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An example of the angle in between two diagonals at a vertex would be angle EBD, wherein diagonals BD and also BE fulfill at crest B.

We will follow the reasonable outlined above.

Triangle BCD is isosceles with BC = CD, and also angle BCD = 108°. The various other two angles room equal: speak to them every x.

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

So, edge CBD = 36°. Well, triangle ABE is in every way equal come triangle BCD, so edge ABE must also equal 36°. Thus, we can subtract from the large angle at vertex B.

(angle EBD) = (angle ABC) – (angle CBD) – (angle ABE)

(angle EBD) = 108° – 36° – 36° = 36°

Answer = (B)

2) If we start at one crest of the 20-sided polygon, climate there’s an nearby vertex on every side. No counting these three vertices, there would be 17 non-adjacent vertices, therefore 17 feasible diagonals can be drawn from any vertex. Twenty vertices, 17 diagonals from every vertex, however this an approach double-counts the diagonals, as mentioned above.

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# of diagonals = (17*20)/2 = 17*10 = 170

Answer = (C)

Editor’s Note: This write-up was initially published in January, 2014, and also has been updated for freshness, accuracy, and also comprehensiveness.