Question 1204641

{{{a = 7}}}

{{{b = 3}}}

{{{beta= 24}}}°

using Law of Cosine, we have

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

{{{3=sqrt(7^2+c^2-2*7*c*cos(24))}}}

{{{9=49+c^2-14*c*cos(24)}}}

{{{9=49+c^2-14*c*0.9135454576426}}}

{{{9=49+c^2-12.7896c}}}

{{{0=49+c^2-12.7896c-9}}}

{{{c^2-12.7896c+40=0}}}

using quadratic formula and got:

{{{c=5.5}}}
or
{{{c=7.3}}}



then angle alpha is

if {{{c=5.5}}}


{{{alpha=cos^-1((b^2+c^2-a^2)/(2bc))}}}

{{{alpha=cos^-1((3^2+5.5^2-7^2)/(2*3*5.5))}}}

{{{alpha=107}}}

or

 if {{{c=7.3}}}

{{{alpha=cos^-1((3^2+7.3^2-7^2)/(2*3*7.3))}}}

{{{alpha=72.3}}}



then angle gamma is

{{{gamma=180-(24+107)}}}

{{{gamma=49}}}°

or

{{{gamma=180-(24+72.3)}}}

{{{gamma=83.7}}}°


so, there are two triangles:

{{{highlight(1). 

{{{a = 7}}}
{{{b = 3}}}
{{{c=5.5}}}
{{{alpha=107}}}°
{{{beta= 24}}}°
{{{gamma=49}}}°


{{{highlight(2). 

{{{a = 7}}}
{{{b = 3}}}
{{{c=7.3}}}
{{{alpha=72.3}}}°
{{{beta= 24}}}°
{{{gamma=83.7}}}°