Question 276147
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I'm going to just assume that you used ' to mean degrees of arc as opposed to minutes of arc which is the traditional meaning of that symbol.


The sum of the interior angles of any *[tex \Large n]-gon is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n\ -\ 2\right)180]


For a pentagon, *[tex \Large n] is 5.  So calculate the sum of the interior angles of a pentagon.


From that calculation result, subtract the sum of 130°, 90°, and 80° leaving you the sum of the measures of the remaining two angles.


Let *[tex \Large x] represent the measure of the smaller angle.  Then the larger angle must measure *[tex \Large 2x\ +\ 30].  Furthermore, the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2x\ +\ 30]


must be equal to the result of the previous calculation.  Putting it all together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2x\ +\ 30\ =\ \left(5\ -\ 2\right)180\ -\ \left(130\ +\ 90\ +\ 80\right)]


Just solve for *[tex \Large x] to get the measure of the smaller of the two remaining angles.  This will turn out to be the smallest of all 5 angles.  The other one will be the largest of all 5.  Discover its measure by multiplying the measure of the smallest angle by 2 and adding 30 degrees.  Now that you know the measure of all 5 angles, it should be an easy matter to add the measures of the two smallest ones.


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