Question

Alguém poderia me ajudar?

Answer 1

a)
Tg30° = x/5
√3/3 = x/5
√3×5 = 3×x
5√3 = 3x
x = (5√3)/3

b)
Cos45° = 10/x
√2/2 = 10/x
√2×x = 2×10
√2x = 20
x = 20/√2
20/√2 × √2/√2 = (20√2)/√4 = (20√2)/2 = 10√2
x = 10√2

c)
sen60° = x/√3
√3/2 = x/√3
√3×√3 = 2×x
√9 = 2x
3 = 2x
x = 3/2

d)
Tg45° = x/7
1 = x/7
x = 1×7
x = 7