Lesson #61 – 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 sin390°.
Find sin(-330°).

Why do all of these angles have the same sine value?

Solve the equation, sin( x) 
1
. Is your initial answer the only solution?
2

Solve the equation, cos( x) 
1
. Find all solutions between 0° and 360°.
2

1
Solve the equation, sin( x)   . Find all solutions between 0° and 360°.
2

1
Solve the equation, cos( x)   . Find all solutions between 0° and 360°.
2
Find sin150°.
Find sin510°.
Find sin(-210°).
~1~
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.
Ex) Solve for x. Round to the nearest
1. Isolate the trigonometric part of the equation
degree. 3cos x  6  8
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°.
Example: Solve for  in the interval 0    360 to the nearest degree.
8cos   2  5  15cos 
1) Solve for x on the interval,
0  x  2
.
sin x  2   sin x
3)
Solve for x on the interval,
0  x  360 to the
2)
Solve for x on the interval,
nearest degree.
0  x  360 to the
Solve for x on the interval,
0  x  360 to the
2 tan x  1  0
4)
~2~
nearest degree.
nearest degree.
3(sin x  5)  4
tan x  2  2
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
The Law of Cosines
a
b
a 2  b2  c2  2bc cos A

sin A sin B
Solve the following equations for x on the interval, 0°<x<180°. Round to the nearest degree.
5
9

sin 30 sin x
10  4  7  2(4)(7) cos x 5   3.2   (6)  2(3.2)(6) cos x
2
2
2
2
2
2
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?
~3~
Lesson #62 - 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.
y  sin 1 ( x)
Graph y=sin(x) using the unit circle and the table below over the interval
x
2  x  2
y
When does sin(x)=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: _________.
~4~
This is called a restricted domain. When you use the the inverse sine function, sin 1 ( x) ,
the calculator will always give you an answer between -90° and 90°, inclusive. In radians
this would be:
Restricted Domain for Sine so that
the inverse will be a function.
Take the points in your table that are in this interval and switch the x’s and y’s. (Use
3

the y-value at
for  since the two angles are coterminal.) Graph y  sin 1 ( x) on your
2
2
calculator, and sketch it on the graph provided. It should match up with the values in the
table.
Note: You must be in
radian mode when
x
y
graphing inverse trig.
functions.
y  cos 1 ( x)
Compare this graph with the
piece of y=sin(x) you circled
on the previous page.
Graph y=cos(x) using the unit circle and the table below over the interval
x
2  x  2
y
When does cos(x)=0?
Convert these values to degrees.
Are these the only places where the cosine 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.
~5~
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,
cos1 ( x) , the calculator will give you an answer between 0° and 180°, inclusive. In radians
this would be:
Restricted Domain for Cosine so
that the inverse will be a function.
Graph y  cos1 ( x) on your calculator and sketch it on the graph provided.
x
y
Compare this
graph with the
piece of y=cos(x)
you circled on the
previous page.
y  tan 1 ( 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?
Convert these values to degrees.
Are these the only places where the
tangent 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.
~6~
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 tangent function,
tan 1 ( x) , the calculator will give you an answer between -90° and 90°, exclusive. In
radians this would be:
Restricted Domain for Tangent so
that the inverse will be a function.
Graph y  tan 1 ( x) , and sketch it on the graph provided.
Compare this
graph with the
piece of y=tan(x)
you circled on the
previous page.
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. y  sin 1 x (also known as y  Arc sin x )
b. y  cos 1 x (also known as y  A rccosx )
c. y  tan 1 x (also known as y  Arc tan 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.
~7~
1. What is the principal value of
?
1)
3)
2)
4)
2 The value of
1) 0
2)
9 What is the principal value of
degrees and radians.
, in
10 What is the smallest positive value of x, in radians,
is
3)
4)
that satisfies
11 Find the value of
3 The value of
?
, in
is
1) 120°
2) 105°
degrees.
3) 90°
4) 75°
4 If
12 If
, what is the value of angle A to
the nearest minute?
1)
3)
2)
4)
, then x is equal to
1)
2)
3)
4)
13 If
, find the value of positive acute
angle A to the nearest minute.
5 If
and
, the measure of angle x
is
1) 45º
2) 135º
3) 225º
4) 315º
14 If
, find the value of positive acute
angle x to the nearest minute.
6 What is the value of x in the equation
1)
3)
2)
4)
7 If
?
15 If
angle
, what is the measure of angle
, in
16
degrees?
8 What is the principal value of
1)
2)
?
3)
4)
~8~
If
angle
, find the value of positive acute
to the nearest minute.
, find the measure of positive acute
to the nearest minute.
Lesson #63- 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.
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, ____.
mABC = 93
B
a
c
C
b
mBCA = 51
A
mBAC = 36
In what types of triangles can you use the Pythagorean
Theorem and SOH-CAH-TOA?
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:
a
b
c


