1. Find the dlat and dep made good if a vessel steams 248 T for 1,936 nm.

A. dlat 725.2’ S; dep. 1,795.1’ B. dlat 725.2’ S; dep. 1,695.1’ C. dlat 725.2’ S; dep. 1,695.1’ D. dlat 700.9’ S; dep. 1,695.1’

• 1. Solution: Find Dlat and Dep.• Sin Co. = Dep / Dist.• Dep. = Sin Co. x Dist.• Dep. = Sin 248º x 1936• Dep = 1,795.0 nm• Cos. Co. = Dlat / Dist.• Dlat = Cos Co. x Dist• Dlat = Cos 248º x 1936• Dlat = 725.2 nm• Dist inv. P–R Co. Dlat inv. x – y Dep• 1936 248º 725.23 1795=

• 2. A vessel steams 026 T. for 435 nm. What was the dlat and departure?

• A. dlat 391.0’ N; dep. 180.7’• B. dlat 380.9’ N; dep. 190.7’• C. dlat 391.0’ N; dep. 190.7’• D. dlat 700.9’ S; dep. 190.7’

2. Solution: Find Dlat and Dep. • P – R ( Polar to Rectangular )• Dist. Inv. P–R Course Dlat inv. x-y Dep• 435 026º = 390.97 190.69

• 3. A vessel steams 215 T for 341 nm. Find dlat and departure made good.

A. dlat 279.3’ S; dep. 196.6’ B. dlat 279.3’ S; dep. 195.6’ C. dlat 297.3’ S; dep. 195.6’ D. dlat 279.3’ N; dep. 195’

• 3. Solution: Find Dlat and Dep.

• P – R (Polar to Rectangular )• Dist inv. P – R Co. Dlat inv. X – Y Dep• 341 215º = 279.33 S 195.58

• 4. A vessel makes a dlat of 289.4’ N. and a departure of 203.2’ nm. Find the course and distance?

• A. Course 324 55.5’; Dist. 353.6’• B. Course 324 55.5’; Dist. 300.0’• C. Course 324 55.5’; Dist. 380.1’

• D. Course 324 55.5’; Dist. 343.0’

• 4. Solution: Find Co. and Dist.• Tan co. = Dep / Dlat• Tan co. = 203.2 / 289.4 inv. Tan• Co. = 35.07º - 360º = 324º 55’• Sin Co. = Dep / Dist.• Dist. = Dep / Sin. Co.• Dist = 203.2 / Sin 35.07º• Dist = 353.65 nm• R – P ( Rectangular to Polar )• Dist R – P Dep Dist inv x – y Co.•

289.4 203.2 = 353.6 35.07º

• 5. A vessel steers a course of 146 T from lat 35 10’ N. to lat 8 46’ N. How far did she steam?

A. 1,900.0 nmB. 1,910.7 nmC. 2,000.0 nm

D. 1,850.0 nm

• 5. Solution: Find Dist.• Lat fr. = 35º 10’ N• Lat in = 08º 46’ N ( - )• Dlat = 26º 24’ S• x 60• Dlat = 1584’ S• Cos.Co = Dlat / Dist • Dist = Dlat / Cos Co.• Dist = 1584 / Cos 146º/ 34º• Dist = 1910.65 nm

• 6. In what latitudes will a departure of 300 nm correspond to a dlong of 6 40’?

A. 46 24’ N or S

B. 39 00’ N or S C. 41 24.6’ N or S D. 43 23’ N or S

• 6. Solution: Find Latitude• Dlo = Dep / Cos Lat.• Cos Lat. = Dep / Dlo • Cos Lat = 300 / 400 • Lat = 0.75 inv. Cos • Lat = 41º 24.6’ N or S

6º 40’ x 60DLO = 400’

• 7. On a certain parallel the distance between two meridians is 250 nm while the dlong between the meridians is 12 30’. What is the latitude?

A. 70 31.9’ N or S B. 71 31.9’ N or S C. 70 00.0’ N or S D. 69 31.8’ N or S

• 7. Solution: Find Latitude• Dlo = 12º 30’• x 60• Dlo = 750’• Cos Lat. = Dist / Dlo• Cos Lat = 250 / 750• Lat = 0.333 inv. Cos• Lat = 70º 31.8’ N / S

• 8. In latitude 50 10’ N. the departure between two meridians is 360 nm. What is the dlong.?

• A. 9 00’ E or W • B. 8 33’ E or W • C. 8 00’ E or W• D. 9 22’ E or W

• 8. Solution: Find Dlo• Dlo = Dep / Cos Lat.• Dlo = 360 / Cos 50º 10’• Dlo = 562.011 / 60• Dlo = 9º 22’ E or W

• 9. A vessel steams on a course of 090 T from A in lat 23 30’ N, long 59 10’ E to B in lat 23 30’ N, long 65 30’ E. How far did she steam?

