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.
Thursday, July 10, 2014
Geometry Problem 1027: Triangle, Double, Quadruple, Angle, Congruence
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Let b= AC=AD
ReplyDeleteWe 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.b
So sin(3α )= ½= sin(30) => α = 10
So x= 90- 4α = 50
There is a geometric solution for this problem ,right?
ReplyDeleteTo Dima
DeleteProblem 1027
Yes, there is a geometric solution, Thanks.
We build
ReplyDeleteAH perpendicular BC
DF perpendicular AB
DF=DH = 1/2ED X =50 grade
Erina New Jersey
Hi Erina,
DeletePlease, send again your complete solution, there is some problem with the system.
Thanks. Antonio
Hi Antonio
ReplyDeletePlease 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
ReplyDelete1. 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
another solution at https://gogeometry.blogspot.com/2014/07/geometry-problem-1027-triangle-double.html
ReplyDelete