Extra Practice

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By: Marc Hensley
Right
Triangles and
Trigonometry
 The Pythagorean Theorem is a relation in Geometry
between the 3 sides of a right triangle
 The theorem states:
In any right triangle, the area of the square whose side is the
hypotenuse (the side opposite the right angle) is equal to the
sum of the areas of the squares whose sides are the two legs
NEXT
Generally speaking, the formula is written as:
2
a
+
2
b
=
2
c
where a and b are the legs, and c is the
hypotenuse.
c
a
See examples
Practice a problem
b
BACK
x
x
7
13
24
72 + 242 = x2
49 + 576 = x2
625 = x2
25 = x
52 + x2 = 132
25 + x2 = 169
x2 = 144
x = 12
5
Practice a Problem
Solve for x.
34
16
x
a) 38
b) 30
c) 40
d) 45
162 + x2 = 342
256 + x2 = 1156
34
16
x
x2 = 900
Try again
 Or click here to go home and start a new section!
You have finished this part of the review.
Now, either go back, and choose a new
topic, or click here for more practice with
this topic!
See solution
Try again
 A special right triangle is a right triangle whose sides
are in a particular ratio.
 Recognizing special right triangles in geometry can
help you to answer some questions quicker.
Types of special triangles
There are 2 main types of special right triangles:
1) The 45-45-90
2) The 30-60-90
45
60
90
90
30
45
Click on the triangle you want to learn about.
The lengths of the sides of a 45°- 45°- 90° triangle are in
the ratio of 1 : 1 : √2
45
x √2
x
90
45
x
Try a practice problem!
Check out the
30-60-90!
The lengths of the sides of a 30°- 60°- 90° triangle are in
the ratio of 1 : √3 : 2
60
2x
x
90
30
x√3
Try a practice problem!
Check out the
45-45-90!
Solve for the missing parts of the triangle. Round your
answers to the nearest tenth.
45
y
7.4
z
90
x
Back to help
Click here
to check
your answers!
Solve for the missing parts of the triangle. Round
your answers to the nearest tenth.
z
13
x
90
30
y
Back to help
Click here
to check
your answers!
 x = 7.4
 y = 10.5
 z = 45
Try a 30-60-90 problem!
Try it again!
Watch a helpful video!
 Or click here to go home and try a new section!
 x = 6.5
 y = 11.3
Try a 45-45-90 problem!
 z = 60
Try it again!
Watch a helpful video!
 Or click here to go home and try a new section!
 Right triangles have 3 special formulas that ONLY
WORK in right triangles.
 They are:
Opposite
 sin = hypotenuse
 cos =
 tan =
adjacent
hypotenuse
opposite
adjacent
How to setup a
trig problem
A
c
b
C
B
a
See an example
Go straight to
try a problem!
 Draw a picture depicting the situation.
 Be sure to place the degrees INSIDE the triangle.
 Place a stick figure at the angle as a point of reference.
 Thinking of yourself as the stick figure, label the
opposite side (the side across from you), the
hypotenuse (across from the right angle), and the
adjacent side (the leftover side).
 Figure out which pair of sides the problem deals with
(for example: opposite and hypotenuse) and choose
the correct equation (in our example, sin)
See an example
Try a problem!
In right triangle ABC, hypotenuse AB=15 and angle A=35º.
Find leg BC to the nearest tenth.
1) First, draw a picture, and
label everything you know.
2) Then, figure out which trig
function we will use. In this
case, we will use sin.
3) Set up the equation. sin 35 = X15
4) Solve for x.
x = 15 sin 35
x = 8.6
See how to set
up a problem
Try a problem!
In triangle RST, angle R is a right angle, angle S has
measure of 65 degrees, and RS = 9. Find the measure
of ST.
A) 3.8
B) .05
C) 27.1
D) 21.3
See solution
Try again
You have finished this part of the review.
Now, either go back, and choose a new
topic, or click here for more practice with
this topic!
 cos 65 =
9
x
 x * cos 65 = 9

