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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Problem 1313Is <COB=360/10=36, then <COQ=36/2=18, so <ABC=144.But <CQB=360/15=24 and <MBC=156, so <ABM=360-144-156=60, therefore triangle ABM is equilateral.Similartriangle CDN is equilateral (OC=OD, CN=DN) so ON is perpendicular bisector of CDthen <NOC=36/2=18 or <NOQ=18+18=36.But <CQN=24, <OQC=24/2=12 (OB=OC, QB=QC)and <NQO=24+12=36=<NOQ.Therefore NO=QN.APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE
‹ABM=360°-(144+156)=60°, AB=BM => ∆ABM equilateralE,F on CD, BC. ‹EOF=180°-144=36°, ‹CQN=180-144=36°, ‹FQC=180-(90+78)=12° => ‹FQN=36° or ∆ONQ isoceles
Exterior angle of decagon = 360/10 =36Exterior angle of pentadecagon = 360/15 =24Hence < ABM = 36+24 = 60.But AB = BC = BMSo ∆ ABM is equilateralSimilarly ∆DCN is equilateral. Hence ∆ OCN ≡ ∆ ODN.Therefore < NOC = ½ < COD = ½ X 360/10 = 18. Similarly < COQ = 18 and so <NOQ = 36.Also < NQO = 1½ X 360/15 = 36Therefore ON = QNSumith PeirisMoratuwaSri Lanka
Number of sides of regular polygons which when placed together so that one side is common and forms an angle of 60 (other than 10,15)7,428,249,1812,12
Good info Sumith.Any two regular polygons that satisfy the equation 6(n1+n2) = n1*n2 form an angle of 60. Where n1 = no of sides of first polygon n2 = No of sides of second polygon.
Yes the above pairs are the only ones that satisfy the equation 360/n + 360/m = 60 which is equivalent to the equation u have given.Those are the only integer solutions