First Picture:
AB and CD are straight lines. Their intersection has formed 4 angles: AOD, COB, AOC and DOB.
AOD and COB are vertically opposite angles, which means that AOD = COB = 152 degrees.
COB is formed by 3 other angles: COE, EOF and FOB.
Which means that COE + EOF + FOB = COB
We already proved that COB = 152 degrees, and on the figure, we have the algebraic measurement of the 3 angles forming COB::
COE = 3x degrees, EOF = x degrees and FOB = x + 12 degrees.
COE + EOF + FOB = COB
3x + x + (x+12) = 152
3x + x + x + 12 = 152
5x + 12 = 152
We need to solve for x.
Subtract 12 from each side:
5x + 12 = 152
5x + 12 - 12 = 152 - 12
5x = 140
Divide both sides by 5:
5x = 140
(5x)/5 = 140/5
x = 140/5
x = 28
Second picture:
Based on the picture, we know that the angle LOM = 90 degrees, MON = x degrees and NOL = 5x degrees.
The angles forms a circle:
LOM + MON + NOL = 360 degrees
So 90 + x + 5x = 360
90 + 6x = 360
We need to solve for x:
Subtract 90 from both sides:
90 + 6x = 360
90 - 90 + 6x = 360 - 90
6x = 270
Divide both sides by 6:
6x = 270
(6x)/6 = 270/6
x = 270/6
x = 45
Hope this helps! :)
