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Solutions to Investigations and Homework
Solutions to lesson 1 “geometric mean”
1. A  C (Students may not list this set), B  B and C  D
2. B  C (Students may not list this set), A  A and C  D
3. B  C , D  D and C  A (You may need to help your groups see these
congruence).
4. ABC ~ CBD , ABC ~ ACD and CBD ~ ACD .
5.
f d
 or equivalent proportion
e f
6. The altitude is the geometric mean of the segments of the hypotenuse created by the altitude.
7.
a c
 or equivalent proportion
e a
8. The leg is the geometric mean of the hypotenuse and segments of the hypotenuse created by
the altitude adjacent to the hypotenuse
9.
b d
 or equivalent proportion.
c b
10. The same definition from 8 applies to the proportion in 9
Solutions to homework for “geometric mean”
1.
X
3
 , X 2 3
4 X
Y 7
 , Y 2 7
4 Y
Z 7
 , Z  21
3 Z
2.
6 4 L

, L  13
4
6
M 13

, M 2 3
4 M
N 17

, Z  221
13 N
Extension:
4 X 6

, after simplification:
X
4
0  X  6 X 16 .
2
Which can be factored to 0=(X + 8) (X - 2), so X = -8 or X = 2. Since a segment cannot have a
negative length, X = 2 is the only valid answer.
Solutions to Lesson 2: Pythagorean Theorem
1) Students compare.
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2) a2 and b2.
3) c2.
4) a2 + b2 = c2.
5) The Pythagorean Theorem.
6) If you have a right triangle, then the sum of the squares of the legs equals the square of the
hypotenuse. Answers will vary.
7) I can see that the areas of the two smaller squares fit exactly into the area of the largest
square.
Solutions to Lesson 2: Pythagorean Homework
1) 13 feet
2) Approximately 127 feet.
3) 1.38 and 4.14.
4) 7
Extension: (3, 3 3 )
Solutions to Lesson 3:
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
Check the students’ constructions for accuracy. These do not need to be constructed.
9+16=25, 36+64=100, 81+144=225, 25+144=169
They are all right triangles (check for initials).
Check for correct angle measurements (and check for intials). CAT is 90, 37, 53; DOG is
90, 37, 53; SIT is 90, 37, 53; RUN is 90, 23, 67.
a2 + b2 =c2 is true only if they are right triangles.
Check students; constructions for tic and construction marks.
a. Triangle CBS is acute.
b. NOT is not a triangle.
c. Triangle ESP is obtuse.
d. Triangle MTV is right (students may need reminded of how to find the decimal of
the square root of 2).
9<9+4, 9>4+1, 36>25+4, 2=1+1.
No. Check to see if the sum of any two sides is greater than the remaining side. Use the
triangle inequality theorem first.
70, 71, 38 acute; not; 70, 18, 92 obtuse; 45, 45, 90 right.
a. If a + b > c, a + c > b, b + c > a then the sides make a triangle.
b. If a2 + b2 = c2 then the triangle is a right triangle.
c. If c2 < a2 + b2 then the triangle is acute.
d. If c2 > a2 + b2 then the triangle is obtuse.
Extension: (3, 4, 5), (6, 8, 10), (9, 12, 15), (5, 12, 13).
Bonus: Check the triples to see if they actually form a right triangle.
Solutions to homework for “Converse to Pythagorean Theorem”
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Check to see that points are plotted and labeled correctly.
1) PR = 3 5 , PQ = 4 5 , and QR =5 5 . Since 45+80=125, this is a right triangle.
2) PQ = 26 , PR = 10 , and QR = 20 . Since 10 + 20 > 26, PQR is an acute
triangle.
Solutions to Lesson 4:
1)
2)
3)
4)
5)
6)
7)
8)
9)
Open the file.
See gsp file for solution.
All 45 degrees
Isosceles triangles.
Sum =90 so <s are complementary.
See gsp file.
All right <s.
