## Monday, May 19, 2008

### Elearn Geometry Problem 44

See complete Problem 44
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

#### 6 comments:

1. Tr. ABC gives us, AC/AB = sin(x)/sin(2α)
Tr. ACD gives us, AC/CD = sin(3α)/sin(2α)
But AB=CD, so sin(x)/sin(2α)= sin(3α)/sin(2α)
Thus, sin(x)=sin(3α)or x=3α. Now in Tr. ABC,
3α+4α+2α =180 or α =20 deg. and x=3α=60 deg.
Ajit:ajitathle@gmail.com

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4. The solution is uploaded to the following link:
https://docs.google.com/open?id=0B6XXCq92fLJJWHh3X19VYWFYNjQ

Sir, My i send my solution in tamil language as a copy, so that my students can easily understand the way.
With regards,
PradheepKannah.

1. To Rengaraj, yes you can. Thanks.

5. Let AD meet BC at E. Drop perpendiculars DH and DJ from D to AC and AB respectively. Drop another perpendicular BFG from B to AF to meet the bisector of < BAD at F and AD at G

Since D is the incentre of Tr. ABC < ABD = x/2 = 90-3@. Since < ABG = AGB = 90-@, < DBG = 2@ and so BD = BG.

Now Tr. s CDH, ABF & AFG all being congruent ASA, BG = BD = 2r where r is the inradius. But JD also = r so in right Tr. BDJ, BD = 2DJ hence this Tr. is 30-60-90.

So x/2 = 30 and x = 60

Sumith Peiris
Moratuwa
Sri Lanka