9
x=
cos 65
S
65
x
9
90
R
 x = 21. 3
Try again?
Back to explanation.
 Click here to go home and try something new!
T
 We use the LAW OF SINES to solve triangles that are
not right triangle.
 The law of sines states the following:
The sides of a triangle are to one another
in the same ratio as the sines of their opposite angles.
What does that mean?
See a diagram
explanation!
B
 We can use this triangle to
set up the equation…..
a
c
sin A = sin B = sin C
a
b
c
A
b
See how a problem is done.
Try one on your own!
C
 The three angles of a triangle are 40°, 75°, and 65°.
When the side opposite the 75° angle is 10 cm, how
long is the side opposite the 40° angle?
 Click to draw the triangle.
 Click again to set up the problem.
 Click a third time to see the answer!
sin 75 = sin 40
10
x
x = 6.7
Now I’m ready to try one!
 The three angles of a triangle are A = 30°, B = 70°, and
C = 80°. If side a = 5 cm, find sides b and c.
Click me to check your answer!
 B = 9.4
 C = 9.9
Did you get it right?
Yes
No
Take me home so
I can try
something new!
Take me back
to the
explanation again.
Or click here for more Law of Sines practice!
 Angles of elevation and depression are angles that are
formed with the horizontal.
 If the line of sight is upward from the horizontal, the
angle is an angle of elevation; if the line of sight is
downward from the horizontal, the angle is an angle of
depression.
 Using these types of angles and some trig, you can
indirectly calculate heights of objects or distances
between points.
 Alternatively, if the heights or distances are known, the
angles can be determined.
OK, I need to see
how one’s done.
Actually, I’m ready to
try one on my own!
 Suppose a flagpole casts a shadow of 20 feet. The angle
of elevation from the end of the shadow to the top of
the flagpole measures 50°. Find the height of the
flagpole.
 Click once to draw the picture.
 Click again to setup the problem.
 Click a third time to see the answer!
tan 50 = x
20
x = 23.8
NOW, I’m ready to
try one on my own!
 Suppose a tree 50 feet in height casts a shadow of
length 60 feet. What is the angle of elevation from the
end of the shadow to the top of the tree with respect to
the ground?
See the solution!
Go back to the example…
 First, we draw the picture
 Then, set up the equation:
tan x = 50
60
• Finally, solve for x.
x = tan-1 (
50
) = 39.8
60
Let me try again.
I need more practice!
 Click here to go home and try something new!
 The Law of Cosines (also known as the Cosine Rule or
Cosine Law) is a generalization of the Pythagorean
Theorem
 Basically, the Pythagorean Theorem requires there to
be a right angle in a triangle
 But, if there is not, the Law of Cosines can be used
Show me the
Law of Cosines
 For a triangle with sides a, b, and c opposite
(respectively) the angles A, B, and C, the Law of
Cosines states:
 c2 = a2 + b2 - 2ab·cos(C)
 a2 = b2 + c2 – 2bc·cos(A)
I need to see a picture
 b2 = a2 + c2 - 2ac·cos(B)
I need to see
an example
Take me straight to
a practice problem!
 The law of cosines should be used when use have SSS




or SAS in a triangle, like the one picture here…
The law of cosines states:
s2 = r2 + t2 – 2rtcos(S), or
r2 = s2 + t2 - 2stcos(R), or
t2 = r2 + s2 – 2rscos(T)
I still need to
see an example.
Great! Let me
try one.
 In triangle ABC, you are given a = 10, B = 32o and c = 15.






Find the measure of side b.
First, write out the equation: b2 = a2 + c2 -2 ac cos B.
So, b2 = 100 + 225 – 2*(10)*(15)*cos 32º
b2 = 325 - 300 (0.848048096)
b2 = 325 - 254.4
b2 = 70.59. Therefore,
b = 8.4.
Take me back to
the explanation.
Great! Let me
try one.
 In triangle ABC, you are given A = 28o, b = 14, c = 10.
Solve for side a.
I can’t do it!
Take me back
to the example.
I think I got it!
Let me see the answer.
 First, write out the equation: a2 = b2 + c2 - 2bc*cos A.
 So, a2 = 296 + 100 – 2*(14)*(10)*cos 28º
 a2 = 396 - 280 (0.8829475929)
 a2 = 396 – 247.2
 a2 = 148.8. Therefore,
 a = 12.2.
I got it! Take me
home so I can try
something different!
I want to
practice more!
I didn’t get it.
I want to try again.
You are now ready to take the
chapter 8 test. Good Luck!
End Show
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