Question

Zadania w załączynikach ; )

Answer 1

6.12 a)  x+√3/√3 = √3

x+√3 = 3

x = 3-√3

 

b) x = 3√3 *2 = 6√3

 

c) x = 8√3/3

 

6.16 a)3 sinβ = 3/4 I : 3
sinβ = 1/4 = 0,25
β ≈ 14⁰

b)2 sinβ = 1,3820 I : 2
sinβ = 0,691
β ≈ 44⁰

c)3 cos β = 3√2/2 I : 3
cos β = √2/2
β = 45⁰

d)3 cos β = 2,511
cos β = 0,837
β ≈ 33⁰

e)cos β = √5/5 = 0,4472
β ≈ 63⁰

f)1 + tg β = 12,85
tg β = 12,85 - 1 = 11,85
β ≈ 85⁰

g) (sin β -0,5)( tg β -1) = 0 <=> sinβ = 0,5 lub tg β = 1
β= 30⁰ lub β = 45⁰

h)(cosβ - 3/4)(sin β - √3/2 ) = 0
cosβ = 0,750 lub sin β = √3/2
β ≈ 41⁰ lub β = 60⁰
i)cos²β - 3 = 0
cos²β = 3 - nie ma rozwiązania , bo cos²β ≤ 1

j)cos²β - 1/4 = 0
cos²β = 1/4
cosβ = 1/2
β = 60⁰

tg²β - 3 = 0
tg²β = 3
tg β = √3
β = 60⁰

 

6.17

a) sin 2α = √2 /2
2α = 45 ⁰
α = 22⁰ 30 minut
b) cos 3α = 1/2
3 α = 60⁰
α = 20⁰

c) 3 + √3 tg 2α = 4
√3 tg 2α = 1
tg 2α = 1/ √3
2α = 30⁰
α = 15⁰

d) tg 2α = √3
2 α = 60⁰
α = 30⁰

e) 4 ( 1 + sin 3α) = 6
1 + sin 3α = 6/4 = 3/2
sin 3α = 3/2 -1 = 1/2
3α = 30⁰
α = 10⁰

6.18
a) sin(α - β) = √2 /2 i cos (α + β) = 1/2
α - β = 45⁰ i α + β = 60⁰
2α = 105⁰
α = 52⁰ 30
β = 60⁰ - 52⁰30 = 7⁰ 30

b) tg(α + β) = √3 i cos (α -β) = √3 /2
α + β = 60⁰ i α - β = = 30⁰
2 α = 90⁰
α = 45⁰
β = α - 30⁰ = 15⁰

c)sin(α +β) = √3 /2 i tg (α - β) = √3 /3
α + β = 60⁰ i α - β = 30⁰
2 α = 90⁰
α = 45⁰
β = α - 30 ⁰ = 15⁰

6.19
a) y = (2sin 60⁰)X + tg 45⁰
a = tg α = 2*√3 /2 = √3 , zatem α = 60⁰

b)y = (2cos 30⁰) x + tg 30⁰
a = tg α = 2*√3 /2 = √3, zatem α = 60⁰

c)y =( cos 45⁰/ sin 45⁰) x + 2
a = tg α =[ (√2/2)/(√2/2)] = 1 , zatem α = 45⁰