Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1028.
Solution:1)Connect B and E Notice that Triangle ABE is isosceles.2)Let AE = AB = a, let DE = BE = EC = b(right triangle).3)AEB ~ DBE (They both isosceles and have the same angles). From here we get that AE/BE = BE/DB =a/b = b/DB --> DB = b^2/a.4)Draw EN_|_BC and EM _|_AC (notice that EN = DB/2 = b^2/2a).5)AEM ~ CEN (Same angles) --> AE/EM = EC/ EN = a / EM = b/(b^2/2a).6)From the proportion in step 5 we get that EM = b/2, in triangle ECM angle ECM = 30 = 90 - 4x.Solution : The measure of x is 15.
Trigonometry solution:Connect BE and let F is the midpoint of BEBDE and BAE are isosceles and similarLet BC= 1we have EC=EB= 1/(2.cos(x)) and BH= 1/(4.cos(x))And AB= BF/sin(x)= 1/(2.sin(2x))AC= AB/cos(3x)=BC/sin(3x) => tan(3x)=2. sin(2x)This equation has 2 solutions: x= 0 and x=15
Prob 1027 We build AH perpendicular BC and DF perpendicular AB CH=HD=1/2DC=1/2ED DF=DH=1/2ED α =10 X=90-4α X=50 Prob 1028 BE=ED=EC EM=EN EN =1/2 EC Triangle ECN < ECN =30 ECN = 90-4X -> X =15 Prob 1029 We note <BDF = 90 BD perpendicular AC Erina New Jersey
construim AN_|_BE si EM _|_AC, N pe BE,M pe BC,avemAEB ~ DEB(EM=NE=EB=BE/2=CE/2=>ECA=30, 90-4X =30=> X =15
Wonderful solution, Prof. Radu. I wish your solutions were in English, though, for Google translation is never adequate or accurate.
Trigonometric Solution 999 and 1028 ( Because they have the same picture and same solution1. EF perpendicular AB ---> BC=2EF 2. Triangle AEF EF= ABsin2X 3 Triangle ABC BC= ABtg3X Tg3X=2sin2X From geometric conditions 0 <3X < 90 ---> 0 <X<30 Equation has this solution X=15 Erina New Jersey
Hi Antonio We have 2 other geometric solutions for problems 1028 . If you want them I can send to you. Have a great night .Erina
Hi Erina,Great. Please send your geometric solutions. Thanks
Let AE = AB = q, BD = r and BE = DE = CE = pBE is tangential to ADE so p^2 = rq ....(1)From similar Tr.s EF altitude of Tr. AEF = qr/2p = p/2 from (1)So < ECA = 30 and hence x = 15Sumith PeirisMoratuwaSri Lanka