Trigonometric Methods

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. (explorar el seno, coseno, las funciones tangents definidas en términos de un punto de la parte terminal de un ángulo en posición estándar en un plano coordenado.)

•

explore properties of the sine, cosine, and tangent ratios of acute angles in right triangles and use those ratios to solve applied problems. (explorar las propiedades y aplicaciones del seno, coseno, y porciones tangentes de ángulos agudos en los triángulos rectos.)

•

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. (derivar la ley senos y la ley cosenos y usar esas leyes para determinar las medidas y ángulos para triángulos que no son rectos.)

•

use the Law of Sines and Law of Cosines to solve a variety of applied problems that involve triangulation. (usar la Ley de Senos de la Ley y Cosenos para resolver una variedad de problemas aplicados que implican la triangulación.)

•

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. (describir las condiciones en las que dos, uno, o cero triángulos son determinados dado las longitudes de dos lados y la medida de un ángulo no incluido entre los dos lados.)

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)

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Trigonometric Methods

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 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. (Para ser completado por el estudiante.)

b. percent grade = 100 sin A = 100 × 0.25 = 12.5% 2 percent grade = 100 sin A = 100 × 0.33 = 16.5% 2

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. (Para ser completado por el estudiante.) 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˚. (Dado que los valores trigonométricas resultan de las porciones de las coordenadas y el radio del círculo, usted puede encontrar los valores trigonométricas de los ángulos en el Cuadrante II (ángulos obtusos) utilizando la imagen del punto en el círculo que se refleja en todo el eje y. Así, por ejemplo, un punto (–x, y) en el Cuadrante II tiene el punto de imagen (x, y) en el Cuadrante I. La coordenada negativa x en el Cuadrante II significa que el coseno y los valores del tangente serán negativos en el Cuadrante II. Entonces, para usar la tabla en la página 465 para 120˚ o P12, refleje P12 en el eje de P6. Utilice la negativa del cos 60˚ y tan 60˚ valores de la función de la tabla para obtener un ángulo de 120˚. El 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. (La pendiente es tan θ. La ecuación de la línea es y = (tan θ)x.)

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Trigonometric Methods

Lesson 1, Investigation 1, Review Task 32 (p. 485) a. x ≈ 1.8 b. x = 5 c. x = 20.25 d. x ≈ –1.3

Lesson 1, Investigation 1, Review Task 33 (p. 485) a. 2 15 b. 5 3 c–e. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 1, Investigation 2, Applications Task 5 (p. 475) a. Using d to represent the distance across the river, tan 31˚ = d and d = 400 tan 31˚ ≈ 240.344 ≈ 400 240 meters. (d representa la distancia de un lado del río al otro.) b. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 1, Investigation 2, Connections Task 13 (p. 479) a. rOA′P is a 30˚-60˚ right triangle. Using the diagram at the right, x =

3 and y = 1 . So, A

2

2

( 23 , 12 ). As seen in

Connections Task 12, cos 30˚ =

3 and sin 30˚ = 1 . So,

2

2

the coordinates of point A can be written as (cos 30˚, sin 30˚). (rOAP es un triángulo recto de 30˚-60˚. Utilizando el diagrama a la derecha, x = Conexiones de Tareas 12, cos 30˚ =

3 y y = 1 . Por lo tanto, A

2

2

( 23 , 12 ). Como se ha visto en

3 y sin 30˚ = 1 . Entonces, las coordenadas del punto A

2

2

pueden ser escritas como (cos 30˚, sin 30˚).) b. OB′ = 1 because OB! 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) →

(

= –1, 2

( 23 , 12 ). In Part b, we identified the coordinates of the image of B as (–sin 30˚, cos 30˚) 3 . These together verify the entries of matrix R . (Las entradas de la primera 30˚ 2 )

columna de la matriz R30˚ son las coordenadas de la imagen de A(1, 0), y las entradas en la segunda columna son las coordenadas de la imagen B(0, 1). En la parte a, verificamos que © 2015 Core-Plus Mathematics Project. All rights reserved.

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Trigonometric Methods

( 23 , 12 ). En la parte b, identificamos las coordenadas de la imagen de B como (–sin 30˚, cos 30˚) = (– 1 , 3 ). Estos juntos verifican las entradas de la matriz R30˚.) 2 2 (1, 0) →

! cos!30˚ – sin!30˚ $ cos!30˚ &%

ii. # " sin!30˚ d.

! # # #"

3

2 1 2

–1$ 2& 3& 2 &%

!– 1 $ !0 $ # 2& # & = # & 1 3 " % #" 2 %&

The result is equivalent to the coordinates of B′. (El resultado es equivalente a las coordenadas de la B′.) d, e. To be completed by the student. (Para ser completado por el estudiante.)

