A bus travels 6 km towards north at an angle of `45^(@)` to the east and then travels 4 km towards north at an angle of `135^(@)` to the east . How for is its final position , due east and due north ? How far is the point from the starting point ? What angle does the straight line joining its initial and final position makes with the east ?

Explanation

Net movement along x-direction (due east)
`= (6 - 4) cos 45^(@)`
`= 2 xx (1)/(sqrt(2)) = sqrt(2) km`
Net movement along y-direction (due north)
`= ( 6 + 4) sin 45^(@)`
`= 10 xx (1)/(sqrt(2)) = 5 sqrt(2) km`
Net movement from starting point
= 6 + 4 = 10 km
Angle which makes with the east direction
`tan theta = ("y-component")/("x-component") = (5 sqrt(2))/(sqrt(2))`
`:. theta = tan^(-1) (5) ` (or)
`theta = 79^(@) ` (approx).