If m AOB = m COD = 90 ,find m COB + m AOD.
no idea where A, B, C, D, and O are
Cannot be answered as is.
To find the sum of the angles m COB and m AOD, we need to know the relationship between the angles in a triangle and the angles formed by a transversal intersecting two parallel lines.
In this case, we are given that m AOB and m COD are both 90 degrees, which means they form right angles. Based on this information, we can infer that lines AB and CD are parallel.
Now, since AB is parallel to CD, angle AOB and angle COD are corresponding angles. Corresponding angles are congruent when two parallel lines are intersected by a transversal.
Therefore, m AOB = m COD = 90 degrees.
Using the fact that the sum of the angles in a triangle is always 180 degrees, we can find angle COB and angle AOD.
In triangle COB:
m COB + m OBC + m COB = 180 degrees
m COB + 90 degrees + m COB = 180 degrees
2m COB + 90 degrees = 180 degrees
2m COB = 180 degrees - 90 degrees
2m COB = 90 degrees
m COB = 90 degrees / 2
m COB = 45 degrees
Similarly, in triangle AOD:
m AOD + m ODA + m ADO = 180 degrees
m AOD + 90 degrees + m AOD = 180 degrees
2m AOD + 90 degrees = 180 degrees
2m AOD = 180 degrees - 90 degrees
2m AOD = 90 degrees
m AOD = 90 degrees / 2
m AOD = 45 degrees
Now that we know that m COB = 45 degrees and m AOD = 45 degrees, we can find the sum of these angles:
m COB + m AOD = 45 degrees + 45 degrees = 90 degrees
So, m COB + m AOD is equal to 90 degrees.