same image but different question

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Very impressive, Alan....!!!

Yes, nice work, thank you :)))

In a triangle ABC, a, b , c are the opposite side of angle A,angle B ,and angle C, and

  $${\frac{{\mathtt{cosB}}}{{\mathtt{cosc}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{b}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}\right)}}$$

(I) find the angle B

$$\boxed{ \dfrac{ \cos{(B)} }{\cos{(C)}}=-\dfrac{b}{(2a+c)} } \small{\text{ $\quad $For any triangle: $ a=b\cos{(C)}+c\cos{(B)}. $ }}\\\\ \small{\text{ $ (2a+c)\cos{(B)} = -b\cos{(C)} \quad | \quad -b\cos{(C)}= c\cos{(B)} - a $ }}\\ \small{\text{ $ (2a+c)\cos{(B)} = c\cos{(B)} - a $ }}\\ \small{\text{ $ (2a)\cos{(B)} = - a $ }}\\ \small{\text{ $ 2\cos{(B)} = - 1 $ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{\cos{(B)} = - \dfrac{1}{2} \qquad B = 120\ensurement{^{\circ}} }$ }}$$

(2) if b=square root 13 ,and a+c=4 what is the area of the triangle ABC

$$\boxed{\small{\text{(1) The shape of the triangle by Heron's formula:$\quad A^2=s(s-a)(s-b)(s-c) \quad s = \frac{a+b+c}{2}$ }} }\\\\ \small{\text{$s =\frac{a+c+b}{2} \quad | \quad a+c=4 $ }}\\ \small{\text{ $s = \frac{4+b}{2} = 2 + \frac{b}{2}\qquad \boxed{s=2 + \frac{b}{2} } $ }}\\\\ \small{\text{ $ A^2=s(s-a)(s-b)(s-c) = \textcolor[rgb]{1,0,0}{s(s-b)} (s-a)(s-c) $ }} }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{s(s-b)}=( 2+\frac{b}{2} )(2+\frac{b}{2}-b) =(2+\frac{b}{2})(2-\frac{b}{2})=4-\frac{b^2}{4} \quad | \quad b^2=13 $ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{s(s-b)} = 4-\frac{13}{4} =\frac{16-13}{4} = \frac{3}{4}$ }}\\\\ \small{\text{ $ A^2=\textcolor[rgb]{1,0,0}{s(s-b)} (s-a)(s-c) = \textcolor[rgb]{1,0,0}{ \frac{3}{4} } (s-a)(s-c) $ }}\\ \small{\text{ $ A^2=\frac{3}{4}(s-a)(s-c)= \frac{3}{4}(s^2-sc-sa+ac) =\frac{3}{4}(s^2-s(a+c)+ac) \quad | \quad a+c=4 $ }}\\ \small{\text{ $ A^2=\frac{3}{4}( \textcolor[rgb]{1,0,0}{s^2-4s} +ac)$ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{s^2-4s} = (2+ \frac{b}{2} )^2-4(2+\frac{b}{2} )=4+2b+ \frac{b^2}{4} -8-2b=-4+\frac{b^2}{4} =\frac{-16+13}{4}= -\frac{3}{4} $ }}\\\\ \small{\text{ $ A^2=\frac{3}{4}( \textcolor[rgb]{1,0,0}{ -\frac{3}{4} } +ac) $ }}\\ \small{\text{ $ A^2*\frac{4}{3} -ac + \frac{3}{4} = 0$ }}\\$$

