Chart work - GC previous questions solved.

 MARCH 2020

A= 30°N            060°E

B= 50°S             155°E

Dlat- 80°S dlong- 95° E

Cos AB = Cos 95°Sin120°Sin40° + Cos 120°Cos 40°

AB        = 115° 33.9’ = 6933.9 miles

Cos A = Cos 40° -Cos 120°Cos 115°33.9’

                    Sin 120° Sin 115°33.9’

A         = 45.3° ~45°13.3’

Course initial = 134.7° T

Cos B =Cos 120° - Cos 40° Cos 115°33.9’

                      Sin 40° Sin 115° 33.9’

B         = 106.9° T final course

Vertex lies outside:

Sin(90-PA)= Tan(90-P) Tan(90-A)

Tan(90-P)= Sin(90-PA) / Tan(90-A)

P               = 116°44.6’

Longitude of vertex = 176° 44.6E

Sin (PO)=Cos(90-PA) Cos(90-A)

PO        = 37°56’

Latitude: 52°4’S

Jan 2020 (PM)

A= 15° S          012°E

B=07°N           053°W

Dlat- 22°N Dlong- 65°W

Cos AB = Cos 65°Sin 97° Sin 75° + Cos 97° Cos 75°

AB         = 68° 3.6’ = 4083.6 miles

Cos A    = Cos 97° -Cos 68°3.6’ Cos 75°

                     Sin 68° 3.6’ Sin 75°

A            = 104.1° T

Course initital =255.9° T

Cos B = Cos 75° - Cos 97°Cos 68° 3.6’

                     Sin 97° Sin 68v 3.6’

B         = 70.7°

Course final = 289.3°T

JAN 2020 (AM)

A= 56° 45’N           065° 32’E 

B= 33° 36’ N          132°20’E

Dlat- 23°09’ S Dlong- 66° 48’E

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\mARCH 16.png

Cos AB =Cos 66°48’ Sin 33°15’ Sin 56° 24’ = Cos 33° 15’ Cos 56°24’

AB         = 50° 0.4’ = 3000.4 miles

 Cos A    = Cos 56°24’ – Cos 33°15’Cos 50°0.4’

                       Sin 33°15’ Sin 50°0.4’

A             = 87.8°T (initial course)~87°49.8’

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\MAR 16.png



Cos B= Cos 33°15’ – Cos 56°24’ Cos 50° 0.4’

                   Sin 56°24’ Sin 50° 0.4’

B     = 41.1°

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\MAR 16(1).png

Course final = 138.9 ° T

Vertex inside:

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\MAR 16(2).png

Sin(90-PA) =Tan(90-P) Tan(90-A)

Tan(90-P) =Sin(90-PA)/ Tan(90-A)

P                 = 2°35.7°

Longitude of vertex = 068°7.7’E

Sin PO = Cos (90-PA) Cos( 90-A)

PO        = 33°13.3’

Latitude of vertex = 56° 46.7’ N

3rd September 2019-

LAT - 38° 30’S

LONG – 025 ° 30’W

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\SEPT 19.png

Cos A = CosPB – CosAB* CosAP 

                 Sin AB *Sin AP

CosPB = Cos A* Sin AB *Sin AP + Cos AB*Cos A

CosPB =  Cos 30 Sin 16°40’ Sin 51°30’ + Cos16°40’Cos51°30’

      PB = 37°44.7’

 Latitude final = 90°-PB = 90° - 37°44.7’

                          = 52°15.3’ S

Cos P =Cos AB –Cos AP Cos PB

                Sin AP Sin PB

Cos P = 13°32.9 ‘ W = Dlong

Longitude final = 039°2.9’ W


Final course 

Cos B = Cos 51°30’ –Cos 16°40’ Cos 37°44.7’

                  Sin 16°40’ Sin 37° 44.7’

