Question

A man walks 1km due east and then 1 km due north His displacement
√2km N 45°E
1km N 30°E
1km N 15°E
√2km N 60°E

Answer 1

A man walks 1km due east and then 1 km due north.

See the attachment.

$displacement = \sqrt{ {1}^{2} + {1}^{2} } = \sqrt{2} \: km$

$angle = { \tan}^{ - 1} ( \frac{1}{1} ) = { \tan}^{ - 1} (1) = {45}^{o}$

So displacement is √2 km N 45° E