SOLUTION: Three angles of a pentagon are 130', 90', and 80'. Of the remaining two angles, one is 30' more than twice the other. What is the sum of the smallest two angles?
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Question 276147: Three angles of a pentagon are 130', 90', and 80'. Of the remaining two angles, one is 30' more than twice the other. What is the sum of the smallest two angles? Answer by solver91311(24713) (Show Source):
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 -gon is given by:
For a pentagon, 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 represent the measure of the smaller angle. Then the larger angle must measure . Furthermore, the sum:
must be equal to the result of the previous calculation. Putting it all together:
Just solve for 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.
