Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 872.
http://img267.imageshack.us/img267/4236/problem872.pngDraw circle center B radius BA.Draw circle O Radius= OB=BA=OADraw common tangent to these 2 circles.This tangent will parallel to BO Circle O cut AD at F and ∠ (AFB)= 30 degreesNote that BD is angle bisector of ∠ (ADE) and DE ⊥CO => ∠(CBF)= 45And ∠(CDF)= ∠(BFA)= 30 degrees so x=15 degrees.Verify that ∠(FBD)=15 and ∠ (CBD)= 45-15=30 degrees= 2.x
My trigonometric solution suggests that x may be 15° or 30°. (In both cases, it is the same 30°-60°-90° triangle). If it were the latter, would Peter's solution hold?A²
Must mention here that if x=30° then /_BCD=90° which is contrary to the given condition & thus x=30° has to be rejected and, therefore, x=15°
Ajithttp://imageshack.us/a/img29/3016/problem8721.pngYes, there is 2nd solution with x= 30 .let 2 circles intersect at A and C' and AD cut circle O at D'note that D'C' tangent to circle O and angle BC'A= 30 , angle AD'B=angle BD'C'=30 and angle C'BD'=60 .Peter
Peter Tran:Why does the common tangent pass through C? You are going by the figure you are building, not proving such a fact. If x=20 degrees, you will see why.....Erina-NJ
ErinaC is any point on circle O such that angle(CBD)=2.angle (CDB). D is the intersection of horizontal line through A and the tangent line from C to circle B.We have 2 solutions :solution 1: C is located on the common tangent line of circles O and B by accident solution 2: C is located on the intersection of circle O and circle B ( not on common tangent line)Peter
We are sure that C is on the Circle O, but are not sure that CD is tangent with Circle B. This is what needs to be clarified. If you can prove this then your 2 solutions are clear.1) C is not the same as E, x=15 degrees2) C is the same as E, x=30 degreesErina-NJ