1. RATIOS IN RIGHT TRIANGLES
Standards 15, 18, 19
INVERSE OF TRIGONOMETRIC RATIOS
END SHOW
PROBLEM 1a
PROBLEM 2a
PROBLEM 3a
PROBLEM 1b
PROBLEM 2b
PROBLEM 3b
PROBLEM 1c
PROBLEM 2c
PROBLEM 3c
USING TABLES AND GRAPHS
ELEVATION VS DEPRESSION
PROBLEM 4
PROBLEM 5
PROBLEM 6
PROBLEM 7
PROBLEM 8
PROBLEM 9
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2. Standard 15:
Students use the pythagoream theorem to determine distance and find missing lengths of sides of right triangles.
Los estudiantes usan el teorema de Pitágoras para determinar distancia y encontrar las longitudes de los lados de teoremas rectángulos.
Standard 18:
Students know the definitions of the basic trigonometric functions defined by the angles of a right triangle. They also know and are able to use elementary relationships between them, (e.g., tan(x)=sin(x)/cos(x), etc.)
Los estudiantes conocen las definiciones de las funciones básicas trigonométricas definidas para los ángulos de triángulos rectángulos. Ellos también conocen y son capaces de usar relaciones básicas entre ellos. (ej., tan(x)=sin(x)/cos(x), etc.)
Standard 19:
Students use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side.
Los estudiantes usan funciones trigonométricas para resolver para una longitud desconocida de un triángulo rectángulo, dado un ángulo y la longitud de un lado.
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3. SINE B Opposite side Sin C= Hypotenuse i Sin C= u u i COSINE
Standard 18
Sin C=
Hypotenuse i
Sin C= u u i
COSINE
Adjacent side
Cos C=
Hypotenuse o
Cos C= C A o u
TANGENT
Opposite side
Tan C=
Adjacent side i
Tan C= o
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4. Can we get the ratio again?X
m X = ?
Standard 18 5
Hypotenuse 3
Opposite side
Sin X= 5
36.86° 4
Sin X=0.6
m X = Sin ( ) -1
0.6 Z Y 3
m X = 36.86°
Can we get the ratio again?
Sin ( ) =
36.86°
0.6
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5. Can we get the ratio again?X
m X = ?
Standard 18 5
Hypotenuse 4
Adjacent side
Cos X= 5
36.86° 4
Cos X= 0.8
m X = Cos ( ) -1
0.8
m X = 36.86° Z Y 3
Can we get the ratio again?
Cos ( ) =
36.86°
0.8
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6. Can we get the ratio again?X
m X = ?
Standard 18 4
Adjacent 3
Opposite Side
Tan X= 5
36.86° 4
Tan X= .75
m X = Tan ( ) -1
.75
53.14°
m X = 36.86° Z Y 3
Can we get the ratio again?
What is the value for the remaining angle?
Tan ( )=
36.86°
.75
90° =
m Z=
36.86°
53.14°
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7. 17 Hypotenuse m R = ?, m Q = ? 17 Q 15 Opposite side R Sin Q= 28.07°
Standard 18 17 Q 15
Opposite side R
Sin Q=
28.07°
61.93° 8 15
Sin Q= .8824 S
m Q = Sin ( ) -1
.8824
90° =
m R=
61.93°
28.07°
m Q =
61.93°
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8. 41 Hypotenuse m R = ?, m Q = ? 41 Q R 9 Adjacent side Cos Q= 12.69°
Standard 18 41 Q R 9
Adjacent side
Cos Q=
12.69°
77.31° 9 40
Cos Q= .2195 S
m Q = Cos ( ) -1
.2195
m Q =
77.31°
90° =
m R=
77.31°
12.69°
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9. m R = ?, m Q = ? 15 Q 9 Opposite Side R Tan R= 12 Adjacent 53.14°