• A. 350.5 nm• B. 349.5 nm• C. 348.5 nm • D. 378.6 nm

• 10. Solution: Find Distance• Long 1 = 59º 10’ E• Long 2 = 65º 30’ E ( - )• Dlo = 06º 20’• x 60• Dlo = 380’ • Dist = Dlo x Cos Lat.• Dist = 380 x Cos 23º 30’• Dist = 349 nm

• 10. From Lat X N. a vessel steams 000 T 50 nm, and then 090 T 100 nm. If the difference of longitude is 185’. Find the Lat of X?

• A. 50 00’ 00” N B. 50 49.5’ N

C. 56 00’ 00” ND. 56 26.75’ N

• Solution: Find Latitude X • •

• Cos Lat. = Dist / Dlo • 100’ / 185’ • Cos Lat = 57º 16.8’ N • ( - ) 50’• Lat X = 56º 26.8’ N• •

X

50’ 000º

100’ 090º dlo

• 11. From lat 44 15’ N, long 10 20’ W. a vessel steamed 270 T for 550 nm, and then 180 T for 753 nm. Find her final position.

• A. Lat. 31 42’ N., Long 20 07.8’ W• B. Lat. 30 42’ N., Long 20 07.8’ W• C. Lat. 30 42’ N., Long. 23 07.8’ W• D. Lat. 31 42’ N., Long 23 07.8’ W

• 11. Solution: Find Final Position• Dlat = 753 / 60’• Dlat = 12º 33’ S• Lat 1 = 44º 15’ N ( - )• Lat 2 = 31º 42’ N• Dlo = Dist. / Cos Lat• Dlo = 550 / Cos 44º 15’ • Dlo = 767.83 / 60’• Dlo = 12º 47.8’ W• Long1 = 10º 20.0’ W ( + )• Long2 = 23º 07.8’ W

• 12. On a certain parallel, the distance between two meridians is 150 nm. On the Equator, the distance between the same two meridians is 235 nm. What is the latitude of the parallel?

• A. 50 00.1’ N or S• B. 48 00.1’ N or S

C. 49 00.1’ N or S• D. 50 20.1’ N or S

• 12. Solution: Find Latitude of Parallel• Cos Lat = Dist / Dlo• Cos Lat = 150’ / 235’• Lat = 0.638297 inv. Cos• Lat = 50º 20.1’ N / S

• 13. The distance between two meridians in lat. 48 12’ N is 250 nm. What is the angle at the pole?

A. 7 15.1’ B. 6 15.1’ C. 5 15.1’

D. 6 50.0’

• 13. Solution: Find Dlo / Angle at the Pole• Cos Lat = Dist / Dlo• Dlo = 250’ / Cos 48º 12’• Dlo = 375.075 / 60• Dlo = 6º 15.1’

• 14. A vessel steams 47nm along the parallel of Y N from long 15 35’ W to the meridian of 27 20’ W. What is the latitude of Y?

• A. 46 11.3’ N• B. 49 11.3’ N• C. 48 11.3’ N• D. 47 11.3’ N

• 14. Solution: Find Latitude• Long 1 = 15º 35’ W• Long 2 = 27º 30’ W ( - )• Dlo = 11º 45’ W• x 60• Dlo = 705’• Cos Lat. = Dist. / Dlo• Cos Lat. = 470’ / 705’• Lat = 0.66666 inv. Cos• Lat Y = 48º 11.3’ N

• 15. From lat 39 00’ N, 33 00’ W a vessel steamed 270 T at 10 knots for 3 days 8 hours. In what D.R. position did she arrive?

• A. Lat 38 00’ N, Long 48 20.4’ W• B. Lat 40 00’ N, Long 50 19.4’ W • C. Lat 39 00’ N, Long 50 19.4’ W• D. Lat 39 00’ N, Long 48 20.4’ W

• 15. Solution: Find D. R. position of Arrival• Dist. = Speed x Time• = 10 kts. X 80 hours• Dist = 800 nm• Dlo = Dist. / Cos Lat.• = 800’ / Cos 39º• Dlo = 1029.407 / 60’• Dlo = 17º 09’ W• Long1 = 33º 00’ W ( + )• Long2 = 50º 09.4’ W

• 16. The distance between two meridians is 427 nm in lat 50 20’ N. What is the angle at the pole?

