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Figura (b).
Lato $b= \dfrac{c·sen(β)}{sen(γ)} = \dfrac{6\sqrt2·sen(45°)}{sen(60°)}=4\sqrt3~cm$;
angolo $α= 180-(β+γ) = 180-(45+60) = 180-105 = 75°$;
area $A= \dfrac{c·b·sen(α)}{2} = \dfrac{6\sqrt2·4\sqrt3·sen(75°)}{2} = 18+6\sqrt3~cm^2$.