       B= 140.3

Final course = 360° - 140.3°

                      = 219.7° T


3rd September 2019-

A=77°08’S                      166° 30’E

B=37° 49’ S                      122° 25’ W

Dlat= 114° 57’ N Dlong- 71°05’ W

Cos P = Cos AB – Cos PB Cos PA

                 Sin PB Sin PA

Cos AB = Cos P Sin PB Sin PA + Cos PB Cos PA

              = Cos 71° 05’ Sin 127° 49’ Sin 12° 52’ + Cos 127° 49’ Cos 12° 52’

              = 122° 43.9’ = 7363.9  miles

Cos A = Cos 127°49’ – Cos 12°52’ Cos 122°43.9’

                        Sin 12°52’ Sin 122°43.9’

         A = 117.3 ° ~117°19.9’

Course = 180° + 117.3° = 297.3°T°

Cos B = Cos 12°52’ – Cos 122°43.9’ Cos 127° 49’

                 Sin 122°43.9’ Sin 127°49’

        B = 14.5° T

Course = 360° - 14.5°

              = 345.5° T

vertex outside 

Sin(90-PA) = Tan (90-P) Tan (90-A)

Tan(90 –P)= Cos PA / Tan( 90-A)

P= 27°55.8’

Long = 138°34.2’ E

Sin P = Cos (90-A) Cos( 90 – PA)

    PO = 11°24.6’

Latitude = 78° 35.4 ‘ S


10th JULY 2019

  1. 10° N             040° E

  2. 30° N             100° E

               Dlat-20°N           Dlong- 60° E 

Cos P= Cos AB- CosAP CosPB

                  Sin AP Sin PB

Cos AB = Cos 60° Sin 80° Sin 60° + Cos 80°Cos 60°

AB        = 59°7.1’ = 3547.1’

Cos A    = Cos 60° -Cos 80°Cos 59°7.1

                  Sin 80° Sin 59° 7.1’

A     = 60.9°

        = 60.9° T

Cos B = Cos 80° - Cos 60°Cos59°7.1’

                Sin 60°Sin 59°7.1’

        B = 96.4°

final course = 180°-96.4° = 83.6° T


6th MAY 2019

A = 25°15’ N                045°12’E

B = 10° 40 ‘N                130°20’E

Dlat- 14°35’ S     Dlong- 085°08’E


Cos AB = Cos 85°8’ Sin 64°45’Sin 79°20’ + Cos 64°45’ Cos 79°20’

AB         = 81°7.2’ = 4867.2’

Cos A    = Cos 79°20’-Cos 64°45’Cos 81°7.2’

                   Sin 64°45’ Sin 81°7.2’

A            = 82.3°T (initial course)

Cos B = Cos64°45’ – Cos79°20Cos 81°7.2’

                       Sin 79°20’Sin 81°7.2’

B         = 65.8° 

Course final = 180° - 65.8°

                       = 114.2° T


6th MARCH 2019

A= 35°40 ‘N     141°00’E

B= 45° 00 ‘N     174° 51.8’ E

Dlat- 9°20’ N Dlong - 44°8.2’ W


Cos P = Cos AB-CosPA Cos PB

           Sin PA Sin PB

Cos AB = Cos 44°8.2’ Sin 54° 20’ Sin 45° + Cos 54°20’ Cos 45°

        AB = 34° 27.3’ = 2067.3

Cos A = Cos 45° -Cos 54°20’ Cos 34° 27.3’

                 Sin 54° 20° Sin 34° 27.3’

A         = 60° 30.1 = 60.5°T (initial course).

      

Cos B = Cos 54° 20’- Cos 45° Cos 34° 27.3°.       

                 Sin 45° Sin 34° 27.3°.                                 

B          = 89.9 ° 

Final course = 180° - 89.9 = 90.1° T.   

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\MAR 19.png

3rd  Jan 2019

A=37° S           019°E

B=56°S            067° W


Cos AB = Cos 86°Sin 34° Sin 53° + Cos 34°Cos 53°

AB        = 57°59.4’ =3479.4 miles

Cos A = Cos 34°- Cos 57°59.4’Cos 53°

                  Sin 57°59.4’ Sin 53°

A          = 41° 8.2’

             = 180° + 41°8.2’ = 221.1 ° T

Cos B = Cos 53° -Cos 57°59.4’ Cos 34°

                  Sin 57° 59.4° Sin 34°

B        = 69° 58.5 = 69.9 ° T. 

