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eroshea
If ABCDE is a regular pentagon and diagonals EB and AC intersect at O, then what is the degree measure of angleEOC?
using the property that diagonals of regular pentagon trisect all the interior angles, we get angle OAB = OBC = 36 degrees (108/3, angle A=angle B =108 degrees) now can u find EOC ?
Can you explain further? Sorry I can't understand :3
diagonals of regular pentagon trisect all the interior angles so, angle ABO will be 1/3 times interior angle ABC so, angle ABO = 36 similarly angle BAO = 36 so angle AOB will be 180 -36-36 (because angle in a triangle sum upto 180) and since EOC is vertical to AOB EOC = AOB
how do you know that the interior angle of ABC is 108?
how do you find the angle A is 108 as well as the angle B? sorry I'm confused