Solutions to Investigations and Homework
Solutions to lesson 1 “geometric mean”
1. (Students may not list this set), and
2. (Students may not list this set), and
3. , and (You may need to help your groups see these congruence).
4. ~, ~ and ~ .
5. or equivalent proportion
6. The altitude is the geometric mean of the segments of the hypotenuse created by the altitude.
7. or equivalent proportion
8. The leg is the geometric mean of the hypotenuse and segments of the hypotenuse created by the altitude adjacent to the hypotenuse
9. or equivalent proportion.
10. The same definition from 8 applies to the proportion in 9
Solutions to homework for “geometric mean”
1. , , ,
2. , , ,
Extension: , after simplification: .
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.
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)
Extension: (3, 3)
Solutions to Lesson 3:
1)Check the students’ constructions for accuracy. These do not need to be constructed.
2)9+16=25, 36+64=100, 81+144=225, 25+144=169
3)They are all right triangles (check for initials).
4)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.
5)a2 + b2 =c2 is true only if they are right triangles.
6)Check students; constructions for tic and construction marks.
- Triangle CBS is acute.
- NOT is not a triangle.
- Triangle ESP is obtuse.
- Triangle MTV is right (students may need reminded of how to find the decimal of the square root of 2).
7)9<9+4, 9>4+1, 36>25+4, 2=1+1.
8)No. Check to see if the sum of any two sides is greater than the remaining side. Use the triangle inequality theorem first.
9)70, 71, 38 acute; not; 70, 18, 92 obtuse; 45, 45, 90 right.
10)
- If a + b > c, a + c > b, b + c > a then the sides make a triangle.
- If a2 + b2 = c2 then the triangle is a right triangle.
- If c2 < a2 + b2 then the triangle is acute.
- 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”
Check to see that points are plotted and labeled correctly.
1)PR = 3, PQ = 4, and QR =5. Since 45+80=125, this is a right triangle.
2)PQ = , PR = , and QR =. Since 10 + 20 > 26, PQR is an acute triangle.
Solutions to Lesson 4:
1)Open the file.
2)See gsp file for solution.
3)All 45 degrees
4)Isosceles triangles.
5)Sum =90 so <s are complementary.
6)See gsp file.
7)All right <s.
8)Isosceles right triangles.
9)Check solutions file for Leg 1, Leg 2, and hypotenuse measurements. For the ratio of the legs (column 4) they should have 1.4/.
10)1.414.
11).
12)Multiply by (or 1.414).
13)Divide the hypotenuse by 1.414 ().
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 .
15)Yes. An example is (1, 1,) because 12 + 12 = ()2 and 2=2.
16)Missing block in order are b, b, c/, and 2c/.
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. Be sure to point out that this (/) is not a division sign.
26)2.
27)1.73.
28).
29)h multiply by 2; h multiply by 2 after divide by 1.73.
30)Yes divide by 2, yes divide by and then multiply by 2.
31)Longer leg divide hypotenuse by 2 and then multiply by 1.73 () and longer leg shorter leg multiplied by.
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 and the smaller leg is half the hypotenuse.
33)Yes. An example is (1, , 2) because 12 + ()2 = 4.
34)In order by missing block: b/, c/2, and c/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.
Teachers solutions to urspecial.gsp
Solutions to homework for “Special Right Triangles”
1)7
2)14
3)8
4)8
5)9
6)8
7)7
8)10
9)11
10)5
11)14
12)6
13)10
Bonus: 96
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.
1)Please calculate the following ratios for the right triangle above. Write your answer as a decimal rounded to four decimal places.
a) .9340
b) .3586
c) 2.6048
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) = .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.
In the above calculations, I noticed that= = .934,
Cosine .358, and Tangent 2.605.
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.
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.
6) Please write a general rule for the relationships that you found.
See definition below.
Definition:
You have discovered SOHCAHTOA or .
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
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.
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 .
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 and. However, this is not true for the tangent function because for .
Extension 3:
Suppose we had another right triangle as shown below. Notice this triangle also has angles measuring 90, 21 and 69 degrees.
Question:
Set up three trigonometric ratios usinginvolving all three sides of the triangle above.
- Do you notice any similarities in the calculations for this triangle and the triangle given in exercise 1?
Yes.
- 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.
Solutions to Lesson 5: Investigation of Trig Ratios Homework
1)AA similarity; the third angle will be the same for both.
2)B
3)X=2.74
4)K=2.94
5)23.09
6)7.71
7)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.
Definition:
We need to know that given, we can find A by taking This means that This definition will also be true for any of the other trigonometric functions.
1)By SOHCAHTOA, we know that
a)Find using the definition above.
(Or close depending on roundoff error).
b)Use the information given in the right triangle above to set up a similar equation for to find angle.
c)Does this also give you the same value for angle ?
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 and so.
1)Now use sine, cosine or tangent, to find the measure of angle .
They could set up either,, or . Then =
37.6 (or close depending on roundoff errors).
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 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).
Solution:
The equation (where H stands for the hypotenuse) should have set up. This implies that feet. This means that the airplane is 17,276.3 feet from the landing strip.
Extension 2:
You are standing 75 meters from the base of the JinMaoBuilding 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.
The student could have set up either or . Then they 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.
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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