Isosceles right triangles.
Check solutions file for Leg 1, Leg 2, and hypotenuse measurements. For the ratio of the
legs (column 4) they should have 1.4/ 2 .
10) 1.414.
11) 2 .
12) Multiply by 2 (or 1.414).
13) Divide the hypotenuse by 1.414 ( 2 ).
14) Check students’ logic. An example could be if you have the hypotenuse of a 45-45-90,
then you find the side by dividing by 2 .
15) Yes. An example is (1, 1, 2 ) because 12 + 12 = ( 2 )2 and 2=2.
16) Missing block in order are b, b 2 , c/ 2 , and 2c/ 2 .
17) Open the file.
18) Check solutions file.
19) They are all either 30 or 60 degrees.
20) Scalene triangle.
21) Again complementary.
22) The measure of all of the angles is 90 degrees.
23) All right.
24) Right scalene triangle.
25) The shorter, longer and hypotenuse measurements are in the gsp answer file (as the other
question of this type). All ratios of the 4th column are 2 to 1. The fifth column is 1.73
or 3 . Be sure to point out that this (/) is not a division sign.
26) 2.
27) 1.73.
28) 3 .
29) h multiply by 2; h multiply by 2 after divide by 1.73.
30) Yes divide by 2, yes divide by 3 and then multiply by 2.
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31) Longer leg  divide hypotenuse by 2 and then multiply by 1.73 ( 3 ) and longer leg 
shorter leg multiplied by 3 .
32) Again answers will vary. Check students’ logic. An example would be, if you have a 30,
60, 90 triangle then the hypotenuse is equal to twice the smaller leg, the longer leg is
equal to 3 and the smaller leg is half the hypotenuse.
33) Yes. An example is (1, 3 , 2) because 12 + ( 3 )2 = 4.
34) In order by missing block: b/ 3 , c/2, and c 3 /2.
35) Students need their name, date and period at the top of paper.
Bonus: No. Check students reasoning to see that they have ratios that include the square root
which will never produce a whole number unless multiplied by another square root.
O
N
NE = 0.58 in.
ON = 0.58 in.
OE = 0.82 in.
mNOE = 45
A
L
B
A
G
E
BA = 0.40 in.
BL = 0.40 in.
LA = 0.56 in.
E
D
mBAL = 45
Y
DY = 1.28 in.
AY = 1.28 in.
AD = 1.81 in.
mYAD = 45
O
GE = 0.68 in.
EO = 0.68 in.
GO = 0.96 in.
mEGO = 45
Teacher Solutions to imspecial.gsp
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O
A
ON = 0.58 in.
NE = 1.00 in.
OE = 1.16 in.
N
L
AB = 0.40 in.
BL = 0.68 in.
AL = 0.78 in.
G
E
mNEO = 30
mNOE = 60
mONE = 90
B
A
mBLA = 30
mLAB = 60
mLBA = 90
GE = 0.68 in.
E
EO = 1.15 in.
OG = 1.33 in.
Y
YA = 1.80 in.
DY = 3.13 in.
DA = 3.61 in.
D
mADY = 30
mYAD = 60
mAYD = 90
O
mEOG = 30
mEGO = 60
mGEO = 90
Teachers solutions to urspecial.gsp
Solutions to homework for “Special Right Triangles”
1) 7
2) 14
3) 8 3
4) 8 3
5) 9
6) 8
7) 7 2
8) 10 2
9) 11
10) 5 2
11) 14
12) 6 2
13) 10
Bonus: 96 3
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Solutions to lesson 5: Investigation of Trig Ratios
Investigation of trigonometric ratios
In this lesson we will derive trigonometric ratios for sine, cosine and tangent. Definitions will be
given for cosecant, secant and cotangent.
Definition:
If we consider angle ABC (labeled  ), we can define the hypotenuse to be side BC, the adjacent
side (next to angle  ) to be side AB and the opposite side (opposite the angle  ) to be side AC.
Opposite 
AC = 16.41 cm
A
mBCA = 21.00
C
mBAC = 90.00
BA = 6.30 cm
Adjacent to 
BC = 17.57 cm
mABC = 69.00
Hypotenuse

1) Please calculate the following ratios for the right triangle above. Write your answer as a
decimal rounded to four decimal places.
a)
Opposite
AC

