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# same image but different question

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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

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

Guest Feb 9, 2015

#1
+26640
+13

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Alan  Feb 9, 2015
Sort:

#1
+26640
+13

.

Alan  Feb 9, 2015
#2
+85809
+5

Very impressive, Alan....!!!

CPhill  Feb 9, 2015
#3
+92221
0

Yes, nice work, thank you :)))

Melody  Feb 9, 2015
#4
+19207
+10

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

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

heureka  Feb 10, 2015
#5
+19207
+8

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

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

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

heureka  Feb 10, 2015
#6
+19207
+8

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

heureka  Feb 10, 2015

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