Trigonometria

ΑΒ = c ; ΒC = a ; ΑC = b = 12

SIN(γ) = SIN(pi - (α + β))----> SIN(γ) = SIN(α + β)

SIN(α + β) = SIN(α)·COS(β) + SIN(β)·COS(α)

α = pi/3

β = pi/4

SIN(pi/3)·COS(pi/4) + SIN(pi/4)·COS(pi/3)

SIN(γ) = √6/4 + √2/4

Th SENI:

c/SIN(γ) = b/SIN(β)---> c = b·SIN(γ)/SIN(β)

c = 12·(√6/4 + √2/4)/SIN(pi/4)

c = (3·√6 + 3·√2)/SIN(pi/4)

c = (3·√6 + 3·√2)/(√2/2)

c = 6·√3 + 6

c = 6·(√3 + 1)

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Angolo $\small \beta= 60°;$

angolo $\small \gamma= 45°;$

angolo incognito $\small \widehat{ACB}= \alpha= 180°-\beta-\alpha= 180-60-45 = 75°;$

lato $\small \overline{AC}= 12;$

quindi calcola il lato $\small \overline{AB}:$

$\small \overline{AB}= \dfrac{\overline{AC}·\sin(\alpha)}{\sin(\gamma)}$

$\small \overline{AB}= \dfrac{12·\sin(75°)}{\sin(45°)} = 6+6\sqrt3 →\, = 6(\sqrt3 +1).$