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

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

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

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

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

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

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

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

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

\(\Leftrightarrow a^2=32\)

hay \(a=4\sqrt{2}\)

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

a) Diện tích \({S_1}\) của tam giác IAB là: \({S_1} = \frac{1}{2}r.AB = \frac{1}{2}r.c\)

Diện tích \({S_2}\) của tam giác IAC là: \({S_2} = \frac{1}{2}r.AC = \frac{1}{2}r.b\)

Diện tích \({S_3}\) của tam giác IBC là: \({S_3} = \frac{1}{2}r.BC = \frac{1}{2}r.a\)

b) Diện tích S của tam giác ABC là:

 \(\begin{array}{l}S = {S_1} + {S_2} + {S_3} = \frac{1}{2}r.c + \frac{1}{2}r.b + \frac{1}{2}r.a = \frac{1}{2}r.(c + b + a)\\ \Leftrightarrow S = \frac{{r(a + b + c)}}{2}\end{array}\)

\(\widehat{B}=180^o-60^o-45^o=75^o\)

Theo định lý sin ta có:

\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\)

\(\Rightarrow AC=\dfrac{AB\cdot sinB}{sinC}=\dfrac{5\cdot sin75^o}{sin45^o}=\dfrac{5+5\sqrt{3}}{2}\) 

Mà: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot5\cdot\dfrac{5+5\sqrt{3}}{2}\cdot sin60^o=\dfrac{75+25\sqrt{3}}{8}\left(dvdt\right)\)

Nửa chu vi tam giác ABC là:

\(P=\dfrac{a+b+c}{2}=\dfrac{12+16+20}{2}=\dfrac{2\left(6+8+10\right)}{2}=24\)(đvđd)Diện tích tam giác ABC là:

\(S=\sqrt{P\cdot\left(P-a\right)\left(P-b\right)\left(P-c\right)}\)

\(=\sqrt{24\cdot\left(24-12\right)\left(24-16\right)\left(24-20\right)}\)

\(=\sqrt{24\cdot12\cdot8\cdot4}\)

\(=2\sqrt{6}\cdot2\sqrt{3}\cdot2\sqrt{2}\cdot2\) 

\(=16\sqrt{36}=96\)(đvdt)