Question 993509

let angles be {{{alpha}}} and {{{beta}}}

if {{{alpha}}} and {{{beta}}} are complementary angles, then {{{alpha+beta=90}}}°

if the second of two complementary angles is {{{beta}}} is {{{57}}}° larger than {{{2}}} times the first angle {{{alpha}}}, we have

 {{{beta=2alpha+57}}}°........substitute it in {{{alpha+beta=90}}}°

{{{alpha+2alpha+57=90}}}°.............solve for {{{alpha}}}

{{{3alpha=90-57}}}°

{{{3alpha=33}}}°

{{{alpha=33/3}}}°

{{{highlight(alpha=11)}}}°

go to {{{beta=2alpha+57}}}° substitute {{{11}}} for {{{alpha}}}

{{{beta=2*11+57}}}°

{{{beta=22+57}}}°

{{{highlight(beta=79)}}}°


so your angles are: {{{highlight(11)}}}° and {{{highlight(79)}}}°