Baixe Exercícios de Trigonometria: Resolvendo Problemas com Seno, Cosseno e Tangente e outras Exercícios em PDF para Matemática, somente na Docsity!
6 × a 2 = 12 ⇔ 6 a = 24 ⇔ a = 4 b × 8 2 = 12 ⇔ 8 b = 24 ⇔ b = 3 Por exemplo, a = 2 e b = 12.
Por exemplo: Triˆangulo 1 Triˆangulo 2 Triˆangulo 3 Base 6 3 12 Altura 4 8 2
6.2. Tem-se a × b = 24 Resposta: (C)
Resposta: (C)
Resposta: (B)
1 − 3 (x^ −^ 2) 4 ≥ − 2 ∧ 1 − 3 (x^ −^ 2) 4 < 1 ⇔ 4 − 3 x + 6 ≥ − 8 ∧ 4 − 3 x + 6 < 4 ⇔ − 3 x ≥ − 18 ∧ − 3 x < − 6
⇔ x ≤ 6 ∧ x > 2 ⇔ x ∈ ]2, 6]
P´ag. 63 1.1.
sin α =^12 ,^5 32 , 5
cos α = 30 32 , 5
=^12
tan α =^12 ,^5 30
sin α =
cos α =
tan α =
sin α =
cos α =
tan α =
sin α =
cos α =
tan α =
2. CE =
42 + 2^2 =
sin
D CEˆ
cos
D CEˆ
tan
D CEˆ
=^2
=^1
sin
C EDˆ
cos
C EDˆ
tan
C EDˆ
3. AC^2 = BC^2 − AB^2
AC^2 = 5^2 − 42 ⇔ AC =
9 ⇔ AC = 3
sin α =^3 5 tan α =^3 4
tan α =
BC
AB
4.2. a) Seja AB = BC = x
x^2 + x^2 = 1 ⇔ 2 x^2 = 1 ⇔ x^2 =
Como x > 0, x =
AB = BC =
e AC = 1
sin α =
BC
AC
√ 2 2 1 =
b)
cos α =
AB
AC
√ 2 2 1
P´ag. 64
5.1. a) [AB ]
b) [AC ]
c) [BC ]
5.2.
6 5
EB^2 + 1, 22 = 2^2
EB =
EB = 1, 6
Os triˆangulos [ABC ] e [BED] s˜ao semelhantes (s˜ao triˆangulos retˆangulos em um ˆangulo agudo comum – crit´erio AA).
A raz˜ao de semelhan¸ca que transforma [BED] em [ABC ] ´e (^) DBBC = 42 = 2. Logo, AC = DE × 2 = 1, 2 × 2 = 2, 4 AB = EB × 2 = 1, 6 × 2 = 3, 2
a)
sin α =
AB
BC
tan α =
AC =
52 + 3^2 ⇔ AC =
cos α =
√^3
sin α = √^5 34
1 − 2cos^2 α + sin^2 α = 1 − 2
= 1 − 2 ×
−1 + (cos x + sin x)^2 cos x
−1 + cos^2 x + 2 sin x cos x + sin^2 x cos x
sin^2 x + cos^2 x
= −1 + 1 + 2 sin^ x^ cos^ x cos x = 2 sin^ x^ cos^ x cos x = 2 sin x
5.1. (1 − sin α) (1 + sin α) + (1 − cos α) (1 + cos α) = 1 − sin^2 α + 1 − cos^2 α = 1 + 1 −
sin^2 α + cos^2 α
1 − cos^2 α tan^2 α
sin^2 α sin^2 α cos^2 α
sin^2 αcos^2 α sin^2 α = cos^2 α
5.3. (sin α + cos α)^2 − (sin α − cos α)^2 = sin^2 α + cos^2 α + 2 sin α cos α −
sin^2 α + cos^2 α − 2 sin α cos α
= 1 + 2 sin α cos α − 1 + 2 sin α cos α = 4 sin α cos α
P´ag. 66
cos x = 0, 2 =^1 5 a^2 + 1^2 = 5^2 a^2 = 24 a =
a = 2
sin x =^2
tan x =
sin x − 2 tan x =
5 −^2 ×^2
5 =^ −^
sin x =
7.1.( sin 2 α + cos^2 α = 1 2 5
- cos^2 α = 1 ⇔ cos^2 α = 1 −
⇔ cos^2 α =
Logo, cos α =
tan α = sin α cos α
2 √^5 21 5
√^2
tan α
- cos α = 3 − x 2 sin α = x − 1 ⇔ sin α = x − 1 2 sin( 2 α + cos^2 α = 1 x − 1 2
- (3 − x)^2 = 1 ∧ 0 < 3 − x < 1 ∧ 0 < x − 1 2
x^2 − 2 x + 1 4
- 9 − 6 x + x^2 = 1 ∧ 2 < x < 3 ∧ 1 < x < 3
⇔ x^2 − 2 x + 1 + 36 − 24 x + 4x^2 − 4 = 0 ∧ 2 < x < 3 ⇔ 5 x^2 − 26 x + 33 = 0 ∧ 2 < x < 3
⇔ x =^26 ±^
262 − 4 × 5 × 33
∧ 2 < x < 3 ⇔
x = 3 ∨ x =^11 5
∧ (2 < x < 3) ⇔ x =^11 5
cos α = 3 − x = 3 −
sin α = x^ −^1 2
=^1
=^1
× 6
=^3
tan 45◦^ = 1 ⇔ sin 45◦ cos 45◦^ = 1 ⇔ sin 45◦^ = cos 45◦
sin^245 ◦^ + cos^245 ◦^ = 1 ⇔ cos^245 ◦^ + cos^245 ◦^ = 1 (sin 45◦^ = cos 45 ◦)
⇔ 2cos^245 ◦^ = 1 ⇔ cos^245 ◦^ =^1 2 ⇔ cos 45◦^ =
(cos 45◦^ > 0)
⇔ cos^245 ◦^ =
Resposta: (B)
tan 60◦^ = sin 60
◦ cos 60◦
sin 60◦^ = cos (90◦^ − 60 ◦) = cos 30◦^ =
tan 60◦^ =
√ 3 2 1 2
2 =^
P´ag. 67
1.1.