      

          = 290.1° T(360° - 69.9°)

Vertex inside

Sin(90-PB)= tan(90-B) tan(90-P).       

Sin(90-PB)/tan(90-B) = tan(90-P)

P= 23°43.9’’ 

Longitude of vertex = 043°16.1’W

Sin PO= cos (90-B) Cos(90-PB)

Sin PO = Cos(90-69.9) Cos( 90-34).        

PO       = 31°41.3’

Lat of vertex = 58°18.7’ S

2ND NOV 2018

A= 49° 50’N     005° 15’W

B= 32°29’N      064° 00 ‘W

Dlat- 17°21’ S Dlong- 58°45’W

Cos AB = Cos 58°45’ Sin 57°31’Sin 40°10’ + Cos 57°31’ Cos 40°10’

AB         = 46° 9.5’ = 2769.5 miles

Cos A =Cos 57°31’- Cos 40°10’ Cos 46°9.5°

                Sin 40°10’ Sin 46° 9.5’

A          = 89.1° 

Course initial = 270.9° T (360-89.1)


Cos B = Cos 40°10° - Cos 57°31’ Cos 46°9.5’

                   Sin 57° 31’ Sin 46°9.5’

B         = 49.9°

 

Final course = 229.9° T (180+ 49.9)

5th September 2018

A= 25°15’ N       045°12’E

B= 10°40’N        130°20’E

Dlat- 14°35’ S dlong- 85°08’ E

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\SEPT 18.png

Cos AB= Cos 85°08’Sin 64°45’Sin 79°20’ + Cos 64°45’ Cos 79°20’

AB       = 81°7.2’ = 4867.2 miles

Cos A = Cos 79°20’ – Cos 64°45’Cos 81°7.2’

                  Sin 64°45’ Sin 81°7.2’

A         = 82.3 ° T

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\SDEPT 18.png

Cos B=Cos64°45’-Cos79°20’Cos81°7.2’

                Sin 79°20’ Sin 81°7.2’

B     = 65.8°

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\SDEPT187.png

Course final = 180° - 65.8° = 114.2° T

3rd July 2018

A= 07°N               053°W

B= 15°S                012°E

Dlat-22°S dlong -65°E

Cos AB =Cos 65°Sin97° Sin 75° + Cos75°Cos 97°

AB        = 68° 3.6’ = 4083.6 ‘

Cos A   = Cos 75° - Cos 68°3.6’ Cos 97°

                        Sin 68°3.6’ Sin 97°

 A         = 70.7° 

Course = 109.3° T

Cos B= Cos 97°- Cos75°Cos 68°3.6’

                    Sin75°sin68°3.6’

B        = 104.1° T = final course


3rd MAY 2018 / JAN 2017

C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\SEPT 19.png

Cos A = Cos PB –Cos PA Cos AB

                    Sin PA Sin AB

Cos 30 = Cos PB –Cos 16°40’ Cos 51°30’

                        Sin 16°40 Sin 51°30’

Cos PB = Cos 30°Sin 16°40’ Sin 51°30’ + Cos 16°40 Cos 51°30’q

PB          = 37° 44.7’

Latitude arrived = 90° -37°44.7’ = 52°15.3’ S

Cos P = Cos 16°40’ – Cos 37°44.7’ Cos 51°30’


                      Sin 37°44.7’ Sin 51°30’

P         = 13°32.9’

Arrival longitude = 39°2.9 ‘ W.   

