If ABCDE is a regular pentagon and diagonals EB and AC intersect at O, then what is the degree measure of angleEOC?

Question

aravindg · Answer

how about drawing a figure?

directrix · Answer

@cap_n_crunch   Why 72? 

OpenStudy values the Learning process - not the ‘Give you an answer’ process

anonymous · Answer

Is there anything like tikz or an other graphic engine, to sketch things in this forum?

aravindg · Answer

there is a draw button below

anonymous · Answer

At first we need an additional line $\overline{FB}$ as shown in the attached graphic.

We know that the pentagon is regular, therefore the angles between the outer edges are all equal to $108^\circ$. As  $\overline{FB} $ is perpendicular to  $\overline{AC}$, it cut's the angle in half which leads to  $54^\circ$.

With the triangle beeing right-angled and the angle at $B$ beeing  $54^\circ$ we can calculate the angle in the corner of  $C$, which is  $180^\circ - 90^\circ - 54^\circ = 36^\circ$.

Another assumption we can make as the pentagon is regular is, that the triangles  $\triangle ABE$ and  $\triangle ABC$ are equal to eachother, which means we can calculate the angle  $\angle EGC$ (it's the angle  $E0C$ you're searching for, i just accidently used another notation) by using the fact, that a triangles angles sum up to  $180^\circ$. So the angle  $\angle BAC = \angle ABG = 36^\circ$ therefore the resulting angle is  $\angle AGB = 180^\circ - 36^\circ - 36^\circ = 108^\circ$ which is the solution to your problem as  $\angle E0C = \angle EGC = \angle AGB$.

I hope that did solve your problem.