Really hard problem with cos rule

[kate]

We have the triangle ABC with angle β = 60°, a + c = 11, where a and c are sides of the triangle and c is bigger than a, and the radius of the inscribed circumference is r = 2/√3. Find a, b, c and h - the height towards the side a.

[Guest]

The task is not difficult .

[Guest]

[tex]\triangle[/tex]ABC c>a ,[tex]\beta=60 ^\circ[/tex] ,r=[tex]\frac{2}{ \sqrt{3} }[/tex]
Let BC=a [tex]\Rightarrow[/tex] AB=c= 11-a
a ,b ,c ,[tex]h_{a }[/tex]= ?

[tex]AC^{2 } =BC^{2 } +AB^{2 }- 2BC.AB.cos \beta[/tex]
[tex]b^{2 }= a^{2 } +(11-a)^{2 } -2a(11-a). \frac{1}{2}[/tex]
[tex]b^{2 }= 3 a^{2 }-33a+121[/tex] (1)

[tex]S_{ABC } =S_{ABC }[/tex]
[tex]\frac{a(11-a).sin60 ^\circ }{2} = \frac{a+b+(11-a)}{2} .\frac{2}{ \sqrt{3} }[/tex]

a(11-a).[tex]\frac{ \sqrt{3} }{2}[/tex]=[tex]\frac{( \sqrt{3 a^{2 } -33a+121}+11)2 }{ \sqrt{3} }[/tex]

33a-3[tex]a^{2 }[/tex]-44 =4[tex]\sqrt{3 a^{2 }-33a+121 }[/tex]

We lay x=[tex]\sqrt{3 a^{2 }-33a+121 }[/tex] (2)

[Guest]

77- [tex]x^{2 }[/tex]= 4x ;[tex]x^{2 }[/tex]+4x-77=0
D=4+77 =81 ;[tex]x_{1,2 } = \frac{-2 \pm 9}{1}[/tex] [tex]\Rightarrow[/tex] [tex]x_{1 }[/tex]=7 ,[tex]x_{2 }[/tex]= -11 is void

(2) [tex]\Rightarrow[/tex] [tex]\sqrt{3 a^{2 }-33a+121 }[/tex]= 7
3[tex]a^{2 }[/tex]-33a+121=49
[tex]a^{2 }[/tex]-11a+24 =0 (3)
D=121-96 =25 ;[tex]a_{1,2 }[/tex]=[tex]\frac{11 \pm5 }{2}[/tex] [tex]\Rightarrow[/tex] [tex]a_{1 }[/tex]=8 ,[tex]a_{2 }[/tex]=3

[tex]c_{1 } =3 ,c_{2 }=8[/tex] Should c>a [tex]\Rightarrow[/tex] The agreement [tex]a_{1 } ,c_{1 }[/tex] is void

(1) [tex]\Rightarrow[/tex] [tex]b^{2 } =3. 3^{2 }-33.3+121[/tex] ; [tex]b^{2 }[/tex]=49 ;b=7
[tex]\triangle[/tex]ABH sin[tex]\beta[/tex]=[tex]\frac{AH}{AB}[/tex] ;sin60[tex]^\circ= \frac{ h_{a } }{c}[/tex] ;[tex]\frac{ \sqrt{3} }{2} =\frac{ h_{a } }{8}[/tex] ;[tex]h_{a }= 4 \sqrt{3}[/tex]

Answers a=3 ,b=7 ,c=8 ,[tex]h_{a }= 4 \sqrt{3}[/tex]