Question 388240
{{{drawing(300,300,-4,12,-4,12,
grid(1),
circle(1,-2,0.3),
circle(3,6,0.3),
circle(5,10,0.3),
line(1,-2,3,6),
line(3,6,5,10),
locate(1.3,-2,A),
locate(3.3,6,B),
locate(5.3,10,C),
graph(300,300,-4,12,-4,12,0))}}}
To go from C to D, you move over 2 and up 4.
To keep it a parallelogram, move that same distance from A to D.
D:({{{1}}},{{{-2}}})+({{{2}}},{{{4}}})=({{{3}}},{{{2}}})

{{{drawing(300,300,-4,12,-4,12,
grid(1),
circle(1,-2,0.3),
circle(3,6,0.3),
circle(5,10,0.3),
circle(3,2,0.3),
line(1,-2,3,6),
line(3,6,5,10),
blue(line(5,10,3,2)),
blue(line(3,2,1,-2)),
locate(1.3,-2,A),
locate(3.3,6,B),
locate(5.3,10,C),
locate(3.3,2,D),
graph(300,300,-4,12,-4,12,0))}}}
.
.
.
{{{drawing(300,300,-4,12,-4,12,
grid(1),
circle(1,-2,0.3),
circle(5,-2,0.3),
circle(5,10,0.3),
line(1,-2,5,-2),
line(1,-2,5,10),
locate(1.3,-2,A),
locate(5.3,-2,X),
locate(5.3,10,C),
blue(line(3,-100,3,100)),
locate(3.2,7,theta),
graph(300,300,-4,12,-4,12,0))}}}
From the diagram,
{{{tan(XAC)=(10-(-2))/(5-1)=12/4=3}}}
{{{XAC=63.4}}}degrees
Also,
{{{XBD=90}}}degrees, since {{{x[B]=x[D]}}}
So then the angle between the diagonals, {{{theta}}}, is,
{{{theta=90-71.6}}}
{{{highlight(theta=18.4)}}}degrees