Examples of Tasks from CCSS Edition Course 2, Unit 7 · (These solutions are for tasks in the CCSS...

Trigonometric Methods

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Examples of Tasks from CCSS Edition Course 2, Unit 7

Getting Started The tasks below are selected with the intent of presenting key ideas and skills. Not every answer is complete, so that teachers can still assign these questions and expect students to finish the tasks. If you are working with your student on homework, please use these solutions with the intention of increasing student understanding and independence. A list of questions to use as you work together, prepared in English and Spanish, is available. Encourage students to refer to their class notes and Math Toolkit entries for assistance. Comments in red type are not part of the solution.

As you read these selected homework tasks and solutions, you will notice that some very sophisticated communication skills are expected. Students develop these over time. This is the standard for which to strive. See Research on Communication.

The Geometry and Trigonometry page might help you follow the conceptual development of the ideas you see in these examples.

Main Mathematical Goals for Unit 7 Upon completion of this unit, students should be able to:

• explore the sine, cosine, and tangent functions defined in terms of a point on the terminal side of an angle in standard position in a coordinate plane.

• explore properties of the sine, cosine, and tangent ratios of acute angles in right triangles and use those ratios to solve applied problems.

• derive the Law of Sines and the Law of Cosines and use those laws to determine measures of size and angles for non-right triangles.

• use the Law of Sines and Law of Cosines to solve a variety of applied problems that involve triangulation.

• describe the conditions under which two, one, or no triangles are determined given the lengths of two sides and the measure of an angle not included between the two sides.

What Solutions are Available? Lesson 1: Investigation 1—Applications Task 1 (p. 474), Connections Task 9 (p. 477),

Reflections Task 19 (p. 481), Review Task 32 (p. 485), Review Task 33 (p. 485) Investigation 2—Applications Task 5 (p. 475), Connections Task 13 (p. 479), Extensions Task 28 (p. 483), Review Task 34 (p. 485) Investigation 3—Applications Task 6 (p. 476), Connections Task 16 (p. 480), Connections Task 17 (p. 480), Review Task 36 (p. 486)

Lesson 2: Investigation 1—Applications Task 1 (p. 503), Connections Task 10 (p. 507), Reflections Task 16 (p. 509), Extensions Task 23 (p. 511) Investigation 2—Applications Task 5 (p. 504), Connections Task 13 (p. 508), Reflections Task 18 (p. 510), Review Task 31 (p. 514) Investigation 3—Applications Task 7 (p. 505), Connections Task 15 (p. 509), Extensions Task 27 (p. 513), Review Task 34 (p. 515)

http://www.wmich.edu/cpmp/parentresource3/questions.html

http://www.wmich.edu/cpmp/parentresource3/questions-s.html

http://www.wmich.edu/cpmp/parentresource3/geometrystrand.html

http://www.wmich.edu/cpmp/parentresource3/geometrystrand.html



Selected Homework Tasks and Expected Solutions

(These solutions are for tasks in the CCSS Edition book. For homework tasks in books with earlier copyright dates, see Helping with Homework.)

Lesson 1, Investigation 1, Applications Task 1 (p. 474)

a, c. To be completed by the student.

b. percent grade = 100 sin A = 100 × 0.252 = 12.5%

percent grade = 100 sin A = 100 × 0.332 = 16.5%

Lesson 1, Investigation 1, Connections Task 9 (p. 477)

a. i. cos 120˚ ≈ –0.5

ii. sin 120˚ ≈ 0.87

iii. tan 120˚ ≈ –1.73

aiv–vi, b, d. To be completed by the student.

c. Since the trigonometric values result from ratios of the coordinates and the radius of the circle, you can find the trigonometric values for angles in Quadrant II (obtuse angles) by using the image of the point on the circle reflected across the y-axis. So, for example, a point (–x, y) in Quadrant II has image point (x, y) in Quadrant I. The negative x-coordinate in Quadrant II means that the cosine and tangent values will be negative in Quadrant II. So, to use the table on page 465 for 120˚ or P12, reflect P12 across the y-axis to P6. Use the negative of the cos 60˚ and tan 60˚ function values from the table for a 120˚ angle. The sin 60˚ = sin 120˚.

Lesson 1, Investigation 1, Reflections Task 19 (p. 481)

The slope is tan θ. The equation of the line is y = (tan θ)x.

