Question 698211
The second angle {{{beta}}} in a triangle is twice as large as the first angle {{{alpha}}}: 

{{{beta=2alpha}}}
The third angle {{{gamma}}} is three-fourths as large as the first angle {{{alpha}}}:

{{{gamma=(3/4)alpha}}} 

we know that {{{alpha+beta+gamma=180}}}...substitute 
{{{beta=2alpha}}} and {{{gamma=(3/4)alpha}}} 


{{{alpha+2alpha+(3/4)alpha=180}}}....solve for {{{alpha}}}


{{{3alpha+(3/4)alpha=180}}}


{{{(3+3/4)alpha=180}}}

{{{(12/4+3/4)alpha=180}}}

{{{(15/4)alpha=180}}}

{{{alpha=180/(15/4)}}}

{{{alpha=(180*4)/15}}}

{{{alpha=720/15}}}

{{{highlight(alpha=48)}}}


now find {{{beta=2alpha}}}


{{{beta=2*48}}}

{{{highlight(beta=96)}}}


{{{gamma=(3/4)alpha}}} 

{{{gamma=(3/4)48}}} 

{{{gamma=(3/cross(4))cross(48)12}}} 

{{{gamma=3*12}}}

{{{highlight(gamma=36)}}}


check:
 
{{{alpha+2alpha+(3/4)alpha=180}}}

{{{48+96+36=180}}}

{{{180=180}}}



{{{drawing( 600,600, -5, 10, -5, 10, locate(0.6,1,"96°"),locate(3.8,1,"48°"),locate(-0.7,8,"36°"),
         grid(0),locate( 0, 0, O( 0, 0 ) ),locate(-1, 10, A( -1, 10 ) ),locate( 4, 0, B( 4, 0 ) ),locate( 5, 0, C( 5, 0 ) ),
         line( 5, 0, -1, 10 ),line( 0, 0, 5, 0 ),line(0, 0, -1, 10 ),line( -1, 10, 5, 0 )
)}}}