See complete Problem 42

Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 42

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## Monday, May 19, 2008

###
Elearn Geometry Problem 42

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 42

Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

draw bisector of ang C

ReplyDeletedraw ang DAE = α, E on bisector of C

join B to D, D to E, draw altitude BH, H on DE

tr BDE is equilateral

ang DBH = ang DAC ( perpendicular sides )

ang KDE = ang DAC ( AD meetBE on K, AD = EC => DE//AC)

ang FED = ang ECA (CE meet BD on F, DE//AC )

BH bisector of DBE, DK bisector of EDB, EF of DEB

=> DBE = 6 ∙ DAC = 180º

=> DAC = 30º

from tr ABC

A + C + B =( α + 30 ) + ( 30 + α ) + x = 180

x = 120° - 2α

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Sir, Did u receive my solution for Problem 42?

ReplyDeleteYes, Thanks. Following your solution.

DeleteDraw a circle having its center B and radius BA,

ReplyDeleteDraw another circle having its center C and radius CB.

Let K be the lower intersection point of two circles.

Connect BK and CK

So, Tri. BKC is an equlateral.

Obviously ang.KBC=60deg

Since ang.BCK=60deg, ang.DCK=60-2α.

Since, ang.DCK is angle on center and ang.DBK is angle on circumferance subtended by same segment: ang.DBK=30-α

similarly BCD=2.BKD

ang.BKD=α

now connect BD

as per the symetrical property in triangles ADB and KBD, ang.ABD also 30-α

so, ang.ABC=X=ang.ABD+ang.KBD+ang.KBC

therefore X=30-α+30-α+60

ie: X=120-2α

............

Drop a perpendicular from B to AD say. BE and from C to BD say CF. Tr.s DFC, BFC and ABE are all congruent ASA. So DF = BF = BE, hence Tr. BDF is 30 60 90 with < BDE = 30.

ReplyDeleteSo < ABD = 30-@ and since < FBC =90-@, x = 120-2@

Sumith Peiris

Moratuwa

Sri Lanka

Hello, I have seen a very similar problem: B and C are known angles (ie 110 and 20), the angle to find is the one in A position.

ReplyDeleteNever could solve that problem, I think now it will possible for me if I try to follow the comments here.

Thank you!