Question 202530
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The sum of the interior angles of any polygon is *[tex \Large (n - 2)\cdot180] degrees where *[tex \Large n] is the number of sides (or vertices, if you will) of the polygon.  You have a pentagon so *[tex \Large n = 5], hence the sum of the interior angles is *[tex \Large 3\cdot180 = 540]


Your angles are given as *[tex \Large 4x], *[tex \Large 4x], *[tex \Large 4x], *[tex \Large 3x], and *[tex \Large 3x], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x + 4x + 4x + 3x + 3x = 540]


Just solve for *[tex \Large x] and then multiply by 3 to get the measure of angle A, or by 4 to get the measure of any one of the larger angles.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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