Cos B= Cos51°30’-Cos 16°40’Cos 37°44.7’

                    Sin16°40’ Sin 37°44.7’

B       = 140°15.8’

Final course = 219.7°T

6th MARCH 2018

A= 10°N          40°E

B= 30°N          100°E

Dlat- 20°N   Dlong- 60°E

CosAB = Cos 60° Sin 80° Sin 60° + Cos 80° Cos 60°

    AB   = 59° 7.1’ = 3547.1 miles

Cos A  = Cos 60° - Cos 80° Cos 59° 7.1’

                      Sin 80° Sin 59° 7.1’

A         = 60.9° T

Cos B   = Cos 80° - Cos 60° Cos 59°7.1’

         Sin 60° Sin 59° 7.1’

B           = 96.4°

Course = 180 – 96.4 = 83.6° T

6Th SEPTEMBER 2017

A= 30° N                085° E

B= 42° N                110° E

Dlat- 12° N  Dlong- 25° E

Cos AB = Cos 25° Sin 48° Sin 60° + Cos 48° Cos 60°

AB        = 23°23.2’ = 1403.18 miles

Cos A = Cos 48° -Cos 60° Cos 23°23.2’

                         Sin 60° Sin 23° 23.2’

A         = 52.3 ° T

Cos B  = Cos 60° - Cos 48° Cos 23°23.2’

                   Sin 48° Sin 23°23.2°

B         = 112.7°

Course final = 180-112.7 = 67.3 °T

7th NOVEMBER 2017

A= 56° S                  067°W

B= 37° S                  019°E

Dlat- 19°N   dlong- 86°E

Cos AB= Cos 86° Sin 34° Sin 53° + Cos 34°Cos 53°

AB        = 57°59.4 ‘ = 3479.4 miles

Cos A =Cos 53° - Cos 34° Cos 57°59.4’

                    Sin 34° Sin 57° 59.4’

A          = 69.9° 

Course initial = 110.1°T

 Cos B = Cos 34° -Cos 53° Cos 57° 59.4’

                      Sin 53° Sin 57° 59.4’

B            = 41.1 ° T = final course

JULY 2017 (PM)

A= 10°S      150° W

B= 40° N     160° E

Dlat- 50°N Dlong- 50°E

Cos AB = Cos 50°Sin 100°Sin50° + Cos 100°Cos 50°

AB         = 68° 4.8’ = 4084.8’

Cos A = Cos 50° -Cos 100° Cos 68° 4.8’.         It's 50 degree not 60 degree

                      Sin 100° Sin 68° 4.8’

A           = 39.2° T


Cos B = Cos 100° - Cos 50° Cos 68° 4.8’

                        Sin 50° Sin 68° 4.8’

B         = 125.6 °

Final course      = 54.4° T

JULY 2017(AM)


A= 49°50’ N             005° 12’W

B= 13°06’N              059°20’W

Dlat-36°44’S     dlong- 54°08’W

Cos AB= Cos 54°08’ Sin 40°10’ Sin 76°54’ + Cos 40°10’Cos 76°54’

AB    = 57° 13.7’ = 3433.8 miles

Cos A = Cos 76°54’ – Cos 40°10’ Cos 57°13.7’

                    Sin 40°10’ Sin 57°13.7’

A        = 110.2° T

Cos B = Cos40°10’ – Cos 76°54’Cos 57°13.7’

B         = 38.4

           = 180° + 38.4° = 218.4 ° T


MARCH 2017 (AM)

A= 19° N          067° W

B= 36° N          6°30’W

Dlat- 17°N Dlong-60° 30’ E

Cos AB = Cos 60°30’ Sin 71° Sin 54° + Cos 71° Cos 54°

AB         = 55°23.2’ = 3323.2 miles.

Cos A    = Cos 54°- Cos 71° Cos 55° 23.2’

                       Sin 71° Sin 55° 23.2’

A            = 58.8° T~58°49.4’

Cos B     = Cos 71° - Cos 54° Cos 55°23.2’

                    Sin 54° Sin 55°23.2’

B             = 90.7 °

Course final = 89.3° T (180-90.7)

Vertex outside:

Sin(90-AP)= Tan(90-A) Tan(90-P)

Tan(90-P)= Sin(90-AP)/Tan(90-A)

                 = Sin(90-71)/tan(90-58.8)

P               = 61°42.9’

Longitude of vertex=005°17.1’W

Sin PC = Cos(90-A)cos(90-AP)

PC        = 53°59.7’

Latitude of vertex = 90° - 53° 58.5’ = 36°0.3’ N

MARCH 2017 (PM).         

