Originally Posted by

**masiboy** Hi,

I am not sure if i have posted this question in rite forum.

it would be great if anyone can help me or give a tip of where to start from.

Refering to the attached doc.

Calculated the distances for boundary lines EF,FG,GH,HI,IJ and JA.

Calculated the bearing and distances for the half anges lines BI,CH and DG.

To your document:

add points x & y

Code:

/
B____x______________________/
/ C /
/ /
_____________________________/
/I y H
/
/
/

Construct point x:

Draw a line perpendicular to line BC, that

passes through point I.

Where the perpendicular line intersects the line BC,

call that point x

To clarify: angle BXI is 90 degrees.

(and,also angle HIX is 90 degrees.)

Line BC and Line JH are parallel, being 10m apart.

The distance between point x and point I is 10m.

Azimuth AB= 065:00:00 (65 deg 0 min 0 sec)

Azimuth BC= 128:30:00 (128 deg 30 min 0 sec)

The angle ABC is 116:30:00 (116 deg 30 min)

Lines AB & JI are parallel and Lines BC & IH are parallel

The angle CBI is half of angle ABC.

Angle CBI is 58deg 15min 00sec ( 58.25 degrees ) [15min is 0.25deg]

Distance BX = 10 Cotan(58.25deg) = 6.188

Construct point y:

Draw a line perpendicular to line IH, that

passes through point C.

Where the perpendicular line intersects the line IH,

call that point y

To clarify: angle BCY is 90 degrees.

(and,also angle IYC is 90 degrees.)

Line BC and Line IH are parallel, being 10m apart.

The distance between point y and point C is 10m.

Azimuth BC= 128:30:00 (128 deg 30 min 0 sec)

Azimuth CD= 074:40:00 (74 deg 40 min 0 sec)

The angle BCD is 126:10:00 (126 deg 10 min 0 sec)

The angle IHC is half of angle BCD.

Angle IHC is 63deg 5min 00sec ( 63.08333333 degrees ) [ 5min is 0.0833333333 deg]

Distance HY = 10 Cotan(63.08333333deg) = 5.077

The external or outside angle of ABC is the

same as the external or large angle of JIH

which is 360deg minus 116deg30min = 243:30:00

Distance BC is 50.29

Distance IH is 50.29 minus distance BT plus distance HY.

IH = -6.188 + 50.29 + 5.077 = 49.179

:::

Use the same process to determine distance HG.

for line JI:

The angle BAJ is 103deg30min

The adjustment along line IJ at point A will be

Tan(13deg30min)*10 = 2.401

for line FG:

the angle DEF is 139deg49min

the adjustment along line FG at point E will be

Tan(49deg49min)*10 = 11.840

As a check:

distance IJ: 56.50 - 6.188 + 2.401 = 52.713

distance FG: 60.50 - 6.766 +11.840 = 65.574

That is the idea. You should be able to complete the other required distances.