## Monday, May 19, 2008

### Elearn Geometry Problem 48

See complete Problem 48
Angles, Triangle, Congruence. Level: High School, SAT Prep, College geometry

1. I've no idea how to solve this by plane geometry alone & would very much appreciate it if someone gave me a hint. In the meanwhile, AB=BC gives us x=7α. Now in Tr. ABC, AC/AB=sin(14α)/sin(7α)=2cos(7α)and in Tr. ABD, AD/AB=AC/AB=sin(5α+30)/sin(30). This gives us, 2sin(5α+30)=2cos(7α)
or sin(5α+30)=sin(90-7α), which in turn, yields(5α+30)=(90-7α) or 12α=60 or α=5 deg.
Hence x = 7α = 35 deg.
Ajit: ajitathle@gmail.com

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3. Solution to Problem 48:

Let BE the perpendicular bisector of AC.
Let F be a point on the extension of DB, so that FD⊥AF, then ∆BFA≅∆BEA.
Therefore: m∠FAB+m∠FBA=7a+5a+30°=90°,12a=60° and a=5°
Since ∆ABC is isosceles, x=7a=7(5°) and x=35°

4. Let ΔABO be equilateral triangle. ∠AOB=60° and ∠ADB=30°, so OA=OB=OD=AB.
---> ∠AOD=7α ---> ∠OAB=12α=60°---> α=5° ---> x=35°

5. The key is to use the 30 degrees.

Draw a 30-60-90 Tr. on AD, Tr. ADE, with AE perpendicular to BD.

If BF is the perpendicular bisector of AC then easily Tr. ABE is congruent with Tr. BCF.

So < EAB = x = 7@ and hence 12@ = 60 and so x = 7@ = 35

Sumith Peiris
Moratuwa
Sri Lanka