Question 1081968
{{{drawing(300,300,-1,11,-1,11,
line(0,0,1.74,9.85),line(1.74,9.85,9.74,9.85),
line(0,0,8,0),line(8,0,9.74,9.85),
locate(-.3,0,A),locate(8,0,B),
locate(1.4,10.3,D),locate(9.9,10.3,C),
arrow(9.74,9.85,2.08,3.42),
arrow(8,0,1.56,7.66),locate(3.7,4.8,O)
)}}}
In a parallelogram ABCD, angles A and C are congruent,
and angles B and D are supplementary to angle A.
In this case, it means that the parallelogram's
angle at B (ABC) measures {{{180^o-80^o=100^o}}} ,
and the angle at C (BCD) measures {{{80^o}}} .
We know how to fight our the measures
of two of the angles in triangle BCO:
Angle OBC is one half of ABC,
so it measures {{{100^o/2=50^o}}} .
Angle OCB is one half of BCD,
so it measures {{{80^o/2=40^o}}} .
Angle BOC is the third angle in the triangle.
It's measure is
{{{180^o-(50^o+40^o)=180^o-90^o=(90^o)}}}