Standard 18 15 Q 9
Opposite Side R
Tan R= 12
Adjacent
53.14°
36.86° 9 12
Tan R= .75 S
m R = Tan ( ) -1
.75
m R =
36.86°
90° =
m Q =
36.86°
53.14°
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10. 51 Hypotenuse m R = ?, m Q = ? 51 Q 45 Opposite side R Sin Q= 28.07°
Standard 18 51 Q 45
Opposite side R
Sin Q=
28.07°
61.93° 24 45
Sin Q = .8824 S
m Q = Sin ( ) -1
.8824
90° =
m R =
61.93°
28.07°
m Q =
61.93°
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11. m R = ?, m Q = ? 75 Hypotenuse 75 Q 72 Adjacent side R Cos R= 16.26°
Standard 18 75 Q 72
Adjacent side R
Cos R=
16.26°
73.31° 21 72
Cos R= .96 S
m R = Cos ( ) -1
.96
m R =
16.26°
90° =
m Q =
16.26°
73.31°
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12. 1296 + 2304 = i i = 3600 i = 3600 |i|=60 i=60 and i=-60 36 + 48 = i
SOLVE QRS: i
= 60
Standards 15, 18, 19 Q R 36
Opposite Side
Tan R=
53.14°
36.86° 36 48
Adjacent 48 S
= i 2
i = 3600 2
Tan R= .75
i = 2
m R = Tan ( ) -1
.75
|i|=60
m R =
36.86°
i=60 and i=-60
90° =
m Q =
36.86°
53.14°
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13. SOLVE FGH: F
Standards 15, 18, 19
36.67°
= u 2 47
Adjacent 47 u G 35
Opposite Side 35
53.32°
Tan F= H
= u 2
Tan F= .7446
u = 3434 2
m F = Tan ( ) -1
.7446
u = 2
m F =
|u|=58.6
90° =
m H =
36.67°
u=58.6 and u=-58.6
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14. SOLVE SRQ: R i Adjacent side Opposite side Sin S= Hypotenuse i u 30
Standards 15, 18, 19 R i
Adjacent side
Opposite side
Sin S=
Hypotenuse i u 30
Hypotenuse u
Sin ( )=
35°
35°
55° 30 S Q
Sin 35° = u 30 30
(30)
Cos 35° =
Cos S =
u=30 Sin 35°
u=30( )
.5736
u=17.2
Cos 35° = i 30
(30)
90° =
m Q=
35°
55°
i=30 Cos 35°
i=30( )
.8192
i=24.57
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15. SOLVE LMK: L Opposite side Sin K= Hypotenuse i u i Sin( )= 55° 35° 55°
Standards 15, 18, 19 L
Opposite side
Sin K=
Hypotenuse i u u
Adjacent side i 45
Hypotenuse
Sin( )=
55°
35°
55° 45 M K
Sin 55° = i 45 45
(45)
Cos 55° =
Cos K =
i=45 Sin 55°
i=45( )
.8191
i=36.86
Cos 55° = u 45
(45)
90° =
m M=
55°
35°
u=45 Cos 55°
u=45( )
.5735
u=25.81
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16. 243 + 81 = e e = 324 e = 324 |e|=18 e=18 and e=-18 15.58 + 9 = e
SOLVE QRS:
Standards 15, 18, 19 R 9 i
Tan ( ) =
30°
15.58= i 9 =
Tan ( 30° ) 9 i 1
60°
30° S Q e
i Tan(30°) = 9
= e 2
Tan(30°) Tan(30°)
i = 9
Tan ( 30° )
= e 2
90° =
m Q =
30°
60°
i = 9
e = 324 2
.5774
i =15.58
e = 2
|e|=18
e=18 and e=-18
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17. X
SOLVE XYZ:
Standards 15, 18, 19 i 6
Tan ( ) =
36°
= a 2
=8.26
36° a =
Tan ( 36° ) 6 i 1 i
= a 2
a = 104.2 2
54°
i Tan(36°) = 6 Z Y
Tan(36°) Tan(36°)
a = 2 6
i = 6
Tan ( 36° )
|a|=10.2
i = 6
a=10.2 and a=-10.2
.7265
90° =
m Z =
36°
54°
i = 8.26
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18. F
SOLVE FGH:
36.46° 17
Opposite Side
Tan F= 23
Adjacent 23 u G
Tan F= .7391 17
53.53° H
m F = Tan ( ) -1
.7391
m F =
36.46°
|u|= 28.6
= u 2
u=28.6 and u= -28.6
= u 2
u = 818 2
90° =
m H=
36.46°
53.53°
u = 818 2
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Standards 15, 18, 19

