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An instance of the angle in between two diagonals in ~ a vertex would certainly be angle EBD, whereby diagonals BD and BE accomplish at crest B.
We will follow the logic outlined above.
Triangle BCD is isosceles v BC = CD, and angle BCD = 108°. The various other two angles are equal: contact them every 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 edge ABE must likewise equal 36°. Thus, we can 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 start at one crest of the 20-sided polygon, then there’s an nearby vertex on every side. Not counting these three vertices, there would be 17 non-adjacent vertices, so 17 possible diagonals can be drawn from any type of vertex. Twenty vertices, 17 diagonals from each vertex, however this method double-counts the diagonals, as mentioned above.
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# that diagonals = (17*20)/2 = 17*10 = 170
Answer = (C)
Editor’s Note: This article was initially published in January, 2014, and has to be updated because that freshness, accuracy, and comprehensiveness.