Lesson 1, Investigation 1, Review Task 32 (p. 485)

a. x ≈ 1.8

b. x = 5

c. x = 20.25

d. x ≈ –1.3


a. 2 15

b. 5 3

c–e. To be completed by the student.

http://www.wmich.edu/cpmp/parentresource3/helpingwithhomework.html




a. Using d to represent the distance across the river, tan 31˚ = d400 and d = 400 tan 31˚ ≈ 240.344 ≈

240 meters.

b. To be completed by the student.


a. OA′P is a 30˚-60˚ right triangle. Using the diagram at the

right, x = 32 and y = 12 . So, A( 3

2 , 12 ). As seen in

Connections Task 12, cos 30˚ = 32 and sin 30˚ = 12 . So,

the coordinates of point A can be written as (cos 30˚, sin 30˚).

b. OB′ = 1 because O ′B is a radius.

i. a = –sin 30˚

ii. b = cos 30˚

c. i. The entries in the first column of matrix R30˚ are the coordinates of the image of A(1, 0), and the

entries in the second column are the coordinates of the image of B(0, 1). In Part a, we verified that

(1, 0) → ( 32 , 12 ). In Part b, we identified the coordinates of the image of B as (–sin 30˚, cos 30˚)

= (– 12 , 32 ). These together verify the entries of matrix R30˚.

ii. cos 30˚ – sin 30˚

sin 30˚ cos 30˚

⎡

⎣⎢

⎤

⎦⎥

d. 32 – 1212

32

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0

1

⎡

⎣⎢

⎤

⎦⎥ =

– 1232

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

The result is equivalent to the coordinates of B′.

d, e. To be completed by the student.

Lesson 1, Investigation 2, Extensions Task 28 (p. 483)

Call x the remaining distance from B to the base of the mountain. Then tan 39˚ = hx , h = x tan 39˚; so, h ≈ 0.8098x. The remainder of the solution is left to the student.




a. The triangles must be congruent by the SSS congruence condition.

b. The triangles could be different. Their shapes would be the same, but their sizes could be different.

c–e. To be completed by the student.


a.

cos 28˚ = 8c

c = 8cos 28˚ ≈ 9.0606 ≈ 9 ft

tan 28˚ = a8

a = 8 tan 28˚ ≈ 4.2538 ≈ 4 ft m∠B = 90˚ – m∠A = 62˚

b, c. To be completed by the student.


a. tan θ = 20; θ ≈ 87˚



a. Using the Pythagorean Theorem: s = 9.52 – 82 = 26.25 ≈ 5.1 in. Perimeter = 2(8) + 2(5.1) = 26.2 in. Area = (8)(5.1) = 40.8 in2



a. (x – 5)(x + 2) = 0 Check: x – 5 = 0 or x + 2 = 0 52 – 3(5) – 10 = 25 – 15 – 10 = 0 x = 5 or x = –2 (–2)2 – 3(–2) – 10 = 4 + 6 – 10 = 0

b. x2 + 9x + 4 = 0 Check:

x = –9 ± 652 (–0.47)2 + 9(–0.47) + 4 ≈ 0

x ≈ –0.47 or x ≈ –8.53 (–8.53)2 + 9(–8.53) + 4 ≈ 0

c, d. To be completed by the student.




a. The intended course is along AC . The actual course is marked by arrows, and the distance AB + BC.

b. AB = 80 sin 10˚sin 135˚ ≈ 19.6460 km; BC = 80 sin 35˚sin 135˚ ≈ 64.8928 km

Extra travel distance ≈ AB + BC – 80 ≈ 5 km


a. Students should verify that parallelograms have 180˚ rotational symmetry. (The inclusion of the diagonals in their analysis of the symmetry will assist them in finding some of the information for Part b.)

b. Students should be able to determine the angle measures and lengths in the diagram shown here.

Students should explain their answers.

c. Use the Law of Sines to find BE ≈ 2.11 cm; AE ≈ 3.81 cm. By symmetry, DE ≈ 2.11 cm and CE ≈ 3.81 cm.

