Question

Cho tam giác ABC có b = 7 ,c = 5 và cos A = \(\dfrac{3}{5}\). Tính a , sin A, diện tích S của ABC, R, r, h_a

Answer 1

\(a^2=b^2+c^2-2bc.\cos A\Rightarrow a=\sqrt{b^2+c^2-2bc.cosA}=\sqrt{7^2+5^2-\dfrac{2.7.5.3}{5}}=4\sqrt{2}\)

\(\sin A=\sqrt{1-cos^2A}=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

\(p=\dfrac{a+b+c}{2}=6+2\sqrt{2}\)

\(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=14\)

\(R=\dfrac{a}{2.sinA}=\dfrac{4\sqrt{2}}{\dfrac{2.4}{5}}=\dfrac{5\sqrt{2}}{2}\)

\(r=\dfrac{S}{p}=\dfrac{14}{6+2\sqrt{2}}=3-\sqrt{2}\)

\(ha=\dfrac{2S}{a}=\dfrac{2.14}{4\sqrt{2}}=2\sqrt{2}\)

Answer 2

\(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\)

\(\Leftrightarrow7^2+5^2-a^2=\dfrac{3}{5}\cdot2\cdot7\cdot5=3\cdot2\cdot7=42\)

\(\Leftrightarrow a^2=32\)

hay \(a=4\sqrt{2}\)

\(\sin A=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)