19. SOLVE LMK: L i Opposite side Sin 57° = Sin = 57° 25 Hypotenuse i u u
Standards 15, 18, 19 L i
Opposite side
Sin 57° =
Sin =
57° 25
Hypotenuse i u u
Adjacent side 25
Hypotenuse
33°
57° M K
Sin 57°= i 25 25
(25)
Cos 57°=
i=25 Sin 57°
i=25( )
.8386
i = 20.96
Cos 57°= u 25
(25)
u=25 Cos 57°
u=25( )
.5446
90° =
m M=
57°
33°
u = 13.61
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20. SOLVE XYZ: X
Standards 15, 18, 19 7
Tan ( ) =
29° i
= 14.4
29° a =
Tan ( 29° ) 7 i 1 i
= a 2
= a 2
i Tan(29°) = 7
61°
Tan(29°) Tan(29°) Z Y
a = 208.5 2 7
i = 7
Tan ( 29° )
a = 2
i = 7
|a|=14.4
a=14.4 and a=-14.4
.5543
i =12.62
90° =
m Z =
29°
61°
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21. Angle to Ratio:
cos 37°=
0.7986
sin 19°=
0.3256
tan 67°=
2.3559

22. Ratio to Angle:
acos( ) =
23°
asin (0.0872) = 5°
atan (9.5144) =
84°

23. TRIGONOMETRIC CIRCLE: First Quadrantx y -1 1 y
Sin =
(x,y) 1 1 y
Sin = y
=Sin x
=Cos x
Cos = 1
Cos = x
Now, we take the first quadrant and we use it to find Trigonometric ratios from angles and angles from trigonometric ratios.
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24. Angle to Ratio:
Sin
Cos
cos 70°=
.34
sin 50°=
.76
0.76
0.34

25. Sin
Cos
52°
Ratio to Angle:
acos( 0.6) =
52°
asin (0.5) =
30°

26. ANGLE OF DEPRESSION AND ANGLE OF ELEVATION
Angles of Depression and Elevation are Alternate Interior Angles and they are congruent.
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27. 45 Tan X= 435 Tan X= .103 m X = Tan ( ) .103 m X = 5.9°
Standards 15, 18, 19
A car road rises vertically 45 feet, over a horizontal distance of 435 feet. What is the angle of elevation of the road? 45
Tan X=
435
45 feet
X=?
Tan X= .103
5.9°
435 feet
m X = Tan ( ) -1
.103
m X =
5.9°
The angle of elevation of the road is about 5.9°.
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28. Standards 15, 18, 19
The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope?
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29. Standards 15, 18, 19
The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope?
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30. Standards 15, 18, 19
The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope?
X=?
600 m
210 m
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31. 210 Cos Y= 600 Cos Y= .35 m Y = Cos ( ) .35 m Y = 69.5° m X= 90° -
Standards 15, 18, 19
The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope?
20.5°
X=?
600 m
210 m
Y=?
210
Cos Y=
600
Cos Y= .35
600 m
Y=?
210 m
m Y = Cos ( ) -1
.35
m Y =
69.5°
We have a right angle, so the angles together are 90°:
90° -
m X=
69.5°
= 20.5°
The angle of depression of the slope is about 20.5°.

32. 15 Tan X= 10 Tan X= 1.5 m X = Tan ( ) 1.5 m X = 56.3°
Standards 15, 18, 19
A glider is flying at an altitude of 15 yards, and starts descending when the distance from the expected landing spot is 10 yards away from one person standing below the glider on the ground. Find the angle of depression for the remaining part of the flying distance.
56.3° X
15 yd
We have alternate interior angles formed by a transversal crossing parallel lines:
X=?
Landind spot
10 yd 15
Tan X= 10
15 yd
Tan X= 1.5
X=?
m X = Tan ( ) -1
1.5
10 yd
m X =
56.3°
The angle of depression of the glider is about 56.3°.

33. So, the Pedro’s eye-sight is at about 6 meters above Javier’s.
Pedro and Javier are looking at each other. Pedro is on top of a brick wall and javier is on the ground at 20 meters from the wall. If the angle of elevation from javier’s eye is 17°, what is the vertical distance from the horizontal passing through his eye and the horizontal passing through Pedro’s eye.
x =?
ANGLE OF ELEVATION
17°
20 m
17°
20 m
x =? x
Tan 17° =
x = 20Tan 17° 20
x = 20( )
.306
Tan 17° = x 20
(20) (20)
x = 6.11 m
So, the Pedro’s eye-sight is at about 6 meters above Javier’s.

34. Rachel is flying a kite, and she has let out 35 yards of string
Rachel is flying a kite, and she has let out 35 yards of string. The angle of elevation for the string and the horizontal is 24°. What is the height of the kite from Rachel’s hand?
x =?
35 yards
ANGLE OF ELEVATION
24°
35 yards x
x =?
Sin 24° =
x = 35Sin 24° 35
24°
x = 35( )
.407
Sin 24° = x 35
(35) (35)
x = yd.
So, Rachel was flying her kite at a height of around yards from her hand.

35. They are approximately 175 feet apart.
Mrs. Xiong is on the top of a building. From that height, she is able to see two pedestrians that walk in opposite directions, going away from the building by the sidewalk. She is standing at the midpoint of the roof and the angles of depression for the pedestrians from that point in the roof are 56° and 43° respectively. The building is 100 feet tall. What is the distance between the pedestrians when they are at those angles of depression. (round final answer to the closest unit). x y
56°
43°
100 ft
Total distance = x + y
100 =
= 174.7
They are approximately 175 feet apart.
Finding x:
Finding y:
100
100
Tan 56° =
Tan 43° = x y
Tan 56° =
100 x
Tan 43° =
100 y
(x) (x)
(y) (y)
(x)Tan 56° =100
(y)Tan 43° =100 x=
100
1.48 y=
100
0.93
Tan 56°
Tan 43° x=
100
Tan 56°
x = ft y=
100
Tan 43°
y = ft