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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 1027.
Let b= AC=ADWe have DC=DE= 2. b. sin(2α )∠(BED)=3α Relation in triangle AED => ED/sin(2α )= AD/sin(3α) = 2.b.sin(2α )/ sin(2α )= 2.bSo sin(3α )= ½= sin(30) => α = 10So x= 90- 4α = 50
There is a geometric solution for this problem ,right?
To Dima Problem 1027 Yes, there is a geometric solution, Thanks.
We build AH perpendicular BC DF perpendicular AB DF=DH = 1/2ED X =50 grade Erina New Jersey
Hi Erina,Please, send again your complete solution, there is some problem with the system.Thanks. Antonio
Hi Antonio Please publish the full solution 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 Erina Jersey
We build 1. AH perpendicular BC --> CH =HD =1/2 DC = 1/2ED 2. DF perpendicular AB ----> DF=DH =1/2 ED 3. triangle DEF α =10 4. X=90-4α X = 50 Erina New Jersey