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4.3
Right Triangle Trigonometry
Objectives
 Evaluate trigonometric functions of acute
angles, and use a calculator to evaluate
trigonometric functions.
 Use the fundamental trigonometric identities.
 Use trigonometric functions to model and solve
real-life problems.
2
The Six Trigonometric Functions
3
The Six Trigonometric Functions
This section introduces the trigonometric functions from a
right triangle perspective.
Consider a right triangle with one acute angle labeled  , as
shown below. Relative to the angle , the three sides of the
triangle are the hypotenuse, the opposite side (the side
opposite the angle  ), and the adjacent side (the side
adjacent to the angle  ).
4
The Six Trigonometric Functions
Using the lengths of these three sides, you can form six
ratios that define the six trigonometric functions of the
acute angle .
Function
sine
cosine
tangent
⟹
⟹
⟹
⟹
Reciprocal
cosecant
secant
cotangent
5
The Six Trigonometric Functions
In the following definitions, it is important to see that
0 <  < 90 ( lies in the first quadrant) and that for such
angles the value of each trigonometric function is positive.
6
The Six Trigonometric Functions
The six trigonometric functions of a right triangle, with an acute angle
, are defined by ratios of two sides of the triangle.
hyp
The sides of the right triangle are:
 the side opposite the acute angle ,
 the side adjacent to the acute angle ,
 and the hypotenuse of the right triangle.
opp
θ
adj
The trigonometric functions are
sine, cosine, tangent, cotangent, secant, and cosecant.
sin  =
opp
hyp
cos  = adj
hyp
tan  =
opp
adj
csc  =
hyp
opp
sec  = hyp
adj
cot  =
adj
opp
7
Example – Evaluating Trigonometric Functions
Use the triangle in Figure 1.20 to find the values of the six
trigonometric functions of .
Solution:
By the Pythagorean Theorem,
(hyp)2 = (opp)2 + (adj)2 it follows that
Figure 1.20
8
Example – Solution
cont’d
So, the six trigonometric functions of  are
5
9
Example – Solution
cont’d
5
10
Your Turn:
Calculate the
trigonometric functions
for  .
10
6

