Lesson 7-4
Right Triangle
Trigonometry
Modified by Lisa Palen
(Reference Angle instead of Angle of Perspective)
With slides by Mr. Jerrell Walker, Lincoln, Nebraska
and Emily Freeman, Powder Springs, Georgia.
1
Anatomy of a Right Triangle
hypotenuse
opposite
hypotenuse

adjacent
adjacent
reference
angle
opposite
If the reference angle is
A
then
A
Hyp
C
Adj
B
then
A
Opp
Opp
Hyp
C
B
Adj
C
AC  Hyp
AC  Hyp
BC  Opp
AB  Opp
AB  Adj
BC  Adj
Lesson 7-4 Right Triangle
Trigonometry
3
13.4 & 13.5
The Trigonometric Functions
we will be looking at
SINE
COSINE
TANGENT
The Trigonometric Functions
SINE
COSINE
TANGENT
SINE
Prounounced
“sign”
COSINE
Prounounced
“co-sign”
TANGENT
Prounounced
“tan-gent”
Greek Letter 
Prounounced
“theta”
Represents an unknown angle
Sometimes called the reference angle or
angle of perspective
Definitions of Trig Ratios
Opp
Sin 
Hyp
Adj
Cos 
Hyp
Opp
Tan 
Adj
hypotenuse
hypotenuse

adjacent
opposite
We need a way
to remember
all of these
ratios…
SOHCAHTOA
Old Hippie
Sin
Opp
Hyp
Cos
Adj
Hyp
Tan
Opp
Adj
Some
Old
Hippie
Came
A
Hoppin’
Through
Our
Old Hippie Apartment
Finding sin, cos, and tan
SOHCAHTOA
4
opp 8


sin  
hyp 10 5
adj
6 3
cos  

5
hyp 10
8
opp
4

tan  

6
adj
3
10
8

6
Find the sine, the cosine, and the tangent of angle A.
Give a fraction and decimal answer (round to 4 places).
10.8
9
A
9
opp

sin A 
10.8  .8333
hyp
adj
6
cos A 

hyp
10.8
 .5555
6
opp
tan A 
adj
9

6
 1.5
Find the values of the three trigonometric functions of .
?
5
4

Pythagorean Theorem:
(3)² + (4)² = c²
5=c
3
opp 4
adj 3
opp 4




sin  
cos  
tan

hyp 5
hyp 5
adj 3
Find the sine, the cosine, and the tangent of angle A
B
Give a fraction and
decimal answer (round
to 4 decimal places).
24.5
8.2
A
23.1
opp  8.2
sin A 
 .3347
24
.
5
hyp
adj
cos A 
hyp
23.1

24.5  .9429
opp
tan A 
adj
8 .2

23.1  .3550
Finding a side
Example: Find the value of x.
Step 1: Identify the “reference angle”.
Step 2: Label the sides (Hyp / Opp / Adj).
Step 3: Select a trigonometry ratio (sin/ cos / tan).
Opp
Sin  = Hyp
Step 4: Substitute the values into the equation.
x
Sin 25 =
A
opp
Hyp
12 cm
x
25
B
12
reference
angle
C
Adj
Step 5: Solve the equation.
sin 25
1
=
x
12
x = 12 sin 25
x = 12 (.4226)
x = 5.07 cm
Lesson 7-4 Right Triangle
Trigonometry
20
Solving Trigonometric Equations
There are only three possibilities for the placement of the variable ‘x”.

sin
A
Opp
=
x
sin  =
A
x
12 cm
25
B
sin 25
=
=
1
x =
C
12
sin 25
x = 28.4 cm
sin 25
sin 25 =
1
25 cm
12 cm
25
B
12
x
12
x
A
12 cm
x
sin 25
x
Hyp
Opp
Hyp
Sin X =
=
x
12
x
12
x = (12) (sin 25)
x = 5.04 cm
x
B
C
sin x
x = sin
=
C
12
25
1
(12/25)
x = 28.7
Note you are looking
21
for an angle here!
Ex.
A surveyor is standing 50 feet from the base of a
large tree. The surveyor measures the angle of
elevation to the top of the tree as 71.5°. How tall
is the tree?
Opp
tan 71.5°
Hyp
y?
71.5°
50
Let y represent the height of
the tree.
Is y opp, adj or hyp?
Is 50 opp, adj or hyp?
y
tan 71.5° 
50
y = 50 (tan 71.5°)
y = 50 (2.98868)
y  149.4 ft
Ex. 5
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
Angle of Elevation and Depression
Example #1
Angle of Elevation and Depression
Suppose the angle of depression from a lighthouse
to a sailboat is 5.7o. If the lighthouse is 150 ft tall,
how far away is the sailboat?