cos 30◦^ × 2 sin 60◦^ + tan^230 ◦^ =
× 2 ×
(3)
5.1. M RAˆ = 180◦^ − 90 ◦^ − 42 ◦^ = 48◦
M R
= tan 42 ⇔ M R = 30 tan 42 ⇔ M R ≈ 27 , 0 cm
30 RA
= cos 42◦^ ⇔ RA = 30 cos 42◦^ ⇔ RA ≈ 40 , 4 cm
M RAˆ = 48◦^ ; M R ≈ 27 , 0 cm; RA ≈ 40 , 4 cm.
5.2. R OIˆ = 180◦^ − 90 ◦^ − 27 ◦^ = 63◦ RI 10 = cos 27◦^ ⇔ RI = 10 cos 27◦^ ⇔ RI ≈ 8 , 9 cm
OI 10 = sin 27◦^ ⇔ OI = 10 sin 27◦^ ⇔ OI ≈ 4 , 5 cm
R OIˆ = 63◦^ ; RI ≈ 8 , 9 cm; OI ≈ 4 , 5 cm.
5.3. RU 2 = 7^2 + 6^2 ⇔ RU =
49 + 36 ⇔ RU =
85 ⇔ RU ≈ 9 , 2 cm tan
U RAˆ
=^6
U RAˆ ≈ 40 , 6 ◦
tan
A U Rˆ
A U Rˆ ≈ 49 , 4 ◦
RU ≈ 9 , 2 cm; U RAˆ ≈ 40 , 6 ◦^ ; A U Rˆ = 49, 4 ◦^.
5.4. 182 = 15^2 + IA 2 ⇔ IA 2 = 18^2 − 152 ⇔ IA =
IA ≈ 9 , 9 cm cos
I T Aˆ
I T Aˆ ≈ 33 , 6 ◦
sin
T AIˆ
T AIˆ ≈ 56 , 4 ◦
IA ≈ 9 , 9 cm; I T Aˆ ≈ 33 , 6 ◦^ ; T AIˆ ≈ 56 , 4 ◦^.
5.5. tan a =^6 6 ⇔ tan a = 1. Logo, a = 45◦
tan b = 6 10
. Logo, b ≈ 30 , 96 ◦
x = a − b ≈ 45 ◦^ − 30 , 96 ◦^ ≈ 14 , 0 ◦
CB
= sin (20◦) ⇔ CB =
sin (20◦) 5 M C = tan (20◦) ⇔ M C =
tan (20◦)
Per´ımetro = 10 + 2 × 5 sin (20◦)
Area = ´
10 × (^) tan(20^5 ◦) 2
tan (20◦)
Per´ımetro 39,24 cm; Area´ 68,69 cm^2.
CM
2 = tan (
◦) ⇔ CM = 2 tan (70◦)
2 AC
= cos (70◦) ⇔ AC =
cos 70◦ CB 2 = CM 2
- M B 2 ⇔ CB 2 = (2 tan (70))^2 + 6^2 ⇔ CB =
4tan^2 (70◦)^2 + 36 Per´ımetro = 8 +
cos (70◦) +
4tan^2 (70◦) + 36 ≈ 21 , 98
Area = ´^8 ×^ 2 tan (
= 8 tan (70◦) ≈ 21 , 98
Per´ımetro 21,98 cm Area´ 21,98 cm^2
P´ag. 69
- (^) a
10 = sin (75◦) ⇔ a = 10 sin (75◦) ⇔ a ≈ 9 , 7
b 10 = sin (50◦) ⇔ b = 10 sin (50◦) ⇔ b ≈ 7 , 7
Altura m´ınima: 7,7 cm; altura m´axima: 9,7 cm.
