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TRIGONOMETRIC
RATIOS
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”
Opp
Sin 
Hyp
Adj
Cos 
Hyp
hypotenus
e
A
Opp
Tan 
Adj
adjacent
opposite
opposit
WE CAN USE TRIGONOMETRIC
RATIOS TO FIND MISSING SIDES
Sine= sinA =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
Cosine = cosA =
A
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
B
Tangent = tanA =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆
WHERE A IS THE ANGLE MEASUREMENT
C
THE RATIOS CAN BE REMEMBERED
BY THE SAYING
SOH
sinA =
CAH
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
cosA =
hypotenuse
A
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
opposite
adjacent
TOA
tanA =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
What is the sine ratio of A?
What is the cosine ratio of A?
A
10
8
What is the tangent ratio of A?
B
6
C
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
6
9
opp

sin A 
hypo 10.8  .8333
adj
6
cos A 

hypo 10.8
opp
tan A 
adj
9

6
 .5555
 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
hypo
23.1
adj

cos A 
24.5  .9429
hypo
opp
tan A 
adj
8 .2

23.1  .3550
HOW CAN WE USE THESE RATIOS TO
FIND SIDE LENGTHS?
A
What is the length of BC?
Ask yourself the following questions
22º
~What angle measurement do I have?
16
A
~From that angle what am I LOOKING FOR
(adjacent, opposite or hypotenuse)
OPPOSITE
~From the angle what do I HAVE?
(adjacent, opposite or hypotenuse)
HYPOTENUSE
~So which ratio should we use?
SINE
B
C
HOW CAN WE USE THESE RATIOS TO
FIND SIDE LENGTHS?
A
What is the length of BC?
Sin22 =
𝒙
𝟏𝟔
22º
16
16Sin22 =x
5.99=x
B
C
Find the length of ED. Which ratio
should we use? Solve.
D
Find the length of DF. Which ratio
Should we use? Solve.
65º
E
14
F
FINDING AN ANGLE.
(FIGURING OUT WHICH RATIO TO USE AND GETTING
TO USE THE 2ND BUTTON AND ONE OF THE TRIG
BUTTONS.)
EX. 1: FIND . ROUND TO FOUR DECIMAL PLACES.
17.2
tan  
9
nd
2
tan 17.2
17.2

9

  62.3789
Make sure you are in degree mode (not
radian).
9
EX. 2: FIND . ROUND TO THREE DECIMAL PLACES.

23
7
7
cos  
23
nd
2
co 7

s
  72.281
Make sure you are in degree mode (not
radian).
23
EX. 3: FIND . ROUND TO THREE DECIMAL PLACES.

200
nd
2
200
sin  
400
sin
200

  30
Make sure you are in degree mode (not
radian).
400
WHEN WE ARE TRYING TO FIND A SIDE
WE USE SIN, COS, OR TAN.
When we are trying to find an angle
we use sin-1, cos-1, or tan-1.
SOLVE THE PROBLEM:
A ladder is leaning against a house so that the top of the
ladder is 12 feet above the ground. The angle with the
ground is 51º. How far is the base of the ladder from the
house?
12
51º
x
An angle of elevation is the angle formed by a
horizontal line and a line of sight to a point above
the line. In the diagram, 1 is the angle of elevation
from the tower T to the plane P.
An angle of depression is the angle formed by a
horizontal line and a line of sight to a point below
the line. 2 is the angle of depression from the
plane to the tower.
Since horizontal lines are parallel, 1  2 by the
Alternate Interior Angles Theorem. Therefore the
angle of elevation from one point is congruent
to the angle of depression from the other point.
The sun hits the top of a tree which creates a shadow 15 feet
long. If the angle of elevation from the ground to the top of
the tree is 42º, how tall is the tree?
𝑥
tan42= 15
x
x= 15tan42
42º
42º
x= 13.51 ft
15
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
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?
tan
71.5°
?
50
71.5
°
Opp

Hyp
y

50
tan
71.5°
y = 50 (tan 71.5°)
y = 50 (2.98868)
y  149.4 ft
HAPPY LANDING
A plane is flying at an altitude of 1.5km. The
pilot wants to descend into an airport so that the
path of the plane makes an angle of 5° with the
ground. How far from the airport (horizontal
distance) should the descent begin?
1.5km
5°
x
RIVER WIDTH
A surveyor is measuring a
river’s width. He uses a tree
and a big rock that are on the
edge of the river on opposite
sides. After turning through an
angle of 90° at the big rock, he
walks 100 meters away to his
tent. He finds the angle from
his walking path to the tree on
the opposite side to be 25°.
What is the width of the river?
Draw a diagram to describe this
situation. Label the variable(s)
RIVER WIDTH
We are looking at the “opposite”
and the “adjacent” from the given
angle, so we will use tangent
d
tan(25 ) 
100
Multiply by 100 on both sides
100 tan(25 )  d
d  46.63 meters
BUILDING HEIGHT
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.
BUILDING HEIGHT
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.
BUILDING HEIGHT
Write an equation and solve.
Building Height
b
o
tan(35 ) 
500
s
500 tan(35o )  b
t
b
38o 35o
500 ft.
BUILDING HEIGHT
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.