same image but different question

.

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

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

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

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