the degrees of a 102 sided shape is 18,000 degrees!
18,000 degrees
The answer will depend on what a and x are and how they relate to one another. Since you have not provided this information, it is not possible to answer the question.
160 degrees
Two adjacent sides of a square and the diagonal joining their ends froms a right angle triangle. The legs of the triangle are 10 ft each and the diagonal is the hypotenuse. By pythagoras, diagonal = sqrt(102 + 102) = 10*sqrt(2) = 14.142 ft (to 3 dp).
102% = 1.02 = 102/100 = 51/50 = 11/50
The interior angles of a 102 sided polygon add up to 18000 degrees
a ikosikaicentagon
a ikosikaicentagon
There are 540 interior degrees in a pentagon So: 540-112-102-118-107 = an angle of 101 degrees
I would call it a 102-gon.
The sum of the interior angles is (102-2)*180 = 18,000 degrees.The sum of the interior angles is (102-2)*180 = 18,000 degrees.The sum of the interior angles is (102-2)*180 = 18,000 degrees.The sum of the interior angles is (102-2)*180 = 18,000 degrees.
Exterior angles add up to 360 degrees Interior angles add up 18000 degrees
Hectodigon
Assuming that in a parallelogram, angle A = 78 degrees angle B = 3y degrees and that angles A and B are adjacent interior angles of the parallelogram. 78 + 3y = 180 so 3y = 102 and so y = 102/3 = 34 degrees.
a ikosikaicentagon
180 degrees minus 102 degrees equals 78 degrees.
It's perfectly acceptable to call it an 102-agon but what's more important is knowing how to work out its properties.