I`d appreciate help with this basic Tranvese question!

30-6shooter

Can someone tell me how to do this please.

Heres a tranverse,, its bearings, (i can do this part ok)

AB 45
+180

---

225
-90

---

BC 135
+180
-

---

315
-90

---

CD 225
+180

---

405
-90

---

DA 315
+180

---

495
-90

---

405 = 45 (405 - 360 = 45)

I can do the above fine.


Line Length WCB X Y

AB 100 45 70.711 -70.711
BA 105 135 74.246 74.246
CD 100 225 -70.711 -70.711
DA 105 315 -74.246 74.246

Heres where im stuck:

I dont follow how to work out X and Y.
I thought x = length Sin(WCB)
which would mean 100 Sin(45) which is 85.090 acc to my calculator.

And to work out y its just length Cos (WBA) which i work to be 52.532.



SO basically,,, im asking is where did the figures of 70.711 and -70.711 come from???

PS: its a 4sided tranverse wit co-ordinates of A 0,0 B 0,60 C 100,60 D100,0


Can anyone please tell/show me.
Please,,,
Thanks.

apeking

30-6shooter wrote:

-
I thought x = length Sin(WCB) which would mean 100 Sin(45) which is 85.090 acc to my calculator.

U have your calc set to radians and u will get 85.090, set it to degrees and u will get 70.11

30-6shooter

Feck feck feck.

Thanks apeking, i cant believe i done that. Took me ages to type that post. :mad:

apeking

no problem man, i'm always doing that aswell thats why i thaught that is what u might have done., It was an impressive post all the same, say it took a while!! Those questions are nice and handy anyway but if u make a mistake in one of the early calculations it can be a pain going back through the whole thing.

30-6shooter

Hey again Apeking,
maybe you can heklp me again.

How do i wok out the x y corrections and the corrected x y coords?
Thanks.

apeking

Sorry mate,
Dont actually know. Isn't there another formula?? Did it last year so cant remeb what to do.