In Fig. 2, draw the known courses a, c, and e. Then introduce the hypothetical course/
to form a closed traverse.
- Calculate the latitude and departure of the courses
Use these relations:
Latitude = L cos a (1)Departure = L sin a (2)Computing the results for courses a, c, e, and/ we have the values shown in the following
table.
Course Latitude, ft (m) Departure, ft (m)
a +93.5 +59.5
c -125.4 -19.5
e +85.4 -29.0
Total +53.5(+16.306) +11.0(+3.35)
/ -53.5(-16.306) -11.0(-3.35)- Find the length and bearing of f
Thus, tan otf= 11.0/53.5; therefore, the bearing off= Sl 1°37'W; length off= 53.5/cos af
= 4.6 ft (16.64m). - Complete the layout
Complete Fig. 2 by drawing line d through the upper end of/with the specified bearing
and by drawing a circular arc centered at the lower end of/having a radius equal to the
length of b. - Find the length of d and the bearing of b
Solve the triangle fdb to find the length of d and the bearing of b. Thus, B = 73°31' -
11°37' = 61°54'. By the law of sines, sin F -/sin BIb = 54.6 sin 61°54V83.6; F= 350 11';
D = !go
0- (61°54' + 35
0
11') = 82°55'; d = b sin D/sin B = 83.6 sin 82°55Vsin 61°54' =
94.0 ft (28.65 m); ab = 180° - (73°31' + 350 Il') = 71°18'. The bearing of b = S71°18'E.
- (61°54' + 35
FIGURE 2