An aircraft travelled from calabar to kano as follows; it flew first to ilorin covering a distance of 300km, 30degree West or North , and then flew 400km, 60degree east of north to kano. what is the resultant displacement?

Answer 1

30° west of north = 120°
60° east of north = 30°
using vectors:
R = (300cos120, 300sin120) + (400cos30, 400 sin30)
= (-150, 150√3) + (200√3 , 200) = (200√3-150,150√3+200)
= (196.41, 459.81)
maginitude = √(196.41^2 + 459.81^2) = 500
angle = tan^-1(459.81/196.41) = appr 66.87°

or
make a sketch and use the cosine law:
R^2 = 300^2 + 400^2 - 2(300)(400)cos90°
= 500
leaving it up to you to find the angle using the sine law in the triangle, it should be the same as my first solution.

Answer 2

Pls show a diagrammatical representation

Answer 3

It's very easy to understand.

Answer 4

Must the two degrees be added together??

Answer 5

An aircraft travelled from calabar to kano as follows; it flew first to ilorin covering a distance of 300km, 300 West or North , and then flew 400km, 600 east of north to kano. what is the resultant displacement?