Download Right Triangle Trigonometry: Solving Triangles Davenport University

Right Triangle Trigonometry: Solving Triangles
Davenport University
There are three trigonometric (“trig” for short) functions that we will be focusing on in the
special case of right triangles. It is important to remember that the definitions given below
are only valid when the angle and measurements are taken from a right triangle. What
should you do if you have a non-right triangle? Learn more about trig, but don’t worry about
that for now because we will be working exclusively with right triangles.
Looking at the right triangle below, we see that we have three sides of lengths x, y, and r.
The side across from the right angle, with length r, is always the hypotenuse and the other
two sides, with lengths x and y, are the legs.
Now we will choose one of the non-right angles (Call it , this is the Greek letter “theta”).
Once we do this, we can now label the legs as opposite and adjacent. Fortunately, the nonmath definitions of opposite and adjacent should help us with labeling the legs.
Notice that the side with length y is on the “opposite” side of the triangle from  and the
side with length x is adjacent to .
Opposite (y)
Hypotenuse (r)

Adjacent (x)
Now that everything is labeled, we can define our trig functions:
Sin ( ) 
opposite
y

hypotenuse r
Cos( ) 
adjacent
x

hypotenuse r
Tan( ) 
opposite
y

adjacent x
Read as “Sine of theta”
Read as “Cosine of theta”
Read as “Tangent of theta”
The value of these functions can be computed by carrying out the division of the lengths of
the appropriate sides.
Also, do not be intimidated by the use of Greek letters. We use Greek letters to represent
the measure of angles so they are not confused with the variables being used to represent
the length of the sides (Honestly, we are trying to make things easier for you to follow).
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Right Triangle Trigonometry: Solving Triangles
Davenport University
Before we move on, we should note that “Opposite” and “Adjacent” are relative to the angle
chosen. So if we choose the other angle (Call it , the Greek letter “alpha”) instead of 
(theta), then the sides of the triangles are now labeled as shown below:

Adjacent (y)
Hypotenuse (r)
Sin ( ) 
opposite
x

hypotenuse r
Cos( ) 
adjacent
y

hypotenuse r
Tan( ) 
opposite x

adjacent y
Opposite (x)
As you can see, the trig functions remain consistent (i.e., sine is still opposite over
hypotenuse). So it really doesn’t matter which of the two angles you choose, but once you
choose your angle, you need to be consistent with your labeling.
Example #1: Compute the sine, cosine and tangent of .
13
5

12
In this case, the leg of length 5 is the opposite side, the leg of length 12 is the adjacent
side, and the side of length 13 is the hypotenuse. So…
opposite
5

hypotenuse 13
adjacent
12
Cos() 

hypotenuse 13
opposite
5
Tan() 

