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An example of the angle between two diagonals at a vertex would be angle EBD, where diagonals BD and BE meet at vertex B.

We will follow the logic outlined above.

Triangle BCD is isosceles with BC = CD, and angle BCD = 108°. The other two angles are equal: call them each x.

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108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

So, angle CBD = 36°. Well, triangle ABE is in every way equal to triangle BCD, so angle ABE must also equal 36°. Thus, we can subtract from the big 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 vertex of the 20-sided polygon, then there’s an adjacent vertex on each side. Not counting these three vertices, there would be 17 non-adjacent vertices, so 17 possible diagonals could be drawn from any vertex. Twenty vertices, 17 diagonals from each vertex, but this method double-counts the diagonals, as pointed out above.

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

Answer = **(C)**

*Editor’s Note: This post was originally published in January, 2014, and has been updated for freshness, accuracy, and comprehensiveness.*

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