Next, in AED, let x = m∠ADE, then m∠DAE = 180˚ – (68˚ + x) = 112˚ – x.

sin x3.81 = sin (112˚ – x)2.11

sin xsin (112˚ – x) = 3.812.11 ≈ 1.81

By letting Y1 = sin xsin (112˚ – x) and using tables or graphs, students can see that when y ≈ 1.81, x is

about 79˚. m∠ADE ≈ 79˚; m∠DAE = 112˚ – x ≈ 33˚

Through further application of the Law of Sines or rotational symmetry, all remaining measures may be found.




a. Using the Properties of Equality, by taking reciprocals, the first proportion can be changed into the second.

b. If you are solving for an angle, it is easier to use I sin Aa = sin Bb = sin Cc . If you are solving for a

side, it is easier to use II asin A = b

sin B = csin C . Students should understand that the proportions can

be solved in either form but that fewer steps are required if the variable for which you are solving is in the numerator.


a. Use the Law of Cosines to determine m∠B. Then ha = sin B and h = a sin B. Or, instead, find m∠A and write h = b sin A.

b. The area of ABC = 12 hc = 12 ac sin B or 12 bc sin A.


a.

b. To be completed by the student.

c. In ABD, the length of the altitude from point A can be found by solving sin 19˚ = h27 . So,

h = 27 sin 19˚, and the area of ABD = 12 (43.5)(27) sin 19˚ ≈ 191 m2. The remainder of the solution is left to the student.


Since by symmetry, m∠ABC = m∠ADC, we have AC2 = 22 + 32 – 2(2)(3) cos 118˚. AC ≈ 4.32 ≈ 4 ft To find BE, you could find CE or AE and use the Pythagorean Theorem. To find CE, you could use

cos ∠1 = CE2 . To find m∠1, use the Law of Cosines

to see that cos ∠1 ≈ 0.7906. The remainder of the solution is left to the student.




a. Use the Law of Cosines to find AC. AC2 = 62 + 42 – 2(6)(4) cos 61˚. One solution is expected.

b–d. To be completed by the student.


a. Using slopes: slope of AB = 2

slope of BC = – 12

So, AB ⊥ BC . Therefore, ΔABC is a right triangle. (Students might choose to find the lengths of the sides as requested in Part b and use the converse of the Pythagorean Theorem.)

b. AB = 20 = 2 5 BC = 80 = 4 5 AC = 10 So, the perimeter is 10 + 6 5 .

c. To be completed by the student.


a. m∠P = 180˚ – 30.0˚ – 136.75˚ = 13.25˚

Use the Law of Sines to determine CP. CP

sin 136.75˚ = 1,500sin 13.25˚

CP = 1,500 sin 136.75˚sin 13.25˚ ≈ 4,484 feet

(Young Glory III’s record throw is listed as 4,483.51 feet on the Punkin’ Chunkin’ Web site: www.punkinchunkin.com)

INSTRUCTIONAL NOTE If students use the Law of Sines in Part b to determine the measure of ∠J, they will get the measure of the supplement of ∠J. You can refer students back to Lesson 1, Connections Task 8, to help them make sense of this situation. From Connections Task 8, they know that two possible angles between 0˚ and 180˚ exist for any given sine value (except for sin 90˚ = 1). Therefore, using sin–1 to determine the measure of an angle in any triangle should always reveal two possible solutions. To avoid this dilemma, students can use the Law of Cosines.





a. iii. Both equations can be true because ∠ACB and ∠AC′B are supplementary. It follows that their sines are equal.

b. i. Let AC = b. By the Law of Cosines, 12 = b2 + ( 3 )2 – 2b( 3 ) cos 30˚. Use this to show that b2 – 3b + 2 = 0. The solutions to this equation represent possible values for AC.

ii.

c. i. To be completed by the student.

ii. Use the Law of Cosines in a manner similar to that shown in the solution to Part bi to find that c ≈ 3.2.


a. When AC = 1.5, m∠ABC will be smallest. Then ABC is equilateral; so, m∠ABC = 60˚. In the

diagram, BFD is a 30˚-60˚ right triangle with hypotenuse of length 4 m. Thus, DF = 2. Alternatively, DF = 4 sin 30˚ = 2. The height of the plane is DG, and DG = 2 + 1.5 = 3.5 m.

b. When the hydraulic cylinder is fully extended, AC = 2.2 m. By the Law of Cosines, 2.22 = 1.52 + 1.52 – 2(1.5)(1.5) cos (∠ABC). cos (∠ABC) ≈ –0.0756; m∠ABC ≈ 94.3˚





a. 236 , or 118

b. 3436 , or 1718


Examples of Tasks from CCSS Edition Course 2, Unit 7 · (These solutions are for tasks in the CCSS...

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