 .9340
Hypotenuse BC
b)
Adjacent
 .3586
Hypotenuse
c)
Opposite
 2.6048
Adjacent
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2) Using your calculator and the right triangle above, please find the following
Trigonometric functions below. Be sure your calculator is in degree mode (type mode
then go over to degree instead of radian).
a) Sin (  ) = .9336
b) Cosine (  )  .3584
c) Tangent (  )  2.6051
3) Do you notice anything about your calculations for question 1 and 2? Explain in full
detail the relationships that you found.
Opposite
= .934,
Hypotenuse
Adjacent
Opposite
Cosine (  ) 
 .358, and Tangent (  ) 
 2.605.
Hypotenuse
Adjacent
In the above calculations, I noticed that Sin (  ) =
4) On graph paper (or with sketchpad), create your own right triangle and see if the
relationships that you found in question 3 are still true.
Opposite 
AC = 14.32 cm
A
mBCA = 21.58
C
mBAC = 90.00
BA = 5.66 cm
Adjacent to 
BC = 15.40 cm
mABC = 68.42
Hypotenuse

B
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5) Were the relationships discovered in question 3 still true. Why or why not? Please explain
with calculations.
Calculations may vary depending on the triangle chosen above. However, they should see that
the relationships are still true and should provide calculations as shown below.
Opposite
 .9299
Hypotenuse
Adjacent
cos(  ) 
 .3678
Hypotenuse
Opposite
tan(  ) 
 2.528
Adjacent
Sin(  ) 
6) Please write a general rule for the relationships that you found.
See definition below.
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Definition:
You have discovered SOHCAHTOA or
opposite
adjacent
opposite
.
sin  
, cos 
, and tan  
hypotenuse
hypotenuse
adjacent
The trig functions above are pronounced “sine, cosine and tangent”, respectively.
There are three additional trigonometric functions “cosecant, secant and cotangent” that are
found by inverting each of the three above functions. Therefore, we also have that
1
hypotenuse
1
hypotenuse
1
adjacent
csc  

, sec  

, and cot  

.
sin 
opposite
cos 
adjacent
tan  opposite
7) Please summarize in detail what you learned in this lesson.
I would expect them to write about the relationships of sine, cosine and tangent given above. In
addition, they should also mention the newly defined cosecant, secant and cotangent. They
should mention that they also learned how to define the side opposite and adjacent to an angle.
They may also mention that they learned how to calculate sine, cosine and tangent of a given
angle.
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Extension 1:
Using the original triangle (from page 1), set angle BCA =  , then find the trigonometric
relations for  . Don’t forget to rename your opposite and adjacent sides based on  .
Opposite
 .9299
Hypotenuse
Adjacent
cos(  ) 
 .3678
Hypotenuse
Opposite
tan(  ) 
 2.528
Adjacent
Sin(  ) 
Extension 2:
Now compare the trigonometric ratios for  and  . (Remember that we found these in
Extension 1 and in problem 3). Do you notice any connection between these trigonometric
ratios?
I notice that sin( )  cos(  ) and cos( )  sin(  ) . However, this is not true for the tangent
a b
function because  for a  b .
b a
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Extension 3:
Suppose we had another right triangle as shown below. Notice this triangle also has angles
measuring 90, 21 and 69 degrees.
Opposite 
A
BA = 3.36 cm
Adjacent to 
AC = 8.75 cm
mBCA = 21.00
C
mBAC = 90.00
mABC = 69.00

BC = 9.38 cm
Hypotenuse
B
Question:
Set up three trigonometric ratios using  involving all three sides of the triangle above.
AC
Sin(  ) 
 .9328
BC
AB
Cos (  ) 
 .3582
BC
AC
tan(  ) 
 2.604
AB
a. Do you notice any similarities in the calculations for this triangle and the triangle
given in exercise 1?
Yes.
b. What are the similarities specifically? Please explain.
I notice that I have the exact same proportion for my sides as shown above.
c) Why do you think this would happen (think back)?
This would happen because both the triangle in exercise one and the triangle given in this
extension have 90, 21 and 69 degree angles. Therefore, the triangles are similar due to
AA similarity. This implies that both of the triangles must have the same proportion
and thus will have the same sine, cosine and tangent values for their angles.
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Solutions to Lesson 5: Investigation of Trig Ratios Homework
1)
2)
3)
4)
5)
6)
7)
AA similarity; the third angle will be the same for both.
B
X=2.74
K=2.94
23.09
7.71
Hypotenuse = 2922.16 km is not at least 4 km. So the ship is in danger.
Solutions to Lesson 6: Investigation of Inverse Trig Ratios
Investigation of Inverse Trig functions
This investigation will rely on previously discovered trigonometric ratios and will focus on
leading students to understand how to use the inverse of trigonometric functions to find angles of
a right triangle. Practical applications will also be studied.
Suppose we do not know the measure of angle A (labeled  ) or the measure of angle C (labeled
 ) in the triangle below. We want to be able to find these angles but we only know the lengths
of the sides of the right triangle. Using SOHCAHTOA (as was previously discovered), we will
find the missing angles.
BC = 6.27 cm
B
mABC = 90.00

