Question

Given: R = 1500.00 ft, Delta = 12 degrees 00'00" Right, and PC Station = 139+92.92. The azimuth from the PC to the PI was 40 degrees 00 minutes 00 seconds.

If a total station was setup at 142+00.00 on the curve, backsighting the PC, and 0 set, determine the angle right necessary to layout the PT of the curve.

Question 10 options:

	183 degrees 57 minutes 18 seconds
	192 degrees 00 minutes 00 seconds
	186 degrees 00 minutes 00 seconds
	3 degrees 57 minutes 18 seconds

Answer 1

- onu 533 R Civen, R: 1500', s. 12° pe station = 139 +92-92 Length of cowe, La 0.01745 33 RA i la 0.01745 33. (1500) (iz) - 314.1594 PT at station & PC +2. (139+ 92.92 ) + (314.15 au) PT cat station - 143+07.08

180 lection L2RR Deflection angle orenc Deflection angle 6 142700 = 180 ( 142700-1399292 (207.09 240 x 150 X (207.08) Po PC e from com Col=314-1594 more 3.955 Deflection angle from to PT PC 180 TISH (314.16) 2x K x 1500 z 6°

PIDA sso 14270 13.955° PC 139 +92.92 6 92 ous - 22.073 -3.955. Tesco pt 143 +07.07 =3.0156

W US Ica 143 07.08 Deflection ale to 142700 (107:08) -14200 QMA 180 2xTx1500 = 107.08 = Tom from PT = 2:045 figure Vom vetov 22.045 Therefore from the figure The angle in the total station backsighting PC asoo should read as 186'0' when sighted to PT from 142 too. T option c ] 142 +00. ption trom