8
The six trig ratios are
6 3
sin  = 
10 5
6 3
tan  = 
8 4
10 5

sec  =
8 4
8 4
cos  = 
10 5
8 4
cot  = 
6 3
10 5

csc  =
6 3
11
The Six Trigonometric Functions
In the Example, you were given the lengths of two sides of
the right triangle, but not the angle .
Often, you will be asked to find the trigonometric functions
of a given acute angle . To do this, construct a right
triangle having  as one of its angles.
12
The Six Trigonometric Functions
In the box, note that sin 30 = = cos 60. This occurs because 30
and 60 are complementary angles.
In general, it can be shown from the right triangle definitions that
cofunctions of complementary angles are equal. That is, if  is an acute
angle, then the following relationships are true.
sin(90 –  ) = cos 
cos(90 –  ) = sin 
tan(90 –  ) = cot 
cot(90 –  ) = tan 
sec(90 –  ) = csc 
csc(90 –  ) = sec 
13
Geometry of the 45-45-90 Triangle
Consider an isosceles right triangle with two
sides of length 1.
45
2
1
12  12  2
45
1
The Pythagorean Theorem implies that the
hypotenuse is of length 2 .
14
Geometry of the 45-45-90 Triangle
Calculate the
trigonometric
functions for a 45
angle.
2
1
45
1
opp
1
sin 45 =
=
=
hyp
2
2
2
adj 1
cot 45 =
= = 1
opp 1
opp 1
tan 45 =
= = 1
1
adj
sec 45 =
hyp
=
adj
2
=
1
1
2
adj
cos 45 =
=
=
2
2
hyp
2
csc 45 =
2
hyp
=
= 2
opp
1
15
Geometry of the 30-60-90 Triangle
Consider an equilateral triangle with
each side of length 2.
The three sides are equal, so the
angles are equal; each is 60.
The perpendicular bisector
of the base bisects the
opposite angle.
30○ 30○
2
2
3
60○
60○
1
2
1
Use the Pythagorean Theorem to
find the length of the altitude, 3 .
16
Geometry of the 30-60-90 Triangle
Calculate the
trigonometric
functions for a 30
angle.
2
1
30
3
opp 1
sin 30 =
=
2
hyp
3
adj
cos 30 =
=
2
hyp
3
1
opp
tan 30 =
=
=
3
3
adj
3
adj
cot 30 =
=
= 3
1
opp
2
2 3
hyp
sec 30 =
=
=
3
3
adj
2
hyp
csc 30 =
=
= 2
opp
1
17
Geometry of the 30-60-90 Triangle
Calculate the
trigonometric
functions for a 60
angle.
2
60○
opp
3
sin 60 =
=
hyp
2
tan 60 =
3
1
3
opp
=
= 3
1
adj
2
hyp
sec 60 =
=
= 2
1
adj
1
adj
cos 60 =
=
2
hyp
3
1
cot 60 = adj =
=
opp
3
3
csc 60 =
2
2 3
hyp
=
=
opp
3
3
18
Trigonometric Identities
19
Trigonometric Identities
Remember an identity is an equation
that is true for all defined values of a
variable.
We are going to use the identities that we have already established to "prove" or
establish other identities. Let's summarize the basic identities we have.
20
Trigonometric Identities
In trigonometry, a great deal of time is spent studying
relationships between trigonometric functions (identities).
21
Trigonometric Identities
Note that sin2  represents (sin  )2,
cos2  represents (cos  )2, and so on.
22
Example – Applying Trigonometric Identities
Let  be an acute angle such that sin  = 0.6. Find the
values of (a) cos  and (b) tan  using trigonometric
identities.
Solution:
a. To find the value of cos , use the Pythagorean identity
sin2  + cos2  = 1.
So, you have
(0.6)2 + cos2  = 1
Substitute 0.6 for sin .
cos2  = 1 – (0.6)2
Subtract (0.6)2 from each side.
cos2  = 0.64
Simplify.
23
Example 5 – Solution
cos  =
Extract positive square root.
cos  = 0.8.
Simplify.
cont’d
b. Now, knowing the sine and cosine of , you can find the
tangent of  to be
= 0.75.
Use the definitions of cos  and
tan  and a right triangle to
check these results.
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Your Turn:
Let  be an acute angle such that sec  = 3. Find the
values of (a) tan  and (b) sin  using trigonometric
identities.
1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃
1 + 𝑡𝑎𝑛2 𝜃 = 9
sec 𝜃 = 3, ∴ cos 𝜃 =
tan 𝜃 =
1
3
sin 𝜃
cos 𝜃
𝑡𝑎𝑛2 𝜃 = 9 − 1
𝑡𝑎𝑛2 𝜃
=8
sin 𝜃
=2 2
1
3
tan 𝜃 = 8 = 2 2
tan 𝜃 = 2 2
sin 𝜃 =
2 2
3
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Verifying or Proving Identities
1.
2.
3.
4.
Learn the fundamental identities.
Try to rewrite the more complicated side of the equation so that it is
identical to the simpler side.
It is often helpful to express all functions in terms of sine and cosine
and then simplify the result.
Usually, any factoring or indicated algebraic operations should be
performed. For example,
sin   2 sin   1  (sin   1) , and
2
5.
6.
2
1
sin 
 cos1 
cos  sin 
sin  cos 
.
As you select substitutions, keep in mind the side you are not
changing, because it represents your goal.
If an expression contains 1 + sin x, multiplying both numerator and
denominator by 1 – sin x would give 1 – sin² x, which could be
replaced with cos² x.
26
Prove the following identity:
Let's sub in here using reciprocal identity
sin  cosec  cos   sin 
2
2
sin  cosec  cos   sin 
 1 
2
2
sin  
  cos   sin 
 sin  
2
We are done!
We've shown the
LHS equals the
RHS
2
1  cos   sin 
2
2
sin   sin 
2
2
We often use the Pythagorean Identities solved for either sin2 or cos2.
sin2 + cos2 = 1 solved for sin2 is sin2 = 1 - cos2 which is our left-hand side
so we can substitute.
In proving an identity you should NOT move things from
one side of the equal sign to the other. Instead substitute
using identities you know and simplifying on one side or
the other side or both until both sides match.
sin

Prove the following identity:
cosec  cot  
Let's sub in here using reciprocal identity and quotient identity
1  cos 
sin 
We worked on cosec  cot  
1  cos 
LHS and then
RHS but never
1
cos 
sin 


moved things
sin  1  cos  FOIL denominator
across the = sign sin 
1  cos   sin
sin  1  cos  
combine fractions



sin 
cos  1  cos  
11cos
Another trick if the denominator
1  cos  sin  1  cos  
is two terms with one term a 1

and the other a sine or cosine,
2
sin 
1  cos 
multiply top and bottom of the
fraction by the conjugate and
1  cos  sin  1  cos  
then you'll be able to use the

2
Pythagorean Identity on the
sin

sin

bottom
1  cos  1  cos 

sin 
sin 
Your Turn: Verifying an Identity
Example
Verify that the following equation is an identity.
cot x + 1 = csc x(cos x + sin x)
Analytic Solution Since the side on the right is more
complicated, we work with it.
cot x 1  csc x(cos x  sin x) Original identity
1
1

