Question

1/Cho tam giác ABC có b=7,c=5,cosA=\(\dfrac{3}{5}\).Tính R.\(\dfrac{3}{5}\)

Answer 1

Ta có: \(a^2=b^2+c^2-2bc.cosA\Rightarrow a=\sqrt{b^2+c^2-2.b.c.cosA}\Rightarrow a=\sqrt{7^2+5^2-2.7.5.\dfrac{3}{5}}=4\sqrt{2}\)
\(cos^2A+sin^2A=1\Rightarrow sin^2A=1-cos^2A=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\Rightarrow sinA=\dfrac{4}{5}\)Ta có:  \(\dfrac{a}{sinA}=2R\Rightarrow\dfrac{4\sqrt{2}}{\dfrac{4}{5}}=2R\Rightarrow R=\dfrac{5\sqrt{2}}{2}\)
\(R.\dfrac{3}{5}=\dfrac{5\sqrt{2}}{2}.\dfrac{3}{5}=\dfrac{3\sqrt{2}}{2}\)