Theo định lí sin: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\quad (*)\)

+) Ta có: \(\hat A = {180^o} - \left( {\hat B + \;\hat C} \right) = {180^o} - \left( {{{60}^o} + {{45}^o}} \right) = {75^o}\)

\( \Rightarrow a = \frac{b}{{\sin B}}.\sin A = \frac{{10}}{{\sin {{60}^o}}}.\sin {75^o} \approx 11,154\)

+) \((*) \Rightarrow R = \frac{b}{{2\sin B}} = \frac{{10}}{{2\sin {{60}^o}}} = \frac{{10}}{{2.\frac{{\sqrt 3 }}{2}}} = \frac{{10\sqrt 3 }}{3}.\)

+) Diện tích tam giác ABC là: \(S = \frac{1}{2}ab.\sin {\mkern 1mu} \hat C\) \( \approx \frac{1}{2}.11,154.10.\sin {45^o}\)\( \approx 39,44\)

+) Lại có: \(R = \frac{c}{{2\sin C}}\)\( \Rightarrow c = 2.\frac{{10\sqrt 3 }}{3}.\sin {45^o} = \frac{{10\sqrt 6 }}{3} \approx 8,165\)

\( \Rightarrow p = \frac{{a + b + c}}{2} \approx \frac{{11,154 + 10 + 8,165}}{2} \approx 14,66\)

\( \Rightarrow r = \frac{S}{p} \approx \frac{{39,44}}{{14,66}} \approx 2,7\)

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow a^2=32\)

hay \(a=4\sqrt{2}\)

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

Từ định lí cosin ta suy ra \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{5^2} + {8^2} - {6^2}}}{{2.5.8}} = \frac{{53}}{{80}}\)

Tam giác ABC có nửa chu vi là:\(p = \frac{{a + b + c}}{2} = \frac{{6 + 5 + 8}}{2} = 9,5.\)

Theo công thức Herong ta có: \(S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)}  = \sqrt {9,5.\left( {9,5 - 6} \right).\left( {9,5 - 5} \right).\left( {9,5 - 8} \right)}  \approx 14,98\)

Lại có: \(S = pr \Rightarrow r = \frac{S}{p} = \frac{{14,98}}{{9,5}} = 1,577.\)

Vậy \(\cos A = \frac{{53}}{{80}}\); \(S \approx 14,98\) và \(r = 1,577.\)

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot BA\cdot BC\cdot sinABC\)

\(=\dfrac{1}{2}\cdot5\cdot7\cdot sin120=\dfrac{35\sqrt{3}}{4}\)

Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{5^2+7^2-AC^2}{2\cdot5\cdot7}=cos120=\dfrac{-1}{2}\)

=>\(25+49-AC^2=-35\)

=>\(AC^2=25+49+35=109\)

=>\(AC=\sqrt{109}\)

Kẻ AH\(\perp\)BC

=>\(h_A=AH\)

\(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC\)

=>\(\dfrac{1}{2}\cdot AH\cdot7=\dfrac{35\sqrt{3}}{4}\)

=>\(AH\cdot3,5=\dfrac{35\sqrt{3}}{4}\)

=>\(AH=\dfrac{10\sqrt{3}}{4}=\dfrac{5}{2}\sqrt{3}\)

Xét ΔABC có \(\dfrac{AC}{sinB}=2R\)

=>\(2R=\dfrac{\sqrt{109}}{sin120}=\sqrt{109}\cdot\dfrac{2}{\sqrt{3}}\)

=>\(R=\sqrt{\dfrac{109}{3}}=\dfrac{\sqrt{327}}{3}\)

Tam giác ABC có:

A + B + C = 180⁰

60⁰ + B + C = 180⁰

B + C = 180⁰ - 60⁰

B + C = 120⁰

B - C = 30⁰

B = (120⁰ + 30⁰) : 2 = 75⁰

C = (120⁰ - 30⁰) : 2 = 45⁰

\(\Delta ABC\) có : \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) ( tổng 3 góc của 1 tam giác )

\(\Rightarrow60^0+\widehat{B}+\widehat{C}=180^0\)

\(\Rightarrow\widehat{B}+\widehat{C}=180^0-60^0=120^0\)

Mặt khác : \(\widehat{B}-\widehat{C}=30\)

\(\Rightarrow\widehat{B}=\left(120^0+30^0\right):2=75^0\)

\(\Rightarrow\widehat{C}=\left(120^0-30^0\right):2=45^0\)

Vậy : \(\widehat{B}=75^0\)

\(\widehat{C}=45^0\)

Xét tam giác ABC có:

A+ B+C=180^o

=> (B+C)= 180^o- A

=> (B+C)180-60=120

Mà, ta lại có:

B-C=30^o

=> C+C+30^o=120^o

=> 2C+30^o=120^o

=>2C=90^O

=> C= 45^O

B= 45^o+30^o=75^o