Dear Gentlemen,

thanks for your kind replies!

So, I've created a simplier model and I've done some steps ahead.

The new model you can see in this picture:

AB = 11 [mm]

CD = 2 [mm]

DE = 2 [mm]

EF = 2,5 [mm]

FG = 3 [mm]

HA = 9 [mm]

ABC = 110 [°]

BCD = 90 [°]

DEF = 90 [°]

DEA = 90 [°]

EFG = 90 [°]

GHA = 150 [°]

HAB = 60 [°]

I need to calculate the distance between AE.

(@Dan: with a CAD program it is too easy..

Actually the distance AE is 11,949 [mm], but I need to calculate it!)

The steps I've made are the followings:

Pythagorean theorem:

EG = 3,91 [mm] (EF^2+FG^2 = EG^2 = 2,5^2 + 3^2)

GEF = ARCCOS ( FG / EG ) = 50,19 [°]

GEA = 180 – GEF = 180 - 50,19 = 129,81 [°]

Law of Cosine:

BH = [AB^2 + HA^2 - 2*AB*HA*COS(HAB)]^2 = 10,15 [mm]

ABH = 50,17 [°]

BHA = 69,83 [°]

Then:

GHB = GHA – BHA = 150 – 69,83 = 80,17 [°]

HBC = ABC – ABH = 110 – 50,17 = 59,83 [°]

CJG = 360 – HAB – GHA – ABC = 360 – 60 – 150 – 110 = 40 [°]

Law of Sines:

HJ = SIN(GHB) / SIN(CJG) * BH = SIN(80,17) / SIN(40) * 10,15 = 15,56 [mm]

BJ = SIN(HBC) / SIN(CJG) * BH = SIN(59,83) / SIN(40) * 10,15 = 13,65 [mm]

Law of Cosine:

DG = [DE^2 + EG^2 – 2*DE*EG*COS(GED)]^2 = [2^2 + 3,91^2 + 2 * 2 * 3,91 * COS(140,19)]^2 = 5,59 [mm]

(GED = 90 + GEF = 90 + 50,19 = 140,19 [°])

IGE = 13,24 [°]

EDG = EDI = 26,57 [°]

Then:

DI = DE / COS(EDI) = 2 / COS(26,57) = 2,24 [mm]

IG = DG – DI = 5,59 – 2,24 = 3,35 [mm]

EI = TAN(EDI) * DE = TAN(26,57) * 2 = 1 [mm]

At this point I'd appreciate a RICH hint (DenisB

) on how to proceed... Any idea?

Thanks a lot, regards,

B.