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

We will certainly follow the reasonable outlined above.

Triangle BCD is isosceles through BC = CD, and angle BCD = 108°. The other two angles room equal: call them every x.

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

So, angle CBD = 36°. Well, triangle ABE is in every means equal come triangle BCD, so edge ABE must also equal 36°. Thus, we can subtract indigenous 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 peak of the 20-sided polygon, climate there’s an adjacent vertex on each side. No counting these three vertices, there would be 17 non-adjacent vertices, for this reason 17 possible diagonals might be drawn from any type of vertex. Twenty vertices, 17 diagonals from every vertex, yet this technique double-counts the diagonals, as pointed out above.

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

Answer = **(C)**

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