• A. 12 58.1’• B. 10 41.7’• C. 10 00.0’• D. 11 08.9’

• 16. Solution: Find Dlo or Angle at the Pole• Dlo = Dist. / Cos Lat• Dlo = 427’ / Cos 50º 20’• Dlo = 668.94 / 60’• Dlo = 11º 08.9’

• 17. Two ships on the parallel of 17 S are 55 nm apart. What would be their distance apart if they were on the parallel of 40 N?

• A. 42.08 nm• B. 44.06 nm • C. 45.01 nm • D. 40.07 nm

17. Solution: Distance at 40º N Parallel• Dlo = Dist / Cos Lat• Dlo = 55’ / Cos 17º• Dlo = 57.51’• Dist. = Dlo x Cos Lat• Dist. = 57.5’ x Cos 40º• Dist. = 44.047 nm

• 18. Two ports, A and B are in the Northern Hemisphere. On the parallel of Port A, the distance between their meridians is 250 nm, on the parallel of Port B it is 350 nm, and on the Equator it is 400 nm. What are the latitudes of the ports?

• A. A 50 19.1’ N; B 26 57’ N• B. A 53 20.0’ N; B 28 57.3’ N• C. A 51 19.1’ N; B 28 57.3’ N• D. A 50 19.1’ N; B 28 57.3’ N

• 18. Solution: Find Lat. A & B• Cos Lat A = Dist. / Dlo• = 250’ / 400’• = 0.625 inv. Cos• Lat A = 51º 19.1’ N• • Cos Lat B = Dist. / Dlo• = 350’ / 400’• = 0.875 inv. Cos• Lat B = 28º 57.3’ N

• 19. A vessel in latitude 48 30’ N, steams 270 T. at 10 knots for 24 hours. By how much is the longitude changed?

• A. Dlo 6 08.7’• B. Dlo 6 00.0’• C. Dlo 6 02.2’• D. Dlo 7 00.1’

• 1919. Solution: Find Dlo• D = S x T• = 10 x 24• D = 240 nm• • Dlo = Dist. / Cos Lat• = 240 / Cos 48º 30’• = 362.19 / 60• Dlo = 06º 02.2’•

• 20. In lat. 50 20’ N. a vessel steams from long 15 46’ W to long 31 18’ W. What distance was made good? A. 554.5 nm

B. 600.0 nm C. 594.9 nm

D. 601.2 nm

20. Solution: Find Dist. Made Good• Long 1 = 15º 46’ W• Long 2 = 31º 18’ W ( - )• Dlo = 15º 32’ W• x 60’• Dlo = 932’• Dist = Dlo x Cos Lat.• Dist = 932 x Cos 50º 20’• Dist = 594.9 nm

• 21. A ship steams 090 T for 200 nm in lat 49 10’ N. By how much will her clocks have to be advanced?

A. 19m 06s B. 21m 44s C. 20m 23s D. 24m 00s

• 21. Solution: Find Time Clock Advance• Dlo = Dist. / Cos Lat• Dlo = 200’ / Cos 49º 10’• Dlo = 305.87 / 60• Dlo = 5º 05.9’ E

• 1º = 4 minutes• 5º 05.9’• x 4• Time = 00h 20m 23s

• 22. The distance between two meridians in the Northern Hemisphere is 240 nm. At the equator it is 400 nm., and in the Southern Hemisphere it is 360 nm. What is the dlat between the two parallels?

A. Dlat 78 58.3’ B. Dlat 69 44.4’ C. Dlat 77 30.6’ D. Dlat 75 15.3’

• 22. Solution: Find Dlat• Cos Lat1 = Dist. / Dlo• = 240 / 400• Lat 1 = 53º 07.8’ N

• Cos Lat 2 = Dist / Dlo• = 360 / 400• Lat 2 = 25º 50.5’ S• Lat 1 = 53º 07.8’ N ( + )• Dlat = 78º 58.3’ S

• 23. In what latitude is the departure in nm be equal to five-sevenths of the Dlo. in minutes?

A. Lat. 43 44’ N or S B. Lat. 44 25’ N or S C. Lat 41 33’ N or S

D. Lat 40 40’ N or S

23. Solution: Find Latitude• Cos Lat = Dep / Dlo• = 5 / 7• = 0.71428 inv. Cos• Lat. = 44º 24.9’ N / S

• 24. Two vessels 200 nm. apart on the same parallel are sailing on course 180 T to the parallel of 20 N, where their dlong is found to be 5 10’. How far did each steam?