5.7o
150 ft.
5.7o

x
Construct a triangle and label the known parts.
Use a variable for the unknown value.
Angle of Elevation and Depression
Suppose the angle of depression from a lighthouse
to a sailboat is 5.7o. If the lighthouse is 150 ft tall,
how far away is the sailboat?
5.7o
150 ft.
5.7o
x
Set up an equation and solve.
Angle of Elevation and Depression
tan(5.7 o ) 
150
x
x tan(5.7o )  150
x
150
tan(5.7o )
Remember to use
degree mode!
x is approximately 1,503 ft.
150 ft.
5.7o
x
Angle of Elevation and Depression
Example #2
Angle of Elevation and Depression
A spire sits on top of the top floor of a
building. From a point 500 ft. from the base
of a building, the angle of elevation to the
top floor of the building is 35o. The angle of
elevation to the top of the spire is 38o. How
tall is the spire?
Construct the required
triangles and label.
38o 35o
500 ft.
Angle of Elevation and Depression
Write an equation and solve.
Total height (t) = building height (b) + spire height (s)
Solve for the spire height.
s
t
Total Height
t
tan(38 ) 
500
o
b
500 tan(38 )  t
o
38o 35o
500 ft.
Angle of Elevation and Depression
Write an equation and solve.
Building Height
b
o
tan(35 ) 
500
s
500 tan(35o )  b
t
b
38o 35o
500 ft.
Angle of Elevation and Depression
Write an equation and solve.
Total height (t) = building height (b) + spire height (s)
500 tan(38o )  t
500 tan(35o )  b
s
500 tan(38 )  500 tan(35 )  s
500 tan(38o )  500 tan(35o )  s
o
The height of the
spire is
approximately 41
feet.
o
t
b
38o 35o
500 ft.
Angle of Elevation and Depression
Example #3
Angle of Elevation and Depression
A hiker measures the angle of elevation to a mountain
peak in the distance at 28o. Moving 1,500 ft closer on a
level surface, the angle of elevation is measured to be
29o. How much higher is the mountain peak than the
hiker?
Construct a diagram and label.
1st measurement
28o.
2nd measurement 1,500 ft closer is 29o.
Angle of Elevation and Depression
Adding labels to the diagram, we need to find h.
h ft
28o
1500 ft
29o
x ft
Write an equation for each triangle. Remember, we can
only solve right triangles. The base of the triangle with
an angle of 28o is 1500 + x.
h
tan 28 
1500  x
o
h
tan 29 
x
o
Angle of Elevation and Depression
Now we have two equations with two variables.
Solve by substitution.
h
tan 28 
1500  x
o
h
tan 29 
x
o
Solve each equation for h.
(1500  x) tan(28o )  h
x tan(29o )  h
Substitute.
(1500  x) tan(28o )  x tan(29o )
Angle of Elevation and Depression
(1500  x) tan(28o )  x tan(29o )
Solve for x. Distribute.
1500 tan(28o )  x tan(28o )  x tan(29o )
Get the x’s on one side and factor out the x.
1500 tan(28o )  x tan(29o )  x tan(28o )
1500 tan(28o )  x  tan(29o )  tan(28o ) 
Divide.
x = 35,291 ft.
1500 tan(28o )
x
o
o
 tan(29 )  tan(28 ) 
Angle of Elevation and Depression
x = 35,291 ft.
However, we were to find the height of the mountain.
Use one of the equations solved for “h” to solve for the
height.
x tan(29o )  h
(35, 291) tan(29o )  19,562
The height of the mountain above the hiker is 19,562 ft.
Ex.
A surveyor is standing 50 feet from the base of a
large tree. The surveyor measures the angle of
elevation to the top of the tree as 71.5°. How tall
is the tree?
Opp
tan 71.5°
Hyp
?
71.5°
50
y
tan 71.5° 
50
y = 50 (tan 71.5°)
y = 50 (2.98868)
y  149.4 ft
Ex. A person is 200 yards from a river. Rather than walk
directly to the river, the person walks along a straight
5
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