Question 756594
let angles be {{{alpha}}},{{{beta}}}, and {{{gamma}}}

if we have a triangle, we know that {{{alpha+beta+gamma=180}}}


if {{{3}}} angles of a triangular flag are {{{alpha=x}}}, 

{{{beta}}} equal {{{4}}} times the complement of {{{x}}}, and we know that complementary angles are angles whose measures sum to {{{90}}}°, then the complement of {{{x}}} is {{{90-x}}}

so, {{{beta=4(90-x)}}}


and if {{{gamma}}} equal to one-third the supplement of {{{x}}} which is {{{180-x}}}, then {{{gamma=(1/3)(180-x)}}}

now substitute all in {{{alpha+beta+gamma=180}}}

{{{x+4(90-x)+(1/3)(180-x)=180}}}....solve for {{{x}}}

{{{x+360-4x+60-x/3=180}}}

{{{420-3x-x/3=180}}}...both sides multiply by {{{3}}}

{{{1260-9x-x=540}}}

{{{1260-10x=540}}}

{{{1260-540=10x}}}

{{{720=10x}}}

{{{720/10=x}}}

{{{72=x}}}

now find angles

{{{alpha=x}}} => {{{highlight(alpha=72)}}}

{{{beta=4(90-x)}}} => {{{beta=4(90-72)}}} => {{{highlight(beta=72)}}}

{{{gamma=(1/3)(180-x)}}} => {{{gamma=(1/3)(180-72)}}} => {{{highlight(gamma=36)}}}


check the sum:

 {{{alpha+beta+gamma=180}}}

 {{{72+72+36=180}}}

 {{{180=180}}}