C
BA = 4.82 cm

AC = 7.90 cm
A
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Definition:
We need to know that given sin A 
a
a
, we can find A by taking sin 1  . This means that
b
b
a
sin 1    A. This definition will also be true for any of the other trigonometric functions.
b
1) By SOHCAHTOA, we know that sin  
BC 6.27

 0.79.
AC 7.90
a) Find  using the definition above.
  52.53 (Or close depending on roundoff error).
b) Use the information given in the right triangle above to set up a similar equation for
cos ine to find angle  .
cos( ) 
4.82
 .6101
7.90
c) Does this also give you the same value for angle  ?
  52.40
d) If the tangent function is used to find angle  , do you think you will find the same
number for  as we previously found? Show the details of your calculation for the
tangent function below.
Yes, I do think I will get the same values as for sine and cosine. We know that
6.27
tan( ) 
and so   52.45 .
4.82
1) Now use sine, cosine or tangent, to find the measure of angle  .
4.82
6.27
4.82
, cos( ) 
, or tan( ) 
. Then  =
7.90
7.9
6.27
37.6 (or close depending on roundoff errors).
They could set up either sin( ) 
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2) What if I had a right triangle but only knew two (of the three) side lengths. For example,
in the triangle above, suppose I know that angle  is 90 degrees, side length BC = 6.27
and side AC = 7.90. Do I need to know the length of side AB or can I find the length of
AB somehow? Please explain below.
I only need two side lengths of a right triangle because I can apply the Pythagorean
Theorem to find side AB. This means that AB  (7.9) 2  (6.27) 2  23.0971  4.8 by
the Pythagorean Theorem. This is what the length of AB was in the original triangle.
3) Summarize what you learned in this lesson.
I would expect them to state that they learned how to find angles inside a right triangle by
using the inverse sine, cosine and tangent functions. They may mention that all three trig
functions gave the same angle measure inside. They also learned that they only need two
side lengths of a right triangle to find the angle inside.
Extension 1:
An airplane flying at an altitude of 30,000 feet is headed toward an airport. To guide the
airplane to a safe landing, the airport’s landing system sends radar signals from the runway to
the airplane at a 10 degree angle of elevation. How far is the airplane (measured along the
ground) from the airport runway? Hint: Set up a trigonometric equation (using
SOHCAHTOA) and solve for the unknown variable. (Be sure you are in degree mode on
your calculator. Type mode, then go over to degree from radians).
A
30,000 ft.
10 degrees
B
C
Solution:
30, 000
(where H stands for the hypotenuse) should have set up. This
H
30, 000
implies that H 
 17, 276.31 feet. This means that the airplane is 17,276.3 feet from
sin(10)
the landing strip.
The equation sin(10) 
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Extension 2:
You are standing 75 meters from the base of the Jin Mao Building in Shanghai, China. You
estimate that the angle of elevation to the top of the building is 80 degrees. What is the
approximate height of the building? Suppose one of your friends is at the top of the building.
What is the distance between you and your friend? (Be sure you are in degree mode on your
calculator. Type mode, then go over to degree from radians).
Solution:
There are a few different ways to answer this question.
Opposite
75
The student could have set up either cos(80) 
or tan(80) 
. Then they
75
Hypotenuse
would find that the Hypotenuse = 431.9 meters or the Opposite side = 425.35 meters. It is
possible that the student only set up one equation and then used the Pythagorean theorem to find
the other side. The hypotenuse gives the distance between my friend and I and the opposite side
gives the height of the building. Therefore, the 431.9 meters is the distance between me and my
friend and 425.35 is the height of the building.
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Solutions to Lesson 6: Investigation of Inverse Trig Ratios Homework
1) 50.98 or 51 degrees.
2) B
3) The third angle is 47 degrees, the side opposite is 23.87 cm, and the adjacent side is 25.60
cm.
4) Answers vary for both.
5) Sin-1(75/200)=22.02 feet
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