100% de satisfacción garantizada Inmediatamente disponible después del pago Tanto en línea como en PDF No estas atado a nada 4,6 TrustPilot
logo-home
Examen

Solution Manual for Trigonometry, 5th Edition by Cynthia Y. Young | 2025–2026 Updated Answers

Puntuación
-
Vendido
-
Páginas
954
Grado
A+
Subido en
02-02-2026
Escrito en
2025/2026

This Solution Manual for Trigonometry, 5th Edition by Cynthia Y. Young provides complete step-by-step solutions for all exercises in the textbook, fully updated for 2026 academic standards. It is an essential resource for students and instructors seeking detailed explanations of trigonometric concepts, problem-solving strategies, identities, equations, and applications. Ideal for exam preparation, homework assistance, and mastering trigonometry.

Mostrar más Leer menos
Institución
Trigonometry
Grado
Trigonometry











Ups! No podemos cargar tu documento ahora. Inténtalo de nuevo o contacta con soporte.

Escuela, estudio y materia

Institución
Trigonometry
Grado
Trigonometry

Información del documento

Subido en
2 de febrero de 2026
Número de páginas
954
Escrito en
2025/2026
Tipo
Examen
Contiene
Preguntas y respuestas

Temas

Vista previa del contenido

Solution manual for trigonometry 5th edition
Solution
by cynthia
manual
y young.pdf
for trigonometry 5th edition by cynthia y young Page 1 of 954




Solution Manual for Trigonometry, 5th Edition by Cynthia Y. Young
CHAPTER 1
Section 1.1 Solutions --------------------------------------------------------------------------------
1 x 1 x
1. Solve for x:  2. Solve for x: 
2 360∘ 4 360∘
360∘  2x, so that x  180∘ . 360∘  4x, so that x  90∘ .

1 x 2 x
3. Solve for x:   4. Solve for x:  
3 360∘ 3 360∘
360∘  3x, so that x  120∘ . 720∘  2(360∘ )  3x, so that x  240∘ .
(Note: The angle has a negative (Note: The angle has a negative
measure since it is a clockwise measure since it is a clockwise rotation.)
rotation.)
5 x 7 x
5. Solve for x:  6. Solve for x: 
6 360∘ 12 360∘
1800∘  5(360∘ )  6x, so that x  300∘ . 2520∘  7(360∘ )  12x, so that x  210∘ .

4 x 5 x
7. Solve for x:   8. Solve for x:  
5 360∘ 9 360∘
1440∘  4(360∘ )  5x, so that 1800∘  5(360∘ )  9x, so that
x  288∘ . x  200∘ .
(Note: The angle has a negative (Note: The angle has a negative
measure since it is a clockwise measure since it is a clockwise rotation.)
rotation.)

9. 10.
a) complement: 90∘ 18∘  72∘ a) complement: 90∘  39∘  51∘
b) supplement: 180∘ 18∘  162∘ b) supplement: 180∘  39∘  141∘

11. 12.
a) complement: 90∘  42∘  48∘ a) complement: 90∘  57∘  33∘
b) supplement: 180∘  42∘  138∘ b) supplement: 180∘  57∘  123∘



1

Page 1 Solution manual for trigonometry 5th edition
Solution
by cynthia
manualyfor
young.pdf
trigonometry 5th edition by cynthia y young.pdf

,Solution manual for trigonometry 5th edition
Solution
by cynthia
manual
y young.pdf
for trigonometry 5th edition by cynthia y young Page 2 of 954


Chapter 1


13. 14.
a) complement: 90∘  89∘  1∘ a) complement: 90∘  75∘  15∘
b) supplement: 180∘  89∘  91∘ b) supplement: 180∘  75∘  105∘

15. Since the angles with measures  4x ∘ and  6x ∘ are assumed to be
complementary, we know that  4x ∘   6x ∘  90∘. Simplifying this yields

10x ∘  90∘ , so that x  9. So, the two angles have measures 36∘ and 54∘ .

16. Since the angles with measures 3x ∘ and 15x ∘ are assumed to be
supplementary, we know that 3x ∘  15x ∘  180∘. Simplifying this yields

18x ∘  180∘ , so that x  10. So, the two angles have measures 30∘ and 150∘ .

17. Since the angles with measures 8x ∘ and  4x ∘ are assumed to be
supplementary, we know that  8x ∘   4x ∘  180∘. Simplifying this yields

12x ∘  180∘ , so that x  15. So, the two angles have measures 60∘ and 120∘ .

18. Since the angles with measures 3x 15 ∘ and 10x 10 ∘ are assumed to be
complementary, we know that 3x 15 ∘  10x 10 ∘  90∘. Simplifying this yields
13x  25 ∘  90∘ , so that 13x ∘  65∘ and thus, x  5. So, the two angles have
measures 30∘ and 60∘ .

19. Since       180∘ , we know 20. Since       180∘ , we know
that that
117∘ 33∘    180∘ and so,   30∘ . 110∘ 45 ∘    180∘ and so,   25∘ .
– – – –
 150∘  155∘



21. Since       180∘ , we know 22. Since       180∘ , we know
that that
 4          180∘ and so,   30∘. 3         180∘ and so,   36∘.
–– –– –– ––
 6   5

Thus,   4   120∘ and     30∘ . Thus,   3  108∘ and     36∘ .


