Question 1123884

 

let’s the largest angle be {{{alpha}}}, the larger {{{beta}}}, and the smallest angle {{{gamma}}}

the sum is: {{{alpha+beta+gamma=180}}}...............eq.1
if the larger is {{{20}}}° greater than the smallest, we have

{{{beta=gamma +20}}}....eq.2

if  the largest is twice the smallest angle, we have 

{{{alpha=2gamma}}}...............eq.3

go to 
{{{alpha+beta+gamma=180}}}...............eq.1, substitute {{{beta}}} and {{{alpha}}} from eq.1 and eq.3

{{{2gamma+gamma +20+gamma=180}}}....solve for {{{gamma}}}

{{{4gamma =180-20}}}

{{{4gamma =160}}}

{{{gamma =160/4}}}

{{{gamma =40}}}°

go to {{{beta=gamma +20}}}....eq.2, plug in {{{gamma}}}

{{{beta=40 +20}}}

{{{beta=60}}}°

go to {{{alpha=2gamma}}}...............eq.3, plug in {{{gamma}}}

{{{alpha=2*40}}}

{{{alpha=80}}}°

so, the largest angle of a triangle is {{{80}}}°