Traverse Question 3?

Got a bearing for BD as 175d21'41" and then worked out angle B' as 34d49'10" and angle C as 64d05'09" not sure if this is right and didnt know where to go next . Any help again would be great.

http://www.imageuploads.net/ims/pic.php?...

Answer:
The azimuth of BC is S 140d32'31" E which places it in quadrant IV at 39d27'29" east of the vertical axis. therefor
Ec= BC(sin 39d27'29")=1000 + .6355BC
Nc=-BC(cos39d27'29") =1000 - .7721BC

I would calculate the eastings and northings counterclockwise from B to D.

The algebraic sum of these will be the distance that D is south and east of B

from thses dat you can then calculate the length and bearing of BD.

from the given bearings of BD and BC the angle CBD can be calculated.

Now you know two sides and an angle in triangle BCD you can calculate angle c with law of sines:
BD/sinBCD = CD/sinCBD

now take the two known angles and subtract their sum from 180 and you have angle CDB. apply the law of sines again to calculate length BC:
BC/sin CDB = CD/sinCBD

now substitute BC into the te equations for Ec and Nc given above and solve for the coordinates.

I have to go now. If you have any problem e mail me.

The answers post by the user, for information only, FunQA.com does not guarantee the right.

FunQA.com

Traverse Question 3?

Topic