, find the

measure of the angles of the triangle to the nearest minute (assume all angles are acute).

10

69

BC

A

7

12

67

5

46

15

5

13

72

A C

B D

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2) Find, to the nearest tenth, the area of a triangle with side lengths, 22, 34, and 50.

3) In , m<A = 50º, m<B = 35º, and a = 12. Find the missing

sides and angle. (nearest tenth, nearest degree)

4)

5) The lengths of the adjacent sides of a parallelogram are 21 cm and 14 cm. The smaller angle

measures 58 . What is the length of the longer diagonal? Round your answer to the nearest centimeter.

6) If the area of a triangle is 14 square feet, one side is 5 units, and another side is 6 units,

find the sine of the included angle.

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7) The side length of a rhombus is 15 feet and the longer diagonal is 23 feet. Find the angles

of the rhombus in degrees-minutes-seconds.

8) A ship at sea heads directly toward a cliff on the shoreline. The accompanying diagram

shows the top of the cliff, D, sighted from two locations, A and B, separated by distance S.

If m DAC m DBC27 50 , , and S = 25 feet, what is the height of the cliff, to the

nearest foot?

9) An angle of a parallelogram has a measure of 145 . If the sides of the parallelogram

measure 9 and 13 centimeters, what is the area of the parallelogram to the nearest tenth?

10) A cross-country trail is laid out in the shape of a triangle. The lengths of the three paths

that make up the trail are 2000 m, 1200 m, and 1800 m. Find to the nearest degree the

measure of the smallest angle formed by the legs of the trail.

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11) If the base angle of an isosceles triangle measures 34 and the base of the triangle is 8

inches, find the length of other sides of the triangle to the nearest tenth.

12) In a triangle, two sides that measure 4 cm and 7cm form an angle of 60°. Find the measure

of the smallest angle of the triangle to the nearest degree.

13) In ,ABC ,18AC ,10BC and .2

1cosC find the area of ABC to the nearest tenth of a

square unit. (Hint: Find angle C first).

14) In an isosceles triangle, the vertex angle is 30º and the base

measures 12 cm. Find the perimeter of the triangle to the nearest integer.

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15) Points A & B are on one side of a river, 100 feet apart, with C on

the opposite side. The angles A and B measure 70º and 60º

respectively. What is the distance from point A to point C, to nearest foot?

16) A triangular field has side lengths of 100 feet, 250 feet, and 300 feet. Find the area of the

field to the nearest square foot.

17) In the accompanying diagram, angle R is an obtuse angle, not a right angle.

Find the length of PQ to the nearest foot.

18)

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Lesson #69 – Forces and Vectors A2.A.74 Determine the area of a triangle or a parallelogram, given the measure of two sides


A2.A.73 Solve for an unknown side or angle, using the Law of Sines or the Law of Cosines

Use that information to find the resulting force to the nearest tenth.

Find the angle between the larger original force and the resultant to the nearest degree.

The forces picture always looks the same!

Think of the parallelogram as two congruent triangles.

You often have to work with the supplementary angle,

not the one you are given.

The resultant is always closer to the larger force.

In other words, there is a smaller angle between them.

Imagine you have an aerial view of a situation.

Two people are pushing on a heavy object in different directions

represented by the circled x.

Each one is exerting a certain amount of force. The first person is

pushing with a force of 25 pounds while the second person is

pushing with a force of 30 pounds.

The angle between the two forces they are exerting is 60°. In

what direction will the object end up moving if they are both

pushing at the same time?

What is the result of their combined forces?

25

pounds

30 pounds

60°

If we form a parallelogram with the two given forces, the

resultant force will be the diagonal from the object to the

other opposite corner of the parallelogram.

Label everything else you know about the parallelogram.

25

pounds

30 pounds

60°

Resultant Smaller

Force

Larger Force

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Set up a diagram and a method for solving the following problems.

1) Two forces of 33 newtons and 80 newtons act on an object with a resultant of 70

newtons. Find to the nearest degree, the angle between two applied forces.

2) If you completely solved the last question, the angle between the two forces is 119°.

Using the same information, find the angle between the resultant and the larger applied

force to the nearest degree.

3) Two forces act on a body so that the resultant is a force of 46 pounds. If the angles

between the resultant and the forces are 20 degrees and 46 degrees, find the magnitude

of the larger applied force to the nearest pound.

4) Two forces act on an object. The first force has a magnitude of 63 pounds and makes an

angle of 35 degrees with the resultant. The magnitude of the resultant is 80 pounds.

Find the magnitude of the second applied force to the nearest tenth of a pound.

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Solve the following problems.

5) If forces of 47 pounds and 52 pounds act on object such that the angle between them is

70 , what is the resultant force to the nearest pound?

6) Two forces of 42 newtons and 57 newtons act on an object with a resultant of 70

newtons.

a. Find to the nearest degree, the angle between two applied forces.

b. Next, find the angle between the resultant and the larger force to the

nearest degree.

7) Two forces act on a body so that the resultant has a force of 135 newtons. If the angles

between the resultant and each of the forces are 72 degrees and 12 degrees, find the

magnitude of the larger applied force to the nearest tenth of a newton.

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8) Two forces act on an object. The first force has a magnitude of 75 pounds and makes an

angle of 34 degrees with the resultant. The magnitude of the resultant is 110 pounds.

c. Find the magnitude of the second applied force to the nearest tenth of a pound.

d. To the nearest tenth of a degree, find the angle the second force makes with the

resultant.

9) Two forces of 80 pounds and 100 pounds act on object such that the angle between them

is 105 .

e. What is the resultant force to the nearest pound?

f. What is the angle between the resultant force and the smaller force to the

nearest minute?

lesson #64 first degree trigonometric equations · lesson #52 - inverse trigonometric functions...

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