(cos x  sin x)
csc
x

sin x
sin x
cos x sin x
Distributive property


sin x sin x
cot x  cossin xx ,1  sinsin xx
 cot x 1
The given equation is an identity
because the left side equals the right side.
Your Turn: Verifying an Identity
Example
Solution
Verify that the following equation is an
identity.
tan t cot t
2
2
 sec t  csc t
sin t cos t
tan t cot t
tan t
cot t


sin t cos t
sin t cos t sin t cos t
1
1
 tan t 
 cot t 
sin t cos t
sin t cos t
sin t
1
cos t
1




cos t sin t cos t sin t sin t cos t
1
1


2
cos t sin 2 t
 sec t  csc t
2
2
Hints for Establishing Identities
• Get common denominators.
• If you have squared functions look for Pythagorean
Identities.
• Work on the more complex side first.
• If you have a denominator of 1 + trig function try
multiplying top & bottom by conjugate and use
Pythagorean Identity.
• When all else fails write everything in terms of sines and
cosines using reciprocal and quotient identities.
• Have fun with these---it's like a puzzle, can you use
identities and algebra to get them to match!
31
Applications Involving Right Triangles
32
Degrees, Minutes and Seconds
If the angle is not exactly to the next degree it can be expressed as a
decimal (most common in math) or in degrees, minutes and seconds
(common in surveying and some navigation).
1 degree = 60 minutes
 = 25°48'30"
degrees
1 minute = 60 seconds
seconds
minutes
To convert to decimal form use conversion fractions. These
are fractions where the numerator = denominator but two
different units. Put unit on top you want to convert to and
put unit on bottom you want to get rid of.
Let's convert the
seconds to minutes
30"  1'
= 0.5'
60"
33
1 degree = 60 minutes
1 minute = 60 seconds
 = 25°48'30" = 25°48.5'
= 25.808°
Now let's use another conversion fraction to get rid of minutes.
48.5'  1
60'
= .808°
Your Turn: Convert to DMS or Decimal
5°40’ 12”
5.67 ̊
16.35°
16 ̊ 21’
16°18’
16.3 ̊
73.56 ̊
73 ̊ 33’ 36”
35
Applications Involving Right Triangles
Many applications of trigonometry involve a process called
solving right triangles.
In this type of application, you are usually given one side of
a right triangle and one of the acute angles and are asked
to find one of the other sides, or you are given two sides
and are asked to find one of the acute angles.
36
Applications Involving Right Triangles
In some applications, the angle you are
given is the angle of elevation, which
represents the angle from the horizontal
upward to an object.
In other applications you may be given the
angle of depression, which represents the
angle from the horizontal downward to an
object.
37
Example – Using Trigonometry to Solve a Right Triangle
A surveyor is standing 115 feet from the base of the
Washington Monument, as shown in figure below. The
surveyor measures the angle of elevation to the top of the
monument as 78.3. How tall is the Washington
Monument?
38
Example – Solution
From the figure, you can see that
where x = 115 and y is the height of the monument. So, the
height of the Washington Monument is
y = x tan 78.3
 115(4.82882)
 555 feet.
39
Example:
A person is 200 yards from a river. Rather than walk directly
to the river, the person walks along a straight path to the
river’s edge at a 60° angle. How far must the person walk to
reach the river’s edge?
cos 60°
x (cos 60°) = 200
200
60°
x
x
X = 400 yards
40
Your Turn:
A six-foot person standing 20 feet from a
streetlight casts a 10-foot shadow. What
is the height of the streetlight?
Height = 18 ft
41
Your Turn:
A ramp 20 feet in length rises to a loading platform that is
3 ⅓ feet off the ground.
– Draw a right triangle that gives a visual
representation of the problem. Show the known
quantities of the triangle and use a variable to
indicate the angle of elevation of the ramp.
– Use a trigonometric function to write an equation
involving the unknown quantity.
– What is the angle of elevation of the ramp?
42
Solution:
3
1
3
𝜃
20
1
33
tan 𝜃 =
20
𝜃 = 9.46 ̊
43
Your Turn:
In traveling across flat land, you notice a mountain
directly in from of you. Its angle of elevation (to the
peak) is 3.5°. After you drive 13 miles closer to the
mountain, the angle of elevation is 9°.
Approximate the height of the mountain.
44
Solution:
tan 3.5 =
h
3.5 ̊
13
ℎ
tan 9 =
𝑥
𝑥=
9̊
x
ℎ
ℎ − 13 tan 3.5
=
tan 9
tan 3.5
ℎ − 13 tan 3.5
tan 3.5
ℎ
𝑥=
tan 9
ℎ(tan 3.5 − tan 9) = − 13 tan 3.5 tan 9
ℎ tan 3.5 = ℎ tan 9 − 13 tan 3.5 tan 9
ℎ=
ℎ tan 3.5 − ℎ tan 9 = − 13 tan 3.5 tan 9
ℎ
13 + 𝑥
− 13tan 3.5 tan 9
tan 3.5 − tan 9
h = 1.3 miles
45
Assignment
Pg. 284 – 287: #1 – 55 odd, 63 – 81 odd
46