2

Page 2 Solution manual for trigonometry 5th edition
Solution
by cynthia
manualyfor
young.pdf
trigonometry 5th edition by cynthia y young.pdf

,Solution manual for trigonometry 5th edition
Solution
by cynthia
manual
y young.pdf
for trigonometry 5th edition by cynthia y young Page 3 of 954


Section 1.1



23.   180 ∘  53.3∘  23.6 ∘   103.1∘ 24.   180 ∘  105.6 ∘ 13.2∘   61.2 ∘

25. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 42  32  c2 , which
simplifies to c2  25, so we conclude that c  5.

26. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 32  32  c2 , which
simplifies to c2  18, so we conclude that c  18  3 2 .

27. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 62  b2  102 , which
simplifies to 36  b2  100 and then to, b2  64, so we conclude that b  8.

28. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes a2  72  122 , which
simplifies to a2  95, so we conclude that a  95 .

29. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 82  52  c2 , which
simplifies to c2  89, so we conclude that c  89 .

30. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 62  52  c2 , which
simplifies to c2  61, so we conclude that c  61 .

31. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 72  b2  112 , which
simplifies to b2  72, so we conclude that b  72  6 2 .

32. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes a2  52  92 , which
simplifies to a2  56, so we conclude that a  56  2 14 .




3

Page 3 Solution manual for trigonometry 5th edition
Solution
by cynthia
manualyfor
young.pdf
trigonometry 5th edition by cynthia y young.pdf

, Solution manual for trigonometry 5th edition
Solution
by cynthia
manual
y young.pdf
for trigonometry 5th edition by cynthia y young Page 4 of 954


Chapter 1


33. Since this is a right triangle, we know from the Pythagorean Theorem that

 7
2
a 2  b 2  c 2 . Using the given information, this becomes a 2   5 2 , which

simplifies to a2  18, so we conclude that a  18  3 2 .

34. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 52  b2  102 , which
simplifies to b2  75, so we conclude that b  75  5 3 .

35. If x  10 in., then the hypotenuse 36. If x  8 m, then the hypotenuse of
of this triangle has length
this triangle has length 8 2  11.31 m .
10 2  14.14 in.

37. Let x be the length of a leg in the given 45∘  45∘  90∘ triangle. If the
hypotenuse of this triangle has length 2 2 cm, then 2 x  2 2, so that x  2.
Hence, the length of each of the two legs is 2 cm .

38. Let x be the length of a leg in the given 45∘  45∘  90∘ triangle. If the hypotenuse
10 10
of this triangle has length 10 ft., then 2 x  10, so that x    5.
2 2
Hence, the length of each of the two legs is 5 ft.

39. The hypotenuse has length 40. Since 2x  6m  x  6 2
 3 2m,
 
2
2 4 2 in.  8 in. each leg has length 3 2 m.

41. Since the lengths of the two legs of the given 30∘  60∘  90∘ triangle are x and
3 x, the shorter leg must have length x. Hence, using the given information, we
know that x  5 m. Thus, the two legs have lengths 5 m and 5 3  8.66 m, and
the hypotenuse has length 10 m.

42. Since the lengths of the two legs of the given 30∘  60∘  90∘ triangle are x and
3 x, the shorter leg must have length x. Hence, using the given information, we
know that x  9 ft. Thus, the two legs have lengths 9 ft. and 9 3  15.59 ft., and
the hypotenuse has length 18 ft.



4

Page 4 Solution manual for trigonometry 5th edition
Solution
by cynthia
manualyfor
young.pdf
trigonometry 5th edition by cynthia y young.pdf
$21.99
Accede al documento completo:

100% de satisfacción garantizada
Inmediatamente disponible después del pago
Tanto en línea como en PDF
No estas atado a nada

Conoce al vendedor

Seller avatar
Los indicadores de reputación están sujetos a la cantidad de artículos vendidos por una tarifa y las reseñas que ha recibido por esos documentos. Hay tres niveles: Bronce, Plata y Oro. Cuanto mayor reputación, más podrás confiar en la calidad del trabajo del vendedor.
Edupac Teachme2-tutor
Ver perfil
Seguir Necesitas iniciar sesión para seguir a otros usuarios o asignaturas
Vendido
4365
Miembro desde
1 año
Número de seguidores
4
Documentos
752
Última venta
17 horas hace

4.7

183 reseñas

5
155
4
12
3
7
2
5
1
4

Recientemente visto por ti

Por qué los estudiantes eligen Stuvia

Creado por compañeros estudiantes, verificado por reseñas

Calidad en la que puedes confiar: escrito por estudiantes que aprobaron y evaluado por otros que han usado estos resúmenes.

¿No estás satisfecho? Elige otro documento

¡No te preocupes! Puedes elegir directamente otro documento que se ajuste mejor a lo que buscas.

Paga como quieras, empieza a estudiar al instante

Sin suscripción, sin compromisos. Paga como estés acostumbrado con tarjeta de crédito y descarga tu documento PDF inmediatamente.

Student with book image

“Comprado, descargado y aprobado. Así de fácil puede ser.”

Alisha Student

Preguntas frecuentes