Question

Determine the measure of the interior angle at vertex A.
It is a pentagon, at vertex A is 3x This is how it is set up
4x
4x 4x
3x 3x vertex A
Answers
A. 150 B. 50 C.90 D.30
The answer is 90 but can't figure out how they got it.

Answer 1

You can start with the formula for the sum (S) of the interior angles of a polygon with n sides: {{{S = (n-2)*180}}} In this problem, you have a pentagon so n =5. The sum of the interior angles can be expressed as: {{{S = 4x+4x+4x+3x+3x}}} Add up the x's. {{{S = 18x}}} Now set this equal to {{{(n-2)*180}}} where n = 5 (for pentagon). {{{(5-2)*180 = 18x}}} {{{3*180 = 18x}}} Divide both sides by 18. {{{3*10 = x}}} so that... {{{x = 30}}} The angle at vertex A is 3*x, so... {{{3x = 3*30}}}={{{90}}} {{{highlight(A = 90)}}}

Answer 2

<font face="Garamond" size="+2"> 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 3cdot180 = 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^{ipi} + 1 = 0] </font>