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Geometry puzzle (13) |  |
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This geometry puzzle was found in Facebook group "Classical Mathematics".
Solutions are partly my own.
Question : calculate angle x.
Several approaches exist.
1.Goniometric + calculator
This is the most simple way.Let BD = 1.
So:
DE = tan(x)
DC = tan(60)........1)
DE = AD.tan(10).......2)
DC = AD.tan(20).......3)
combining:
Calculator:
DE = 0,839099631
x = arctan(0,839099631) = 40
2.Goniometric without calculator
DE = tan(40)Multiply in the last equation 1) numerator and denominator by tan(40)
Now this must be true:
tan(60).tan(10) = tan(20).tan(40), after division tan(40) remains.
Geometric picture:
Look at quadrilateral ABCD above.
Known angles are pictured in black.
Triangles ASD, ASB, CSD may be constructed.
But a question remains: what are the angles of triangle BCS: 40 and 50 degrees?
Who helps?
3.Geometric solution
Look again at the first picture.Edges of lines AC, AE are extended to make isosceles triangle ABG.
AF is perpendicular bisector of BG.
LFEB = x+10..........{exterior angle of ΔABE}
For reasons of clarity a part is lifted out and enlarged:
2DB = BG....{because LDBG=60}
2BF = BG....{because F is center of BG}
so:
BD = BF
LDBF=80
LBDF=LBFD=50
Construct the circumscribed circle of quadrilateral DBFE....{BE is centerline because 90 degree angles D and F}
LFEB=LFDB....{both on arc BF}
x+10 = 50
x = 40
