Question 642241
Let {{{x}}} be the measure, in degrees, of the smaller acute angle.
If the measure of the larger acute angle is {{{15^o}}} more than twice that,
the measure of the larger acute angle, in degrees, is
{{{2x+15}}}
In a right triangle, the measures of both acute angles add up to {{{90^o}}}, so
{{{x+2x+15=90}}}
first we solve for {{{x}}}, the measure, in degrees, of the smaller acute angle:
{{{x+2x+15=90}}} --> {{{3x+15=90}}} --> {{{3x+15-15=90-15}}} --> {{{3x=75}}}
 --> {{{3x/3=75/3}}} --> {{{x=25}}}
The measure of the larger acute angle, in degrees, is
{{{2x+15=2*25+15=50+15=highlight(65)}}}
or {{{90-25=highlight(65)}}}
The larger acute angle measures {{{highlight(65^o)}}} .