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 andfinal position makes with the east ?

A bus travels 6 km towards north at an angle of 45^(@)to the east and

1.	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 andfinal position makes with the east ?
Answer» Solution :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` Netmovement 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).