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

Labels:
congruence,
double angle,
quadruple,
triangle

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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