Question

TRIGONOMETRIA, AYUDENMEEE

Answer 1

Sen(∅) = Cos(π/2 - Ø)

Tenemos:

Sen(Ø) = Cos(Ø/2)

Cos(π/2 - Ø) = Cos(Ø/2)

Igualando ángulos:

π/2 - Ø = Ø/2

π/2 = Ø + Ø/2

π/2 = 3Ø/2

Ø = π/3

Tenemos:

Tg[(Ø+a)/3] = Ctg[(Ø+a)/2]

Tg[(π+3a)/9] = Ctg[(π+3a)/6]

Tg(x) = Ctg(π/2 - x)

Entonces:

Ctg[π/2-(π+3a)/9] = Ctg[(π+3a)/6]

Igualando ángulos:

π/2-(π+3a)/9 = (π+3a)/6

27π = 15π +45a

27π-15π = 45a

12π = 45a

a = 12π/45

a = 4π/15

M = Sen(Ø/2) - Cos(Ø) + Tg(36°)Tg[(Ø+a)/2]

M = 1.18