In the given figure lines AB and CD intersect ato. If angle AOC + angle BOE = 70 and angle DOB
=40°, find angle BOE and reflex angle EOC.

reflex angle EOC = 250°

Step-by-step explanation:

reflex angle EOC

= angle COA + angle DOA + angle DOB + angle BOE

angle DOB = 40° (given)

angle DOB + angle DOA = 180° (linear pair)

angle DOA = 180° - 40°

angle DOA = 140°

angle COA + angle DOA = 180° (linear pair)

angle COA = 180° - 140°

angle COA = 40°

angle COA + angle BOE = 70° (given)

angle BOE = 70° - 40°

angle BOE = 30°

Now,

we have,

reflex angle EOC = 40° + 140° + 40° + 30°

= 250°

160

step-by-step explanation:

idk

i hope i

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.