Question

Can anyone please solve this one?​

Answer 1

splitting the components of 30√2

Answer 2

from c it turns south west

So it would make 45 angle with BC

In ABC

<C= 45

So AB = BC = 20

AC= √20^2 + 20^2 = √400+400 = √800

= 20√2

AS = 30√2 -20√2 = 10√2

AD= 30-20= 10

<BAC = <SAD = 45

In ASD

Cos <SAD = SA^2 + AD^2 - SD^2)/2×SA× AD

cos 45 = 200 + 100 - SD^2)/200√2

1/√2 = 300 - SD^2 )/200√2

200 = 300 - SD^2

SD^2 = 300-200= 100

SD= 10

SD= 10

So displacement= 10

now direction

SD= 10

AS= 10√2

AD= 10

It clearly forms right angled triangle( 10^2 + 10^2 = 10√2)^2))

So AD= DS= 10

So AD perpendicular to DS

As AD towards north

So DS would be towards west

So its

1) 10 m towards west