See complete Problem 43
Angles, Triangle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 43
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See complete Problem 43
Angles, Triangle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
This comment has been removed by a blog administrator.
ReplyDeleteThe solution is uploaded to the following link:
ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJdTVoS3RUbVp4dVE
Reference my proof for Problem 42
ReplyDeleteLet BD meet AC at G.
Since Tr. ABC is isoceles and < ABC = 120-2@, < BCA = 30 +@
In Tr. BCG, < GBC = 90-@ and < GCB = 30+@. So x must be 60
Sumith Peiris
Moratuwa
Sri Lanka
[For the sake of easy typing, I will use "a" instead of alpha]
ReplyDelete<CBD=<CDB=90-a
<CDE=90+a
<DCE=90-x-a
<BCE=90-x+a=<BAC
<ABC=180-(90-x+a)-(90-x+a)=2x-2a
With the help of problem 42, 2x-2a=120-2a
2x=120
x=60