Question 155475
Let's bisect the  problem carefully.
</pre><font size=4><b> First illustration:
{{{A}}} and {{{B}}} are supplementary angles, so {{{A+B=180}}} -----> eqn 1
It shows {{{B}}} is a supplement of {{{A}}} and a {{{highlight(condition)}}} exist that the supplement which is {{{B}}}, is 12 degrees greater than 3 times the complement.
2nd illustration:
We make again {{{A}}} and {{{B}}} complementary angles this time, so
{{{A+B=90}}} ----> eqn 2. we make {{{A}}} a complement of {{{B}}}. To show the condition on the 1st illustration: {{{highlight(B=12+(3A))}}}----> 12 greater than 3 times the complement, being {{{A}}} a complement of {{{B}}} right?
Substitute the {{{highlight(condition)}}} to eqn 1:
{{{A+12+3A=180}}}
{{{4A=180-12=168}}}
{{{cross(4)A/cross(4)=cross(168)42/cross(4)}}}
{{{A=42degrees}}}
For {{{B}}} in question via eqn 1:{{{42+B=180}}}
{{{B=180-42=138degrees}}}
To check, go back to the {{{highlight(condition)}}}:
{{{B=12+(3A)}}}
{{{138=12+(3*42)}}}
{{{138=12+126}}}
{{{138deg=138deg}}}
Thank you,
Jojo