Question

Sa se afle aria triunghiului ABC daca b=2rad6,a+c=6+rad12 si m(B)=pi/3.

Answer 1

$\it a+c = 6+\sqrt{12} \Rightarrow (a+c)^2 = (6+\sqrt{12})^2 \Rightarrow \\ \\ \Rightarrow a^2+c^2+2ac =36+12+12\sqrt{12} \Rightarrow \\ \\ \Rightarrow a^2+c^2= 48+12\sqrt{12} -2ac \ \ \ \ \ (1)$

Aplicăm teorema cosinusului pentru unghiul B:

$\it cos \dfrac{\pi}{3} = \dfrac{a^2+c^2-(2\sqrt6)^2}{2ac} \Rightarrow \dfrac{1}{2} = \dfrac{a^2+c^2-24}{2ac}|_{\cdot2ac} \Rightarrow \\ \\ \\ \Rightarrow ac = a^2+c^2 -24 \Rightarrow a^2+c^2=ac+24 \ \ \ \ \ (2) \\ \\ \\ (1), (2) \Rightarrow ac+24 = 48+12\sqrt{12} -2ac \Rightarrow 3ac = 24+12\sqrt{12} |_{:6} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{ac}{2} = 4+2\sqrt{12} \ \ \ \ \ (3)$

$\it \mathcal{A} = \dfrac{ac}{2} \cdot sinB \stackrel{(3)}{\Longrightarrow} \mathcal{A} = (4+2\sqrt{12})\cdot sin\dfrac{\pi}{3} = (4+2\sqrt{12})\cdot \dfrac{\sqrt3}{2} = \\ \\ \\ = 4\cdot\dfrac{\sqrt3}{2} +2\sqrt{12} \cdot\dfrac{\sqrt3}{2} = 2\sqrt3+\sqrt{36} =2\sqrt3+6$