we know that
In a parallelogram
1) Opposite angles are equal
2) Consecutive angles are supplementary
3)The diagonals of a parallelogram bisect each other
so
problem N 21
see the attached figure to better understand the problem
∡A=2*32°----> ∡A=64°
∡A+∡B=180°-----> ∡B=180-64-----> ∡B=116°
∡1=∡B/2-----> ∡1=116°/2------> ∡1=58°
∡C=∡A
so
∡2=∡A/2-----> ∡2=32°
in the triangle BOC
∡BOC=90°
∡BOC +∡3=180--------> supplementary angles
∡3=180-90-----> ∡3=90°
the answers problem N 21 are
a) ∡1=58°
b) ∡2=32°
c) ∡3=90°
problem N 22
see the attached figure to better understand the problem
∡1+56°=180°---------> supplementary angles
∡1=180-56---------> ∡1=124°
triangle OBC (an isosceles triangle)
∡1+∡2+∡2=180--------> ∡2=(180-∡1)/2-----> ∡2=(180-124)/2---> ∡2=28°
∡2+∡3=90--------> complementary angles
∡3=90-∡2------> ∡3=90-28----> ∡3=62°
the answers problem N 22 are
a) ∡1=124°
b) ∡2=28°
c) ∡3=62°
