Answer:
[tex]m\angle BOC=30^o[/tex]
Step-by-step explanation:
step 1
Find the measure of angle COD
we know that
[tex]m\angle COF=m\angle COD+m\angle DOF[/tex] ---> by addition angle postulate
we have
[tex]m\angle COF=150^o[/tex] ----> given problem
[tex]m\angle DOF=90^o[/tex] ----> because AD is perpendicular to BF
substitute the given values
[tex]150^o=m\angle COD+90^o[/tex]
[tex]m\angle COD=150^o-90^o[/tex]
[tex]m\angle COD=60^o[/tex]
step 2
Find the measure of angle BOC
we know that
[tex]m\angle BOC+m\angle COD=90^o[/tex] ---> by complementary angles
we have
[tex]m\angle COD=60^o[/tex]
substitute
[tex]m\angle BOC+60^o=90^o[/tex]
[tex]m\angle BOC=90^o-60^o[/tex]
[tex]m\angle BOC=30^o[/tex]
