Trig-trouble

May 2019
5
0
milano
Hi everyone,

I've got a little trig-trouble here..

Immagine.png

What I know:

AB = 8 [mm]
CD = 1 [mm]
DE = 2 [mm]
EF = 4 [mm]
FG = 5 [mm]
HA = 10 [mm]
ABC = 100 [°]
BCD = 90 [°]
DEF = 90 [°]
EFG = 90 [°]
GHA = 130 [°]
HAB = 60 [°]

I need to calculate the distance AE. Any idea?

Thanks in advance!

B.
 
Feb 2015
2,255
510
Ottawa Ontario
Well, too many "points...angles...lines" for me to refer to without a schoolroom blackboard!

Here's a hint:
Draw perpendicular HK (K on AB) to AB;
right triangle AHK = 30-60-90, so AK = 5.

Good luck...
 
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topsquark

Forum Staff
Jan 2006
11,565
3,452
Wellsville, NY
Well, too many "points...angles...lines" for me to refer to without a schoolroom blackboard!

Here's a hint:
Draw perpendicular HK (K on AB) to AB;
right triangle AHK = 30-60-90, so AK = 5.

Good luck...
Angle HAB is 60 degrees, not angle HAE.

But you're right. This thing needs a blackboard to do it properly. If I had a CAD program I might try it.

-Dan
 
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Reactions: 1 person
May 2019
5
0
milano
Thanks a lot for your kind replies!

@DenisB: thanks for your hint! So we are 1 step ahead ;)
@Dan: it would be too easy con CAD... ;) actually the distance is 6,99 [mm], but I need to calculate it..

I've made a simplier and clearer model, where:

AB = 8 [mm]
AK = 5 [mm]
KB = 3 [mm]
CD = 1 [mm]
DE = 2 [mm]
EG = 6,40 [mm]
HK = 8,66 [mm]

ABC = 100 [°]
BCD = 90 [°]
DEG = 218,66 [°]
AED = 90 [°]
GHK = 100 [°]
AKH = 90 [°]

Immagine.png

Any other hint? Thanks a lot!

B.
 
May 2019
5
0
milano
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:

1.png

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:

2.png

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 :p ) on how to proceed... Any idea?

Thanks a lot, regards,

B.
 
May 2019
5
0
milano
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:

1.png

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:

2.png

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 :p ) on how to proceed... Any idea?

Thanks a lot, regards,

B.