\(sinA\cdot cos\left(B+C\right)+cosA\cdot sin\left(B+C\right)\)

\(=sinA\cdot cos\left(180^0-A\right)+cosA\cdot sin\left(180^0-A\right)\)

\(=sinA\cdot\left(-cosA\right)+cosA\cdot sinA\)

\(=sinA\left(-cosA+cosA\right)=0\).

Ta có : \(S=\dfrac{abc}{4R}=\dfrac{abc}{4\cdot8}=\dfrac{abc}{32}\)

\(\Rightarrow\dfrac{1}{2}\cdot a\cdot c\cdot sinB=\dfrac{abc}{32}\)

\(\Rightarrow\dfrac{ac}{2}\cdot sin50^0=\dfrac{abc}{32}\)

\(\Rightarrow sin50^0=\dfrac{abc}{32}:\dfrac{ac}{2}\)

\(\Rightarrow sin50^0=\dfrac{abc}{32}\cdot\dfrac{2}{ac}=\dfrac{b}{16}\)

Từ đó , ta được : \(b=16\cdot sin50^0\approx12,257\left(cm\right)\)

Vậy độ dài cạnh AC là xấp xỉ \(12,257cm\).

Ta có : \(\dfrac{1006}{1007}< 1,\dfrac{1007}{1008}< 1,\dfrac{1008}{1009}< 1\)

\(\Rightarrow\dfrac{1006}{1007}+\dfrac{1007}{1008}+\dfrac{1008}{1009}< 1+1+1=3\)

Do đó \(\dfrac{1006}{1007}+\dfrac{1007}{1008}+\dfrac{1008}{1009}< 3\)