sin A sin B sin C
1. Given: a=12, mA  25 , mC  58 . Find side c.
Connection to Proofs:
Use when given:
ASA or AAS
“Trick”
CIRCLE The PAIRS
You must have 1 Angle/Side
Pair where you know the
values.
2. Example: Solve for x.
22
B
43
C
70
x
A
~9~
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.
7
103
41
x
Law of Cosines:
a  b  c  2bc cos A
2
2
2
4. Given: c=12, b=15, mA  84 . Find a.
Connection to Proofs:
Use when given:
SSS or SAS
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.
5. Given: a=10, b=15, and c=20, find mA .
~ 10 ~
The letters are less important in the formula than the actual placement of the sides.
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.
36
5
7
x
7. Find the measure of angle B.
C
11
15
B
12
A
8. In triangle RST, what is the value of r in terms of R, T, and t?
1)
3)
2)
4)
~ 11 ~
Lesson #64 –The Ambiguous Case & the Donkey Theorem (SSA)
A2.A.75 Determine the solution(s) from the SSA situation (ambiguous case)
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 x  .5678
2.
cos x  .5678
3.
cos x  .5678
When working with triangles, what is the only trigonometric function that can give us
two answers for the angle?
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 the last lesson 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.
~ 12 ~
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.
No triangles
One triangle
Two triangles
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.
~ 13 ~
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
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?
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, two possible values for angle B, to the nearest degree are
a. 42 and 132
b. 42 and 138
c. 42 and 108
d. 41 and 139
5. In triangle ABC, if A=30, a=5, and b=10, triangle ABC is
a. acute
b. obtuse
c. acute or obtuse
d. right
~ 14 ~
Lesson #65 – 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.
~ 15 ~
Area of a Triangle
There is one more trigonometric formula that you will be given to you on the regents. You
1
already know that the area of a triangle can be calculated with the formula 𝐴 = 2 𝑏ℎ. 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:
K
1
ab sin C
2
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 mC  100 . Round to the nearest
square unit.
Area of a Triangle
You need a known “corner”
(SAS).
2. Find the area of a triangle where: b=20, c=30, and mA  34 . Round to the nearest
square unit.
~ 16 ~
3. A ranch in the Australian Outback is shaped like triangle ACE, with
and
miles. Find the area of the ranch, to the nearest square mile.
,
,
4. 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.
7
104
9
~ 17 ~
Lesson #66 – Using Trig. Apps. in Word Problems
A2.A.74 Determine the area of a triangle or a parallelogram, given the measure of two sides
and the included angle
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
Parallelograms
 2 sides congruent (called the legs)
 Opposite Sides Parallel
 Base angles congruent
 Opposite Sides Congruent
 Opposite Angles Equal
A
 Adjacent Angles Supplementary
 Can be cut into two congruent triangles
10
12
69
C
B
7
67
If you are given a ASA, SAS, SSS, or AAS pattern on a triangle, you can find all of the sides
and angles of the triangle. You might not be able to find the side or angle you want with one
step. The key is to solve for other sides/angles until you have enough information.
Ex) Find the measure of the angles of the triangle to the nearest degree (assume all angles
are acute).
For this triangle, you cannot find either of the angles first, but you
can find the other side.
20
75◦
15
Now you have a side-angle pair allowing you to use the law of sines
to find either of the unknown angles. Do not round the side you just
found; use the whole decimal.
~ 18 ~
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).
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)
~ 19 ~
4) A boat starts at point A and finds the angle of elevation to the lighthouse measures 28 .
After traveling 3.2 miles towards the light house, the angle of elevation now measures 36 .
Determine the distance, d, from the boat to the lighthouse to the nearest thousandth of a
mile. Convert your answer to the nearest hundred
feet. (There are 5280 feet in a mile).
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.
~ 20 ~
Lesson #67 – Forces and Vectors
A2.A.74 Determine the area of a triangle or a parallelogram, given the measure of two sides
and the included angle
A2.A.73 Solve for an unknown side or angle, using the Law of Sines or the Law of Cosines
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?
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
60°
30 pounds
25
pounds
60°
30 pounds
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.
~ 21 ~
Smaller
Force
Resultant
Larger Force
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.
~ 22 ~