In the given parallelogram ABCD (Fig. 3), AC and BD are its diagonals intersecting
If ZCAB - 25. ZCAD - 35° and ZCOD -110", find
ZODC
LDCB
ZOBC
Fig. 3
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i dont lnow because i am weak in mathematics i am sorry
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angle AOB = 110°(vertically opposite)
angle ABO
25 + 110 + ABO = 180°
ABO = 45°
ABO = ODC (alternative angle)
OAB = OCD (alternative angle)
OAD = OCB (alternative angle)
AOD = 70° (linear pair)
AOD = COB (vertically opposite)
OCB + COB + OBC = 180°
OBC = 75°
DCB = OCD + OCB
DCB = 60°
angle ABO
25 + 110 + ABO = 180°
ABO = 45°
ABO = ODC (alternative angle)
OAB = OCD (alternative angle)
OAD = OCB (alternative angle)
AOD = 70° (linear pair)
AOD = COB (vertically opposite)
OCB + COB + OBC = 180°
OBC = 75°
DCB = OCD + OCB
DCB = 60°
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