Examples of Tasks from 2008 Course 2, Unit 7

Trigonometric Methods

© 2008 Core-Plus Mathematics Project. All rights reserved. 1

Examples of Tasks from ©2008 Course 2, Unit 7

Getting StartedThe tasks below are selected with the intent of presenting key ideas and skills. Not every answer iscomplete, 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 increasingstudent 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 entriesfor assistance.

As you read these selected homework tasks and solutions, you will notice that some very sophisticatedcommunication skills are expected. Students develop these over time. This is the standard for which to

strive. See Research on Communication.

The Geometry page or the Scope and Sequence (2nd edition) might help you follow the conceptual

development of the ideas you see in these examples.

Main Mathematical Goals for Unit 7Upon 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 and applications of the sine, cosine, and tangent ratios of acute angles in right

triangles.

• determine measures of sides and angles for nonright triangles using the Law of Sines and Law of

Cosines.

• 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)

http://www.wmich.edu/cpmp/parentresource2/questions.html

http://www.wmich.edu/cpmp/parentresource2/questions-s.html

http://www.wmich.edu/cpmp/parentresource2/researchoncommunication.html

http://www.wmich.edu/cpmp/parentresource2/geometrystrand.html

http://www.wmich.edu/cpmp/parentresource2/scopeandsequence2ndEd.pdf



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)

Selected Homework Tasks and Expected Solutions

(These solutions are for tasks in the 2nd edition book—2008 copyright.For homework tasks in books with earlier copyright dates, see Helping with Homework.)

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

a. 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%

c. To be completed by the student.

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 thepoint 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 andtangent 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 tablefor a 120˚ angle. The sin 60˚ = sin 120˚.

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

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

http://www.wmich.edu/cpmp/parentresource2/helpingwithhomework.html



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. 9 3

d. 13


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. OAP 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 OB is a radius.

i. a = -sin 30˚

ii. b = cos 30˚

c.cos 30˚ – sin 30˚

sin 30˚ cos 30˚

d.32 – 1212

32

0

1 =

– 1232

The result is equivalent to the coordinates of B .



e.cos – sin

sin cos

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 thestudent.


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˚ = a8a = 8 tan 28˚ 4.2538 4 ftm B = 90˚ – m A = 62˚

b, c. To be completed by the student.


a. tan = 20; 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 thediagonals 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 andCE 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 intothe 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.

If your student has done Connections Task 9, the area formula for a triangle given in that task can be usedhere. The solution given below does not use that formula. This solution uses the standard area formula for

a triangle, area = 0.5(base height).

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. In CBD, the length of the

altitude from point B can be found by solving sin 53˚ = h24 . So, h = 24 sin 53˚, and the area of

CBD = 12 (53.5)(24) sin 53˚ 513 m2. Area ABCD 191 m2 + 513 m2 = 704 m2.


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 thePythagorean 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. This gives CE 1.6 ft. Use CE to find

BE 1.2 ft and thus BD 2.4 ft.


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 = – 12So, AB BC . Therefore, ABC is a right triangle. (Students

might choose to find the lengths of the sides as requested inPart b and use the converse of the Pythagorean Theorem.)

b. AB = 20 = 2 5

BC = 80 = 4 5

AC = 10So, the perimeter is 10 + 6 5 .

c. To be completed by the student.


a. m P = 180˚ – 30.0˚ – 135.6˚ = 14.4˚

Use the Law of Sines to determine CP.CP

sin 135.6˚ = 1,500sin 14.4˚

CP = 1,500 sin 135.6˚sin 14.4˚ 4,220 feet

(Old Glory’s record throw is listed as 4,224.00 feet on the Punkin’ Chunkin’ Web site:

www.atbeach.com/punkinchunkin/)

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, ConnectionsTask 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. 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.

All other parts of this task are to be completed by the student.


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

diagram, BFD is a 30˚-60˚ right triangle withhypotenuse of length 4 m. Thus, DF = 2.

Alternatively, DF = 4 sin 30˚ = 2. The height ofthe 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 2008 Course 2, Unit 7

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