MK10
1) BDC= BAC( angles lying on the same segment) BAC= 42° AC is diameter. So ABC is 90°( Angle subtended by a semicircle on the circum-circle is 90°) ABC+BAC+X°= 180° x= 48°
2)BOC+AOB+AOC= 360° BOC= 160° here BAC= 1/2BOC( angle subtended by the circle is double the angle subtended by any point on the circle) 1/2×160°= 80°= BAC Hope it helps!
Answers & Comments
BAC= 42°
AC is diameter.
So ABC is 90°( Angle subtended by a semicircle on the circum-circle is 90°)
ABC+BAC+X°= 180°
x= 48°
2)BOC+AOB+AOC= 360°
BOC= 160°
here BAC= 1/2BOC( angle subtended by the circle is double the angle subtended by any point on the circle)
1/2×160°= 80°= BAC
Hope it helps!