Question

Trigonometrie clasa a 9-a

Answer 1

$\it \dfrac{b^3+c^3-a^3}{b+c-a}=a^2 \Rightarrow b^3+c^3-a^3=a^2b+a^2c-a^3 \Rightarrow\\ \\ \\ \Rightarrow b^3+c^3=a^2b+a^2c \Rightarrow (b+c)(b^2-bc+c^2)=a^2(b+c)|_{:(b+c)} \Rightarrow\\ \\ \\ \Rightarrow b^2-bc+c^2=a^2 \Rightarrow b^2+c^2-a^2=bc\ \ \ \ \ (1)$

Aplicăm Th. cos pentru unghiul A:

$\it cosA=\dfrac{b^2+c^2-a^2}{2bc}\ \stackrel{(1)}{=}\ \dfrac{bc}{2bc}=\dfrac{1}{2} \Rightarrow m(\hat A) =60^o\ \ \ \ (2)\\ \\ \\ \dfrac{1}{2}=cosA =cos(180^o-(B+C))=-cos(B+C)\ \ \ \ (3)$

$\it sinBsinC =\dfrac{3}{4} \Rightarrow \dfrac{1}{2}[cos(B-C)-cos(B+C)]=\dfrac{3}{4}|_{\cdot2} \Rightarrow \\ \\ \\ \Rightarrow cos(B-C)-cos(B+C)=\dfrac{3}{2}\ \stackrel{(3)}{\Longrightarrow} cos(B-C) +\dfrac{1}{2}=\dfrac{3}{2}|_{-\frac{1}{2}} \Rightarrow\\ \\ \\ \Rightarrow cos(B-C) =1 \Rightarrow B-C = 0 \Rightarrow m(\hat B)=m(\hat C)\Rightarrow\Delta ABC-isoscel\ \ \ (4)$

$\it (2),\ (4) \Rightarrow \Delta ABC-echilateral$