adjacent 12
Sin () 
OK, that looks too easy, so I bet you are wondering where we’re going with this. You are
going to use these functions to help find unknown values in a given right triangle.
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Right Triangle Trigonometry: Solving Triangles
Davenport University
You are probably asking yourself, “Can’t I just use the Pythagorean Theorem?” The answer
to that is yes, but only when the question gives you sufficient information; otherwise, you
will have no choice but to use trig functions. Remember, the Pythagorean Theorem only
deals with the sides of the right triangle; it neither gives you nor uses information about the
other two angles.
Now you are probably freaking out because there are three trig functions and you are
wondering if you will have to use all three every time, or, if you only need one of the three,
how will you know which one to use. Remain calm and keep in mind what we always say
about solving for an unknown: we need a single equation with a single variable. Therefore,
your choice of a trig function is guided by setting up an equation with a single unknown.
Example #2: Find the length of z.
10 cm
z
30
Solution: First, note that there is not enough information about the three sides to use the
Pythagorean Theorem. Because side z is opposite of our given angle, we will want to use a
trig function that uses the opposite side (either sine or tangent). Also, because we are given
the length of the hypotenuse (and we don’t know about the length of the adjacent side), we
should use the sine function. Hence:
Sin (30) 
z
10
 z 
10Sin (30)   10 
 10 
10  Sin (30)  z
 Now, solve for z
 Multiply both sides by 10 to clear denominators
 Compute the sine of 30 degrees
10  0.5  z
 Multiply
5 z
 The length of z is 5 cm.
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Right Triangle Trigonometry: Solving Triangles
Davenport University
Example #3: Find the length of z.
18 cm
Solution: Because side z is adjacent to our angle, we will want to use a
trig function that uses the adjacent side (either cosine or tangent).
Because we are given the length of the hypotenuse (and we don’t
know about the length of the opposite side), we should use the cosine
function. Hence:
60
z
Cos(60) 
z
18
 Now, solve for z
 z 
18Cos(60)   18 
 18 
18  Cos(60)  z
 Multiply both sides by 18 to clear denominators
 Compute the cosine of 60 degrees
18  0.5  z
 Multiply
9z
 The length of z is 9 cm.
Example #4: Find the length of z.
z
30
Solution: Because side z is opposite of our angle, we will
want to use a trig function that uses the opposite side
(either sine or tangent). Because we are given the length
of the adjacent side (and we don’t know about the length
of the hypotenuse), we should use the tangent function.
12 cm
Tan(30) 
z
12
 z 
12Tan(30)   12 
 12 
12  Tan(30)  z
 Now, solve for z
 Multiply both sides by 12 to clear denominators
 Compute the tangent of 30 degrees
12  0.577350...  z
 Multiply
6.928203...  z
 The length of z is about 6.926 cm.
Note: Rounding the final answer to 3 decimal places is fine for “generic” questions. Once we
begin actual application questions, we will follow the rounding rules given on page 602 of
the text: Calculation with Significant Digits. Also, for our initial discussion, we will not worry
about converting angular measures to minutes and seconds.
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Right Triangle Trigonometry: Solving Triangles
Davenport University
Solving Application Problems:
As with any application problem, you should read the problem carefully to identify not only
the question but also any additional information, implicit or explicit, that will guide you to
finding the solution. For any problem that describes a physical situation (such as things that
form right triangles) you will be well served to draw a picture (note that the text does this
on the majority of the exercises). Additionally, there may also be some vocabulary that is
specific to the context. Generally speaking, it is faster and more effective to look up any
unknown words/phrases then it is to continue on with the question by ignoring them.
Once you have identified a question as being a “right triangle” problem (for those of you
that do draw the pictures, your drawing of a right triangle should serve as a fair indicator),
your next step is to take inventory of what value(s) you are looking for and what values you
are given. Your choice of Sine, Cosine, Tangent, their inverses, or the Pythagorean Theorem
as the means of finding the solution are strictly a function of the values that are already
known.
Example #5: A 12.5 foot ladder leaning against a building forms a 60 angle with the
ground. At what height does the ladder touch the building?
Solution: After reading the question carefully, it would make sense to draw a picture. With
the exception of the leaning tower of Pisa, it is a fair assumption that buildings form right
angles to the ground.
Building
Ladder
12.5’
”
60
z
Ground
Now that we have the picture, the problem now looks like our previous three examples. In
this case, we are looking for the length of the side opposite of the given angle (z = the
height, in feet, along the building where the ladder touches) and we’ve been given the
hypotenuse. Therefore, sine is the trig function we should use.
Sin (60) 
z
12.5
 Set up and solve for z
 z 
12.5Sin (60)   12.5

 12.5 
12.5  (0.8660254)  z
10.8253175  z
10.8  z
 Clear denominators and compute the sine of 60 degrees
 Multiply
 Now round and interpret
 The ladder hits the building at about 10.8 feet above the ground.
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Right Triangle Trigonometry: Solving Triangles
Davenport University
Example #6: Fictional Bob is building a shed for his backyard. To comply with the local
zoning ordinance, Fictional Bob will build a roof with a pitch, or slope, of 0.3. What is the
angle of elevation of his roof? (Round answer up to the next whole degree)
Solution: After reading the question carefully, it would make sense to draw a picture as well
as review what is meant by a “pitch of 0.3”. Also, we define  as the angle of elevation (in
degrees) of the shed’s roof.
Pitch = Slope: Hence, the roof has a rise of 3 units for every run of 10 units (0.3 = 3/10).
Because we are not interested in the actual lengths (at least right now), we are able to label
the sides as shown below.
3

10
Now, because we are looking for the angle when given the opposite and adjacent sides, we
should use the tangent function.
Tan() 
3
 0.3
10
 Set up (carry out division)
Tan 1 Tan()  Tan 1 0.3
 Take the inverse tangent of each side to isolate 
  16.699244
  17
 Compute the inverse tangent of 0.3
 Round up to the next whole degree
The angle of elevation for the roof on Fictional Bob’s shed is 17.
Final Note: The text does not explicitly show the “ Tan
1
Tan()  Tan 1 0.3 ” step. This is
where we “get rid of” the tangent. Just like we add to “get rid of” a subtraction, divide to
“get rid of” a multiplication, square to “get rid of” a square root, and so on, we have to undo
whatever is being done to our variable in order to isolate it. Inverse trig functions “get rid
of” their corresponding trigonometric functions (and vice versa too!).
Tan 1 Tan()   
Cos 1 Cos()   
Sin 1 Sin ()   


CosCos ( x)   x
Sin Sin ( x)   x
Tan Tan 1 ( z )  z
1
1
(To make the mathematicians happy: 0    90 ,
0  x  1 , and z  0 )
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