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Regular pentagon P has all five diagonals drawn. What is the angle between two of these diagonals where they meet at a

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by BTGmoderatorDC » Wed Aug 26, 2020 6:26 pm

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Regular pentagon P has all five diagonals drawn. What is the angle between two of these diagonals where they meet at a vertex of the pentagon?

(A) 12°
(B) 36°
(C) 54°
(D) 60°
(E) 72°



OA B

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Re: Regular pentagon P has all five diagonals drawn. What is the angle between two of these diagonals where they meet at

by Scott@TargetTestPrep » Sun Aug 30, 2020 7:23 am
BTGmoderatorDC wrote: ↑
Wed Aug 26, 2020 6:26 pm
 Regular pentagon P has all five diagonals drawn. What is the angle between two of these diagonals where they meet at a vertex of the pentagon?

(A) 12°
(B) 36°
(C) 54°
(D) 60°
(E) 72°



OA B

Source: Magoosh
Solution:
Pentagon.png
The sum of the interior angle measures of a polygon with n sides is calculated by the formula: sum = 180 x (n - 2). Thus, for a pentagon, which has 5 sides, the sum of all the interior angles is 180 x (5 - 2) = 540. Thus, each angle of a regular pentagon is 540/5 = 108 degrees.

When a diagonal is drawn, it creates an isosceles triangle of 36, 36 and 108 degrees. When two diagonals are drawn from the same vertex, they divide the angle of the pentagon into 3 angles, two of which are 36 degrees. If we let x = the measure of the angle between two diagonals, then we have:

36 + x + 36 = 108

x + 72 = 108

x = 36

Answer: B

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