A= 20°30’ S       120° 15’E

B= 36°20’N       179°40’ E

Dlat- 56°50’N  Dlong- 59° 25’E

Cos AB = Cos 59°25’Sin 110°30’ Sin 53°40’ + Cos 110°30’ Cos 53°40’

AB        = 79°50.3’ = 4790.3 miles

Cos A   = Cos 53°40’- Cos 110°30’ Cos 79°50.3’

                          Sin 110°30’ Sin 79° 50.3’

A           = 44.8° T

Cos B =Cos 110°30’ – Cos 53°40 Cos 79° 50.3’

                    Sin 53°40’ Sin 79° 50.3’

B         = 124.9° ~ 124°59.6’

Course = 55.1°T (180-124.9)

Vertex lies outside:




In small traiangle b= 180-124°59.6’=55°0.4

Tan(90-P) = Sin(90-PB) / Tan(90-B) C:\Users\AMAL\AppData\Local\Microsoft\Windows\INetCache\Content.Word\MARCH PM(1).png

P                = 49°45.4’

Longitude of vertex= 130°34.6’W

Sin PO = Cos(90-PB) Cos(90-B)

PO       = 41°17.8’

Latitude of vertex = 48°42.2’N

MAY 2017 (AM)

A= 30° N          060°E

B= 50° S           155°E

Dat- 80°S Dlong-95°E

Cos AB =Cos 95° Sin 120° Sin 40° + Cos 120° Cos 40°

AB        = 115° 33.9’ = 6933.9 miles

Cos A = Cos 40° - Cos 120° Cos 115° 33.9°

                   Sin 120° Sin 115° 33.9’

A        = 45.2°


Course = 134.8° T

Cos B =Cos 120° - Cos 40° Cos 115 ° 33.9’

                  Sin 40° Sin 115° 33.9’

B         = 106.9 ° T

Vertex outside:

Sin (90-AP) =Tan (90-p) tan(90-A)

Tan(90-P)   = Sin(90-AP)/ Tan( 90-A)

P =116° 43.5’

Longitude of vertex = 176° 43.5’ E

Sin PC = Cos (90-AP) Cos (90-A)

AC       = 37° 54.9 ‘

Latitude of vertex = 52° 5.1’S


MAY 2017 (PM)

A=  08° 15’N        078° 00’E

B=  33°50’S          025°30’E

 Dlat- 42°05’ S Dlong- 52°30’W

Cos AB =Cos 52°30’ Sin 56°10’ Sin 98°15’ + Cos 56°10’ Cos 98°15’

AB        = 65° 7.9’ = 3907.9 miles

Cos A   = Cos 56° 10’ – Cos 98°15’ Cos 65° 7.9’

                    Sin 98° 15’ Sin 65° 7.9’

A           = 46.6 ° 

Course initial = 226.6° T

Cos B= Cos 98°15’ –Cos 56°10’ Cos 65°7.9’

                     Sin 56° 10’ Sin 65° 7.9’

B        = 120.1° 

Course final = 239.9° T (360-B)

NOVEMBER 2016

A= 56° 45’N           065° 32’E 

B= 33° 36’ N          132°20’E

Dlat- 23°09’ S Dlong- 66° 48’E

Cos AB =Cos 66°48’ Sin 33°15’ Sin 56° 24’ = Cos 33° 15’ Cos 56°24’

AB         = 50° 0.4’ = 3000.4 miles

 Cos A    = Cos 56°24’ – Cos 33°15’Cos 50°0.4’

                       Sin 33°15’ Sin 50°0.4’

A             = 87.8°T (initial course)~ 87°49.8’


Cos B= Cos 33°15’ – Cos 56°24’ Cos 50° 0.4’

                   Sin 56°24’ Sin 50° 0.4’

B     = 41.1°

Course final = 138.9 ° T






Vertex inside:

Sin(90-PA) =Tan(90-P) Tan(90-A)

Tan(90-P) =Sin(90-PA)/ Tan(90-A)

P                 = 2°35.7’

Longitude of vertex = 068° 7.7’ E

Sin PO = Cos (90-PA) Cos( 90-A)

PO        = 33°13.3’

Latitude of vertex = 56° 46.7’ N

SEPTEMBER 2016

A= 06° N      079° W.                

