Ruide RCS - Page 69

66
Then CL=200 * 95°52′11″*
180


=334.648 m
Tangent length
 )
2
tan(
IA
RTL
200 * tan(95°52′11″/ 2) =221.615 m
Each coordinates are computed:

22
cos TL
NN
IPBC

2
2
sin TL
EE
IPBC

32
cos TL
NN
IPEC

3
2
sin TL
EE
IPEC
Here:

2
(Bearing from IP1 to IP2) = 322°07′30.1″

3
(Bearing from IP2 to EP) = 57°59′40.6″

N
BC
1750 - 221.615 * cos322°07′30.1″ =1575.068 m

E
BC
1400 - 221.615 * sin322°07′30.1″ =1536.058 m

N
EC
1750 –(-221.615) * cos57°59′40.6″=1867.456 m

E
EC
1400 –(-221.615) * sin57°59′40.6″=1587.929 m
The calculated results display as below: