Question 873858
Since {{{80^o+10^o=90^o}}} and {{{25^o+65^o=90^o}}} ,
those pairs of angles could be the acute angles in right triangles.
The same ratio {{{leg/hypotenuse}}} that is the sine of one angle
is also the cosine off the other angle.
When two angles add up to {{{90^o}}} (when they are complementary),
the sine of one angle is the cosine of the other angle.
In formulas,
{{{sin(90^o-x)=cos(x)}}} and {{{cos(90^o-x)=sin(x)}}}
 
{{{drawing(300,150,-0.1,1.1,-0.1,0.5,
triangle(0,0,0.906,0,0.906,0.423),
rectangle(0.906,0,0.866,0.04),
locate(0.43,-0.02,x),locate(0.92,0.25,y),
locate(0.453,0.21,1),locate(0.14,0.09,25^o),
arc(0,0,0.5,0.5,-25,0),locate(0.8,0.4,65^o),
arc(0.906,0.423,0.4,0.4,90,155)
)}}} {{{x=x/1=sin(65^o)=cos(25^o)}}} {{{y=y/1=sin(25^o)=cos(65^o)}}}