1. By: Lee Lukasik and Bailey Potrzebowski
TRIGONOMETRY TEST REVIEW
2. Ohio Academic Content Standards
Grade 9
“Define the basic trigonometric ratios in right
triangles: sine, cosine, and tangent.”
“Apply proportions and right triangle trigonometric
ratios to solve problems involving missing lengths and
angle measures in similar figures.”
3. Parts of a Triangle
Consider angle A
Side a- side opposite of A
Side c- side adjacent to A
Side b- the hypotenuse
(always opposite of the 90
degree angle)
5. Sine
Ratio of the length of the side opposite the given angle to the length
of the hypotenuse of a right-angled triangle
Sine= Opp/Hyp
Practice
Problems
8. Practice Problem #1 Answer
We are given that angle A=40 degrees, angle B=90
degrees, angle C=50 degrees and side a=6.
We know that sine is opposite over hypotenuse so
we can take the sine of 40 degrees and make it
equal to 3 divided by side b (sin40=3/b). Then we
would multiply both sides by b to get rid of the
denominator leaving us with b(sin40)=3. Finally we
would divide both sides by sin40 to get b by
itself, b=3/sin40, giving us the answer that side
b=4.67.
Next we can find side c by taking the sine of 50
degrees and set it equal to c divided by 4.67
(sin50=c/4.67). We multiply both sides by 4.67 to
get c by itself, giving us 4.67(sin50)= c. c= 3.58.
9. Practice Problem #2
A ship travels 10 km on a course heading 50º east of north.
How far north, and how far east has the ship travelled at this point?
Answer
10. Practice Problem #2 Answer
To find the distance traveled north we
would take the sine of 40 degrees and set it
equal to y divided by 10 km (sin40=y/10km).
Next we would multiply both sides by 10 km
to get y by itself giving us 10km(sin40)=y. y=
6.43km
To find the distance traveled east we would
take the sine of 50 degrees and set it equal
to x divided by 10km (sin50=x/10km). Next
we would multiply both sides by 10km to get
x by itself giving us 10km(sin50)=x.
x=7.66km
11. Practice Problem # 3
Given that the sine of angle A=
0.6, calculate the length of side x.
Answer
12. Practice Problem #3 Answer
Since we know the sine of angle A is 0.6
we can set that equal to 12cm divided by
side AB (0.6=12cm/AB). We can conclude
that AB= 12cm/0.6, giving us the side of
AB=20cm.
Now using Pythagoras’ theorem we see
that x= the square root of 20 squared
minus 12 squared. Which equals 16cm.
13. Cosine
Ratio of the adjacent side to the
hypotenuse of a right-angled triangle.
Cos=Adj/Hyp
Practice
Problems
15. Practice Problem #1
In triangle ABC, angle C= 42 degrees, a=19 and b=26. Find the length of side
b.
A
c=17
Answer
42
B C
a= 19
16. Practice Problem #1 Answer
We are given that angle C=42 degrees
and side a=19, so we can conclude that
the cosine of 42 degrees is equal to 19
divided by side b(cos42=19/b). This gives
us that b=19/cos42. b=25.6
17. Practice Problem #2
Find the cosine of angle A, and angle C.
A
c=4
Answer
B C
a= 3
18. Practice Problem #2 Answer
We know that cosine is adjacent over
hypotenuse. For angle A, side c=4 is
adjacent, and the hypotenuse= 5; Cos
A=4/5
For angle C, side a=3 is adjacent, and the
hypotenuse=5; Cos B=3/5.
20. Practice Problem #3 Answer
We know that cosine is adjacent over
hypotenuse. So we can agree that the cos
36 degrees= x/17.
To get x by itself we must multiply both
sides by 17 (36x17=x). Resulting with
x=13.75
21. T a n g e n t
Tangent=Opposite Look at This
video of
Adjacent how we use
tangent
Tan B = o
a
Click Here For a Practice
Graph of Tangent Problems
23. Notice:
Vertical Asymptotes- the graph approaches a value but
never reaches it
The graph of tangent has vertical asymptotes at
multiples of 90°
Tangent has NO value
at these asymptotes!
24. Practice Problems
Get Out Your Calculator!
• Going for • Back to
A Hike Basics
• At the
Airport
25. Going on a Hike
While on a hike, you come up to hill. The map you
have says its 300 ft. until you reach the top of the
hill. If you know that the incline of the hill is (10
degrees) determine the height of the hill (x).
Tan 10 = x
Click Here 300 Check Your
For a Hint Answer
Tan (10)
52.898 ft
= .17632
x
10°
26. Basic Tangent Problem
HINT
Find The Tangent of
the angle labeled F 3050 is the side opposite
angle F and 2000 is
adjacent
Check Your
Answer
1.525
27. At The Airport
An airplane takes off at an angle of 25°. Before
the plane changes its angle, it is at a height of
5000ft. How far has the airplane traveled along
the ground?
Click Here
For Answer!
Click the airplane for
a hint! 10722.535 ft
Solve for x
Tan(25) = 5000
x
5000 ft
x 25°
28. Triangles that have angle measurements of 30-60-90 or 45-
45-90 have side length ratios that are easy to remember. If
you know one side length the other two on the triangle will
follow
30 60 90 Triangles 45 45 90 Triangles
29. Solving With Special Triangles
• Length of
the longer
• The length leg is √3
of the legs times the
are the smaller leg
same
• The
hypotenuse is • The
√2 times the hypotenuse
length of a is 2 times
leg the length of
the smallest
leg
30. SOH CAH TOA
Q: What is SOH CAH TOA?
A: an easy way to remember what makes
up the trigonometry functions
Said as "so - cuh - toe – uh”
31. References
1st slide Picture from Powerpoint clip art
2nd slide Picture from powerpoint clip art
Ohio standards from
http://www.ode.state.oh.us/GD/Templates/Pages/ODE/ODEDetai
l.aspx?Page=3&TopicRelationID=1704&Content=86689
Tangent http://www.onlinemathlearning.com/tangent-problems.html
info/vid
Tangent All created information by self as well as pictures created by self
problem in program “paint”
slides
Special http://www.onlinemathlearning.com/image-files/special-rt-
triangles triang-454590-1.gif
Video 2 http://www.yourteacher.com/geometry/306090triangle.php
All other Created using powerpoint tools
pictures
