Question 1131378
given:
{{{A=36}}}°
{{{C=105}}}°
{{{c=52}}}


solution:

{{{B=180-(A+C)}}}

{{{B=180-(36+105)}}}

{{{B=180-141}}}

{{{highlight(B=39)}}}°



use law of cos 

{{{a^2=b^2+c^2-2bc*cos(A)}}}
{{{b^2=a^2+c^2-2ac*cos(B)}}}

=>
{{{a=sqrt(b^2+c^2-2bc*cos(A))}}}

{{{b=sqrt(a^2+c^2-2ac*cos(B))}}}


if given {{{c=52}}}, {{{A=36}}}°, {{{B=39}}}°, we have



{{{a^2 = b^2 + 52^2 -2b*52 cos(36)}}}

{{{a^2 = b^2 + (-26 - 26 sqrt(5)) b + 2704}}}


{{{a^2 = b^2 - 84.14b + 2704 }}}..............eq.1


=>{{{a =sqrt( b^2 - 84.14b + 2704) }}}....eq.1a


now,

{{{b^2=a^2+52^2-2a*52*cos(B)}}}

{{{b^2 = a^2  - 80.8a + 2704}}}....substitute {{{a^2}}}-eq.1 and {{{a}}}-eq.1a

{{{b^2 = b^2 - 84.14b + 2704 + 2704 -80.8sqrt( b^2 - 84.14b + 2704) }}}

{{{b^2 - b^2 =- 84.14b  -80.8sqrt( b^2 - 84.14b + 2704) +5408}}}

{{{0 =- 84.14b  -80.8sqrt( b^2 - 84.14b + 2704) +5408}}}

{{{ 84.14b- 5408 =-80.8sqrt( b^2 - 84.14b + 2704) }}}...square both sides

{{{( 84.14b- 5408)^2 =(-80.8sqrt( b^2 - 84.14b + 2704) )^2}}}

{{{7079.54b^2 - 910058*b + 29246464=6528.64( b^2 - 84.14b + 2704) }}}

{{{7079.54b^2 - 910058*b + 29246464=6528.64 b^2 - 549320. b + 17653442.56 }}}

{{{7079.54b^2 - 910058*b + 29246464-6528.64 b^2 + 549320. b -17653442.56 =0}}}

{{{550.9b^2 - 360738*b + 11593000=0}}}

{{{550.9(b - 620.925)(b - 33.89) = 0}}}

{{{highlight(b=33.89)}}}


substitute in eq.1a

{{{a =sqrt( b^2 - 84.14b + 2704) }}}....eq.1a

{{{a =sqrt( (33.89)^2 - 84.14*33.89 + 2704) }}}

{{{a =sqrt( 1148.5321 - 2851.5046+ 2704) }}}

{{{a =sqrt(1001.0275) }}}

{{{highlight(a =31.64) }}}