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˚ = h , x h = x tan 39˚; so, h ≈ 0.8098x. The remainder of the solution is left to the student. (Digamos que “x” es la distancia restante de la letra "B" hasta la base de la montaña. Luego tan 39˚ = h , h = x tan 39˚; x así, h ≈ 0.8098x. El resto de la solución es para ser completado por el estudiante.)

Lesson 1, Investigation 2, Review Task 34 (p. 485) a. The triangles must be congruent by the SSS congruence condition. (Los triángulos deben ser congruentes por la condición de congruencia de “SSS.”) b. The triangles could be different. Their shapes would be the same, but their sizes could be different. (Los triángulos podrían ser diferentes. Sus formas serían iguales, pero sus tamaños podrían ser diferentes.) c–e. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 1, Investigation 3, Applications Task 6 (p. 476) a.

cos 28˚ = 8

c 8 c = cos!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. (Para ser completado por el estudiante.)

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Trigonometric Methods

Lesson 1, Investigation 3, Connections Task 16 (p. 480) a. tan θ = 20; θ ≈ 87˚ b, c. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 1, Investigation 3, Connections Task 17 (p. 480) 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 b, c. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 1, Investigation 3, Review Task 36 (p. 486) a. (x – 5)(x + 2) = 0 x – 5 = 0 or x + 2 = 0 x = 5 or x = –2

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

b. x2 + 9x + 4 = 0

Check:

x = –9!±! 65 2

(–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. (Para ser completado por el estudiante.)

Lesson 2, Investigation 1, Applications Task 1 (p. 503)

!##" a. The intended course is along AC . The!##actual course is marked by arrows, and the distance AB + BC. " (El curso está destinado a lo largo de AC . El curso actual está marcado por flechas, y la distancia AB + BC.)

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

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

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Trigonometric Methods

Lesson 2, Investigation 1, Connections Task 10 (p. 507) 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.) (Los estudiantes deben verificar que los paralelogramos tienen una simetría rotacional de 180˚. (La inclusión de las diagonales en su análisis de la simetría les ayudará en la búsqueda de la información para la Parte b.)) b. Students should be able to determine the angle measures and lengths in the diagram shown here. (Los estudiantes deben ser capaces de determinar las medidas de los ángulos y las longitudes en el diagrama que se muestra aquí.)

Students should explain their answers. (Los estudiantes deben explicar sus respuestas.) 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 rAED, let x = m∠ADE, then m∠DAE = 180˚ – (68˚ + x) = 112˚ – x. sin!x = sin!(112˚!–!x) 2.11 3.81 3.81 ≈ 1.81 sin!x = 2.11 sin!(112˚!–!x) sin!x By letting Y1 = sin!(112˚!–!x) and using tables or graphs, students can see that when y ≈ 1.81, x is sin!x about 79˚. m∠ADE ≈ 79˚; m∠DAE = 112˚ – x ≈ 33˚ (Al dejar Y1 = sin!(112˚!–!x) y por usando tablas

o gráficos, los estudiantes pueden ver que cuando y ≈ 1.81, x es de aproximadamente 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 través de una mayor aplicación de la Ley de Senos o la simetría de rotación, todas las medidas restantes se podrían encontrar.)

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Trigonometric Methods

Lesson 2, Investigation 1, Reflections Task 16 (p. 509) a. Using the Properties of Equality, by taking reciprocals, the first proportion can be changed into the second. . (Usando las propiedades de la igualdad, por tomando los inversos, el primer porcentaje se puede cambiar al segundo.) b. If you are solving for an angle, it is easier to use I sin!A = sin!B = sin!C . If you are solving for a b c a a b c side, it is easier to use II sin!A = sin!B = sin!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. (Si usted está resolviendo para un ángulo, es más fácil usar I sin!A = sin!B = sin!C . b c a a b c Si está resolviendo para un lado, es más fácil usar II sin!A = sin!B = sin!C . Los estudiantes deben entender que las proporciones pueden ser resueltos, ya sea en su forma, pero un menor número de pasos son necesarios si la variable para la que está resolviendo está en el numerador.)

Lesson 2, Investigation 1, Extensions Task 23 (p. 511) a. Use the Law of Cosines to determine m∠B. Then h = sin B and h = a sin B. Or, instead, find m∠A a and write h = b sin A. (Utilice la Ley de cosenos para determinar m∠B. Entonces h = sin B y a h = a sin B. O, en su lugar, encontrar m∠A y escribir h = b sin A.) b. The area of rABC = 12 hc = 12 ac sin B or 12 bc sin A.

Lesson 2, Investigation 2, Applications Task 5 (p. 504) a.

b. To be completed by the student. (Para ser completado por el estudiante.) c. In rABD, the length of the altitude from point A can be found by solving sin 19˚ = h . So, 27 h = 27 sin 19˚, and the area of rABD = 12 (43.5)(27) sin 19˚ ≈ 191 m2. The remainder of the solution is left to the student. (En rABD, la longitud de la altitud del punto A puede ser encontrada con la solución sin 19˚ = h . Por lo tanto, h = 27 sin 19˚, y el área de rABD = 12 (43,5)(27) sin 19˚ ≈ 191 27 m2. El resto de la solución es para ser completado por el estudiante.)

