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See complete Problem 42Angles, Triangle. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
draw bisector of ang Cdraw ang DAE = α, E on bisector of Cjoin B to D, D to E, draw altitude BH, H on DE tr BDE is equilateralang 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 = 30ºfrom tr ABCA + C + B =( α + 30 ) + ( 30 + α ) + x = 180 x = 120° - 2α-----------------------------------------------
Sir, Did u receive my solution for Problem 42?
Yes, Thanks. Following your solution.
Draw a circle having its center B and radius BA,Draw another circle having its center C and radius CB. Let K be the lower intersection point of two circles.Connect BK and CKSo, Tri. BKC is an equlateral.Obviously ang.KBC=60degSince 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-αsimilarly BCD=2.BKDang.BKD=αnow connect BDas per the symetrical property in triangles ADB and KBD, ang.ABD also 30-αso, ang.ABC=X=ang.ABD+ang.KBD+ang.KBCtherefore X=30-α+30-α+60ie: 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. So < ABD = 30-@ and since < FBC =90-@, x = 120-2@Sumith PeirisMoratuwaSri 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.Never could solve that problem, I think now it will possible for me if I try to follow the comments here.Thank you!