Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 44Angles, Triangle. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
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:email@example.com
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The solution is uploaded to the following link:https://docs.google.com/open?id=0B6XXCq92fLJJWHh3X19VYWFYNjQSir, My i send my solution in tamil language as a copy, so that my students can easily understand the way.With regards,PradheepKannah.
To Rengaraj, yes you can. Thanks.
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 GSince 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 = 60Sumith PeirisMoratuwaSri Lanka