a, a=9 ; \(\widehat{A}=180^0-60^0-45^0=75^0\)
\(\frac{a}{sinA}=2R\Rightarrow R=\frac{a}{2.sinA}=\frac{9}{2.sin75}=4,7\)
\(\frac{b}{sinB}=\frac{AC}{sin60}=2R=2.4,7\Rightarrow AC=8,1\)
\(c=2R.sinC=2.4,7.sin45=6,6\)
b, \(S_{ABC}=\frac{1}{a}ab.sinC=\frac{1}{2}.9.8,1.sin45=25,8\)
\(r=\frac{S}{p}=\frac{25,8}{\frac{9+8,1+6,6}{2}}=2,2\)
\(BN^2=m_b^2=\frac{a^2+c^2}{2}-\frac{b^2}{4}=45,9\Rightarrow BN=6,8\)