tHE MEASURES OF THEANLGES OF A CERTAIN POLYGON FORM AN ARITHMETIC SEQUENCE. tHE LEAST MEASUREMENT IN THE SEQUENCE IS 128 DEGREES. THE GREATEST MEASUREMENT IS 172 DEGREES. FIND THE NUMBER OF SIDES IN THIS POLYGON.
HOW DO YOU DO THIS??? PROVE IT!!!
2007-02-21 23:18:05 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If these are internal angles - due to it having 2 different angle sizes, you have an irregular polygon.
I think it is a decagon - your two angles add up to 300 degrees, for the other pairs of angles this would mean 5 sets of 300 degree angles - but the other angles could be just in-between these regions (For a Decagon the Total internal angles = 1440 degrees) see this link for help on irregular polygons - you can even play with the shapes so that you can see how the angle systems work.
I'm pretty sure it's a decagon though
http://www.mathopenref.com/polygoninteriorangles.html
2007-02-21 23:50:57 · answer #1 · answered by Doctor Q 6 · 0⤊ 0⤋
Ok, several ways to do this - the 'easiest' is probably by trial and error, but will try and give you a more mathematical solution.
Let n= no. of sides or no. of angles.
In any polygon the sum of the angles = 180(n-2)
The average angle in any polygon is 180(n-2)/n
The average angle in your polygon is 150 degrees.
Hence 150 = 180(n-2)/n
150n=180(n-2)
150n=180n - 360
360=30n
12 = n
Your polygon has 12 sides.
2007-02-22 07:51:58 · answer #2 · answered by mustafa 2 · 0⤊ 0⤋
OK I AM NOT SURE BUT TRY THIS AS U SAID THE ANGLES ARE IN ARTHMETIC PROGRESSION SO THERE IS A FORMULA FOR SUM TO N TERMS.Sn=n/2(A+An)
here a=a72 and An IS 128 SUM OF ALL ANGLES THERE IS SUM FORMULA FOR SUM OF ANGLESOF CLOSED BODY OF N SIDES.SUBSTITUTE THAT AND HURRAY HERE IS UR ANSWER
2007-02-22 07:50:47 · answer #3 · answered by priyaprayag1989 1 · 0⤊ 0⤋