$$\boxed{\small{\text{(2) The shape of the triangle :$\quad A= \dfrac {ac \sin{(B)} } {2} \qquad ac = \dfrac{2A}{\sin{(B)}} $ }} }\\\\ \small{\text{ $ A^2*\frac{4}{3} -ac + \frac{3}{4} = 0$ }} }\\ \small{\text{ $ A^2*\frac{4}{3} -\frac{2A}{\sin{(B)}} + \frac{3}{4} = 0$ }} }\\\\ \small{\text{ $ A_{1,2}= \dfrac { \frac{2}{\sin{(B)}} \pm\sqrt{ \left( \frac{2}{ \sin{(B)} } \right)^2-4*\frac{4}{3}*(\frac{3}{4} ) } } { 2*\frac{4}{3} } = \dfrac { \frac{2}{\sin{(B)}} \pm 2\sqrt{ \frac{1}{ \sin^2{(B)} }-1 } } { \frac{8}{3} } = \dfrac { \frac{2}{\sin{(B)}} \pm 2 \frac{ \cos{(B)} } { \sin{(B)} } } { \frac{8}{3} } $ }}\\ \small{\text{ $ A_{1,2} = \frac{3}{4} \left( \frac{1}{\sin{(B)}} \pm \frac{ \cos{(B)} } { \sin{(B)} } \right) = \frac{3}{4} \left( \frac{1\pm\cos{(B)} }{\sin{(B)}} \right) $ }}\\\\ \small{\text{ $ A_{1} = \frac{3}{4} \left( \frac{1+\cos{(B)} }{\sin{(B)}} \right) \quad | \quad \cos{(B)} = - \frac{1}{2} \quad \sin{(B)} =\sqrt{ 1-\cos^2{ (B) }} = \frac{\sqrt{3}}{2} $ }}\\ \small{\text{ $ A_{1}=\frac{3}{4}*\frac{1+ (-\frac{1}{2}) }{ \frac{\sqrt{3}}{2} } =\frac{3}{4}*\frac{2}{ \sqrt{3} } *\frac{1}{2} =\frac{3}{4}*\frac{2*\sqrt{3}}{3} *\frac{1}{2} = \frac{1}{4}*\sqrt{3} $ }}\\ \small{\text{ $ \boxed{A_{1}= \frac{1}{4}\sqrt{3} $ no solution!$} $ }}\\\\ \small{\text{ $ A_{2} = \frac{3}{4} \left( \frac{1-\cos{(B)} }{\sin{(B)}} \right) \quad | \quad \cos{(B)} = - \frac{1}{2} \quad \sin{(B)} =\sqrt{ 1-\cos^2{ (B) }} = \frac{\sqrt{3}}{2} $ }}\\ \small{\text{ $ A_{2}=\frac{3}{4}*\frac{1- (-\frac{1}{2}) }{ \frac{\sqrt{3}}{2} } =\frac{3}{4}*\frac{2}{ \sqrt{3} }*\frac{3}{2} =\frac{3}{4}* \frac{2*\sqrt{3}}{3} *\frac{3}{2} = \frac{3}{4}*\sqrt{3} $ }}\\ \small{\text{ $ \boxed{A_{2}= \frac{3}{4}\sqrt{3}} $ }}$$

   $$\small{\text{ c and a ? $ \boxed{ \; ac = \dfrac{2A}{\sin{(B)}} \qquad a=4-c \qquad A= \frac{3}{4}\sqrt{3} \qquad \sin{(B)}=\frac{\sqrt{3}}{2}\; } $ }}\\\\\\ \small{\text{ $ (4-c)c = 2*\dfrac{3\sqrt{3}}{4}*\dfrac{2}{\sqrt{3}}= 3$ }}\\\\ \small{\text{ $ c^2-4c+ 3 =0 \qquad $ }} \small{\text{ $ c_{1,2}=\dfrac{ 4\pm\sqrt{16-4*3 } } {2} = \dfrac{ 4\pm \sqrt{4} } {2} = \dfrac{ 4\pm 2 } {2} =2\pm 1 $ }}\\\\ \small{\text{ $ c_1= 2+1 \quad \boxed{c_1=3} \qquad c_2= 2-1 \quad \boxed{c_2= 1}$ }}\\\\ \small{\text{ $ a_1=4-c_1 \qquad a_1=4-3 \qquad \boxed{a_1=1}$ }}\\ \small{\text{ $ a_2=4-c_2 \qquad a_2= 4-1 \qquad \boxed{a_2=3}$ }}$$