• A. 1,789.3 nm• B. 1,800.0 nm• C. 1,825.0 nm• D. 1,625.0 nm

24. Solution: Find Distance• ’ Cos Lat = Dist. / Dlo• = 200’ / 310’ • Lat = 49º 49.3’ N

• Lat 1 = 49º 49.3’ N• Lat 2 = 20º 00.0’ N• Dlat = 29º 49.3’ • x 60’• Dist. = 1789.3 nm

DLO = 5º 10’ x 60DLO = 310’

• 25. Your vessel receives a distress call from a vessel reporting her position as lat 35 01.0’ S, long 18 51.0’ W. Your position is lat 35 01.0’ S, long 21 42.0’ W. Determine the true course and distance from your vessel in distress by parallel sailing.

• A. 090 T, 140.0 nm• B. 090 T, 189.2 nm• C. 270 T, 140.0 nm • D. 270 T, 189.2 nm

25. Solution: Find T/Co. and Distance• Long 1 = 18º 51’ W• Long 2 = 21º 42’ W ( - )• Dlo = 2º 51’ W• x 60• Dlo = 171’• Dist. = Dlo x Cos Lat.• = 171 x Cos 35º 01’• Dist = 140 nm Course = 090º

• 26. A vessel in latitude 55 12’ N. steamed on a course 270 T and made a dlong. of 21 36.6’. If the time taken was 3 days 2 hours, find the vessel’s speed.?

A. 10 knots B. 11 knots

C. 12 knots D. 14 knots

• 26. Solution: Find Speed• Dlo = 21º 36.6’• x 60’• Dlo = 1296’• Dist. = Dlo x Cos Lat.• Dist = 1296 x Cos 55º 12’• Dist = 740 nm• Speed = Dist. / Time• Speed = 740 nm / 74 Hrs.• Speed = 10 Knots

• 27. Two ships A and B, which are 40 miles apart on the Equator, steamed due North to the 20th parallel. What is the distance between them in this latitude?

• A. 37.6 miles• B. 38.2 miles • C. 36.0 miles• D. 39.2 miles

EQUATORVSL “A” VSL “B”

20TH PARALLEL

DISTANCE = ?

DISTANCES 40 NM

27. Solution: Find Distance• Dist. = Dlo x Cos Lat.• = 40’ x Cos 20º • Dist. = 37.6 nm

• 28. Two vessels, A and B, are on the parallel of 49 50’ N, A steering 090 T and B 270. At noon Local Apparent Time by B’s clock they were 349 nm apart, when their clocks were set to the apparent time of their respective meridians. At 10:15 p.m. by B’s clock they collide. What was the time by A’s clock, neither having been altered since noon?

• A. 21 h 38 m 56 s • B. 21 h 59 m 00 s• C. 22 h 00 m 00 s • D. 22 h 00 m 00 s

• 28. Solution: Find Time of Vessel A

VSL A Co. 090º VSL B Co. 270º

Dist 349 nm at 12 noon

Lat 49º 50’ N

At 1015 Hrs at Vsl. “B” time both vessel collide

Find: Time of Vsl. “A” during collision ?

• 28. Solution:• Dlo = Dist / Cos Lat.• = 349’ / Cos 49º 50’• Dlo = 541.07 / 60• Dlo = 09º 01’ ( Convert to Time )• x 4’ Longitude in Time = 00h 36m 04s• Time of Collision = 22h 15m 00s ( - )• Time of Vessel A = 21h 38m 56s

• 29. On certain parallel, a vessel must steam one nm to alter her longitude by two minutes. What is the latitude of the parallel?

A. Lat 60 N or S B. Lat 64 N or S C. Lat 55 N or S D. Lat 61 N or S

• 29. Solution: Find Latitude• Cos Lat. = Dist. / Dlo• = 1 / 2• = 0.5 inv Cos• Lat = 60º N / S

• 30. A vessel sails due West from Greenwich at noon of 1st January 2003, and makes a departure of 216 miles in 24 hours. Assuming it is possible for this vessel to circumnavigate the world along the parallel of Lat 60 N., find the time and date she would regain the meridian of Greenwich.

• A. Noon, Feb. 20, 2003• B. Midnight, Feb. 20,2003 • C. Noon, Feb.19,2003• D. Midnight, Feb.19,2003

• 30. Solution: Find ETA• Speed = Dist. / Time• = 216 / 24• Speed = 9 Knots• Dlo = 360º x 60’• Dlo = 21,600’• Dist = Dlo x Cos 60º• = 21,600 x Cos 60º• Dist = 10,800 nm

• 30. Solution:• Steaming Time = Dist. / Speed• = 10,800 nm / 9 knots• Steaming Time = 50d 00h 00m• Departure Time = 01d 12h 00m ( + )• 51d 12h 00m• ( - ) 31 ( Jan. )• E.T.A. = 20 Feb. 1200h 2003