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Trigonometric Methods

Lesson 2, Investigation 2, Connections Task 13 (p. 508) 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 = CE . To find m∠1, use the Law of 2

Cosines to see that cos ∠1 ≈ 0.7906. The remainder of the solution is left to the student. (Por la simetría, m∠ABC = m∠ADC, tenemos AC2 = 22 + 32 – 2(2)(3) cos 118˚. AC ≈ 4,32 ≈ 4 ft. Para encontrar BE, usted puede encontrar CE o AE y utilizar el teorema de Pitágoras. Para encontrar CE, se puede usar cos ∠1 = CE . Para encontrar m∠1, use la 2

Ley de cosenos para ver que cos ∠1 ≈ 0,7906. El resto de la solución es para ser completado por el estudiante.)

Lesson 2, Investigation 2, Reflections Task 18 (p. 510) a. Use the Law of Cosines to find AC. AC2 = 62 + 42 – 2(6)(4) cos 61˚. One solution is expected. (Una solución es esperada.) b–d. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 2, Investigation 2, Review Task 31 (p. 514) a. Using slopes: slope of AB = 2 slope of BC = – 1 2

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.) (Por lo tanto, AB!!!BC . Entonces, rABC es un triángulo recto. Los estudiantes podrían encontrar las longitudes de los lados como solicitado en la parte b y usar el opuesto del teorema de Pitágoras.) 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. (Para ser completado por el estudiante.)

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Trigonometric Methods

Lesson 2, Investigation 3, Applications Task 7 (p. 505) a. m∠P = 180˚ – 30.0˚ – 136.75˚ = 13.25˚ Use the Law of Sines to determine CP. 1,500 CP sin!136.75˚ = sin!13.25˚ CP = 1,500!sin!136.75˚ ≈ 4,484 feet sin!13.25˚

(Young Glory III’s record throw is listed as 4,483.51 feet on the Punkin’ Chunkin’ Web site: www.punkinchunkin.com) (La lanza más larga de “Young Glory III” aparece como 4,483.51 pies en el sitio de Web “Punkin’ Chunkin”: 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. (NOTA DE ENSENANZA: Si los estudiantes usan La ley de senos en la parte b para determinar la medida de ∠J, encontrarán la medida de suplemento de ∠J. Los estudiantes pueden hacer referencia a Lesson 1, Connections Task 8, para ayudarles entender la situación. De Connections Task 8, saben que existen dos posibles ángulos entre 0˚ y 180˚ para cualquier valor del seno (menos 90˚ = 1). Así que, usando sin–1 para determinar la medida de un ángulo en cualquier triángulo debe revelar dos soluciones posibles. Para evitar esta dilema, los estudiantes pueden usar La ley de cosenos.) b, c. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 2, Investigation 3, Connections Task 15 (p. 509) a. iii. Both equations can be true because ∠ACB and ∠AC′B are supplementary. It follows that their sines are equal. (Ambas ecuaciones pueden ser verdaderas porque ∠ACB y ∠AC′B son complementarios. De ello se deduce que sus senos son iguales.) 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. (Que AC = b. Por la Ley de Cosenos, 12 = b2 + ( 3 )2 – 2b( 3 ) cos 30˚. Use esta para demostrar que b2 – 3b + 2 = 0. Las soluciones a esta ecuación representan los posibles valores de AC.) ii.

c.

i. To be completed by the student. (Para ser completado por el estudiante.) 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. (Utilice la Ley de Cosenos de una manera similar a la que se muestra en la solución de la Parte bi para encontrar que c ≈ 3,2.)

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Trigonometric Methods

Lesson 2, Investigation 3, Extensions Task 27 (p. 513) a. When AC = 1.5, m∠ABC will be smallest. Then rABC is equilateral; so, m∠ABC = 60˚. In the diagram, r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. (Cuando AC = 1.5, m∠ABC será la más pequeña. A continuación, rABC es equilátero, así, m∠ABC = 60˚. En el diagrama, rBFD es un triángulo recto de 30˚-60˚ con una hipotenusa de la longitud 4 m. Por lo tanto, DF = 2. Por otra parte, DF = 4 sin 30˚ = 2. La altura del plano es DG y 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˚ (Cuando el cilindro hidráulico está completamente extendida, AC = 2.2 m, por la ley de cosenos, 2.22 = 1.52 + 1.52 – 2(1.5)(1.5) cos (∠ABC). cos (∠ABC) ≈ –0.0756; m∠ABC ≈ 94.3˚) c, d. To be completed by the student. (Para ser completado por el estudiante.)

Lesson 2, Investigation 3, Review Task 34 (p. 515) a.

2 , or 1 18 36

b.

34 , or 17 36 18

c, d. To be completed by the student. (Para ser completado por el estudiante.)

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