2.1. h 8 , 5 = tan (
◦) ⇔ h = 8, 5 tan (42◦) ⇔ h ≈ 7 , 7
s = tan (42◦) ⇔ s = 3 ,^2 tan (42◦) ⇔ a ≈ 3 , 6 m
cos α =^1 ,^5 2 , 5 ⇔ cos α =^3 5 Logo, α = 53, 1 ◦^.
4.1. OA = OB
360 ◦^ : 5 = 72◦^ ; 72◦^ : 2 = 36◦^ ; A OMˆ = 36◦
M A
r = tan (36◦) ⇔ M A = r tan (36◦)
Per´ımetro = 5 × BA = 5 × 2 r tan (36◦) = 10r tan (36◦)
A = Apent´agono − Ac´ırculo = 5 × AB × r 2 − πr^2 = (r = 1)
=
× 2 tan (36◦) − π × 12 = 5 tan (36◦) − π A ≈ 0 , 5 cm^2
P´ag. 70
AC N C
= sin (28◦) ⇔ 5 N C
= sin (28◦) ⇔ N C = 5 sin (28◦)
N D = N C − DC =
sin (28◦)
AB =
cm = 2 cm 1 OM
= tan (22, 5 ◦)
OM = 1 tan (22, 5 ◦)
Abase = AB^ ×^ OM 2
× 8 =^2 ×^ OM
× 8 = 8OM = 8
tan (22, 5 ◦)
V =
× Abase × altura =
×
tan (22, 5 ◦)
× 6 =
tan (22, 5 ◦) V ≈ 38 , 6 cm^3
- (^) x
100 = tan 5◦^ ⇔ x = 100 tan 5◦
x ≈ 8 , 7 m
- 90 ◦^ − 30 ◦^ = 60◦ OM 6
= tan 30◦^ ⇔ OM = 6
3 ×
⇔ OM =
6 × 3
⇔ OM = 6
OP
= tan 60◦^ ⇔ OP = 6
3 ×
3 ⇔ OP = 18
Maria: 6 m; Pedro: 18 m.
- 52 = AB^2 + 4^2 ⇔ AB =
25 − 16 ⇔ AB = 3
sin α =
tan α =
5 sin α − 3 tan α = 5 ×
− 3 ×
- (sin α − cos α) (sin α + cos α) + 2cos^2 α =
[
(a − b) (a + b) = a^2 − b^2
]
= sin^2 α − cos^2 α + 2cos^2 α = = sin^2 α + 2cos^2 α = 1
P´ag. 73 1.1. OC r = cos α ⇔ OC = r cos α
BC r = sin α ⇔ BC = r sin α
Area de [^ ´ OAB] = 2 × Area de [´ OCB] = 2 × OC^ ×^ BC 2 =^ OC^ ×^ BC^ =^ r^ cos^ α^ ×^ r^ cos^ α^ =^ r
(^2) sin α cos α
1.2. r = 1 e α = 45◦
A[OAB] = 1^2 × sin 45◦^ × cos 45◦^ =
×
- (^) x
2 , 5 = cos (42◦) ⇔ x = 2, 5 cos (42◦)
h = 2, 5 − x = 2, 5 − 2 , 5 cos (42◦) h ≈ 0 , 64 m
3.1. Seja a o afastamento do topo da torre: a 56 = sin (5, 5 ◦) ⇔ a = 56 × sin (5, 5 ◦)
a ≈ 5 , 37 56 × sin (5, 5 ◦) 6 = 5, 2
3.2. 3 , 9 h = sin (3, 93 ◦)
h =
sin (3, 93 ◦) h ≈ 56 , 9 m
P´ag. 74
- πr^2 = 72π ⇔ r^2 = 72 R^2 = 3^2 + r^2 ⇔ R^2 = 9 + 72 ⇔ R = 9 V =
π × 93 = 972π
V = 972π cm^3
5.1. 120 × 30 = 60 × 60 = 80 × 45
5.2. 120 × 30 = 3600. Representa o n´umero de unidades de ra¸c˜ao que o Sr. Joaquim tem em armaz´em (uma unidade ´e a quantidade de ra¸c˜ao que uma vaca consome num dia).
5.3. 3600 : 90 = 40 A ra¸c˜ao daria para 40 dias.
5.4. xy = 3600 ⇔ y =^3600 x
x − 1 4 (1)
(2)
(3x − 1) < (^2) (4)
∧ 2 x − 1 −^ x 2 ≥ 2 ⇔ x − 1 − 2 (3x − 1) < 8 ∧ 4 x − 1 + x ≥ 4 ⇔
⇔ x − 1 − 6 x + 2 < 8 ∧ 5 x ≥ 5 ⇔ − 5 x < 7 ∧ x ≥ 1 ⇔ x > −
∧ x ≥ 1 ⇔ x ≥ 1
S = [1, +∞[
7.1. O sistema define o conjunto de pontos de interse¸c˜ao da reta de equa¸c˜ao y = x + 1 com a par´abola de equa¸c˜ao y = 2x^2.
