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An instance of the angle in between two diagonals at a vertex would be angle EBD, whereby diagonals BD and also BE fulfill at peak B.

We will follow the reasonable outlined above.

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

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

So, edge CBD = 36°. Well, triangle ABE is in every means equal come triangle BCD, so angle ABE must additionally equal 36°. Thus, we have the right to subtract native the huge angle in ~ vertex B.

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

(angle EBD) = 108° – 36° – 36° = 36°

Answer = (B)

2) If we begin at one vertex of the 20-sided polygon, then there’s an nearby vertex on each side. No counting these 3 vertices, there would certainly be 17 non-adjacent vertices, therefore 17 feasible diagonals could be attracted from any kind of vertex. Twenty vertices, 17 diagonals from each vertex, yet this technique double-counts the diagonals, as pointed out above.

See more: 1987 Donruss The Rookies Set Value, 1987 Donruss

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

Answer = (C)

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