B= 38° S       175° E

DLAT-44°S DLONG-106°E


Cos AB = Cos 106° Sin 96° Sin 52° + Cos 96° Cos 52°.          

AB        = 106°16.9’ = 6376.9 miles

Cos A = Cos 52°- Cos 96° Cos 106°16.9’

                Sin 96° Sin 106°16.9’

A          = 52.1°

Initial course = 127.9 ° T

Cos B = Cos 96°- Cos 52° Cos 106° 16.9’

                      Sin 52° Sin 106° 16.9’

B        = 84.8° 

Final course = 84.8° T

JULY 2016

A= 20° N          075°W

B= 45°N           050°W

Dlat- 25°N Dlong- 25°W

Cos AB= Cos 25°Sin 70°Sin 45°+ Cos 70° Cos 45° 

AB        = 32°25.7’ = 1945.7’

Cos A   = Cos 45°- Cos 70° Cos 32°25.7’

                         Sin 70° Sin 32°25.7’

A           = 33.9°T

Cos B    = Cos 70°- Cos 45°Cos 32°25.7’

                        Sin 45° Sin 32°25.7’

B            = 132.2°

Final course= 47.8° T                      

 MAY 2016    

A= 24° N       074° E

B= 48° N        145° 18’E

Dlat - 20° N Dlong- 70°48’E

Cos AB= Cos 78° 48’ Sin 66° Sin 42° + Cos 66° Cos 42°

AB       = 59° 46.9’ =3586.9 miles

Cos A = Cos 42° - Cos 66° Cos 59° 46.9’

                    Sin 66° Sin 59° 46.9’

A         = 46.9 ° T(initial course)

 Cos B= Cos 66° -Cos 42° Cos 59° 46.9°

                  Sin 43° Sin 59° 46.9’

B         = 86.8°

Final course = 93.2° T


MARCH 2016 

A= 56° 45’N           065° 32’E 

B= 33° 36’ N          132°20’E

Dlat- 23°09’ S Dlong- 66° 48’E


Cos AB =Cos 66°48’ Sin 33°15’ Sin 56° 24’ = Cos 33° 15’ Cos 56°24’

AB         = 50° 0.4’ = 3000.4 miles

 Cos A    = Cos 56°24’ – Cos 33°15’Cos 50°0.4’

                       Sin 33°15’ Sin 50°0.4’

A             = 87.8°T (initial course)~ 87°49.8°


Cos B= Cos 33°15’ – Cos 56°24’ Cos 50° 0.4’

                   Sin 56°24’ Sin 50° 0.4’

B     = 41.1°

Course final = 138.9 ° T

Vertex inside:


Sin(90-PA) =Tan(90-P) Tan(90-A)

Tan(90-P) =Sin(90-PA)/ Tan(90-A)

P                 = 2°35.7°

Longitude of vertex = 068°7.7’E

Sin PO = Cos (90-PA) Cos( 90-A)

PO        = 33°13.3’

Latitude of vertex = 56° 46.7’ N

JAN 2016

A= 24° N          074°15’W

B= 46°N            053°45’W

Dlat=22°N  Dlong- 20°30’ E

Cos AB= Cos 20°30°Sin 66° Sin 44° + Cos 66° Cos 44°

AB        = 27° 30.1’ = 1650.1 miles


Cos A = Cos 44°- Cos 66° Cos 27°30.1’

                  Sin 66° Sin 27°30.1’

A         = 31.8° T

Cos B = Cos 66° -Cos 44° Cos 27° 30.1’

                   Sin 44° Sin 27° 30.1’

B          = 136.1 °

Course final =43.9° T

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