Question

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

Answer 1

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 \(108°\). As \(\overline{FB}\) is perpendicular to \(\overline{AC}\), it cut's the angle in half which leads to \(54°\).

With the triangle beeing right-angled and the angle at \(B\) beeing \(54°\) we can calculate the angle in the corner of \(C\), which is \(180°−90°−54°=36°\).
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 \(\angle E0C\) you're searching for, i just accidently used another notation) by using the fact, that a triangles angles sum up to \(180°\).

So the angle \(\angle BAC = \angle ABG = 36°\) therefore the resulting angle is \(\angle AGB = 180° - 36° - 36° = 108°\) which is the solution to your problem as \(\angle E0C = \angle EGC = \angle AGB\).

I hope that did solve your problem.

Answer 2

WOW! :D thank you so much!

Answer 3

You're welcome

Answer 4

Did you make a sketch of it? It usally helps me to understand what's going on. Then i try to divide in shapes, that are easy to calculate, eg. right-angled triangles.

Answer 5

I have here one last problem..can you also help me with that?
here it is:
Find the area of a triangle if its two sides measures 6 in. and 9 in. and the bisector of the angle between the sides is 4 square root of 3. in.

Answer 6

Let me quickly sketch it

Answer 7

Or could you make a quick drawing?

Answer 8

can't i'm not good with drawings or sketching stuffs

Answer 9

just try it.

Answer 10

i think it goes like this..i just tried it though..