A man starts walking from home and walks 3 miles at north of west, then 5 miles at west of south, then 4 miles at north of east. If he walked straight home, how far would he have to the walk, and in what direction

Step-by-step explanation:

We shall find the solution of this problem with the help of vector notation of i , j , which show east and  north direction .

The first displacement can be represented by the following

D₁ = - 3 cos 45 i + 3 sin45 j = - 3 / √2 i + 3 / √2 j

The second  displacement can be represented by the following

D₂ = - 5 cos 45 i - 5 sin45 j = - 5 /√2 i - 5 /√2 j

The third  displacement can be represented by the following

D₃ =  4 cos 45 i + 4 sin45 j =  4 /√2 i + 4 /√2 j

Total displacement D =

D₁ +D₂ + D₃

= i ( -3 -5 + 4 ) / √2 + j ( 3 - 5 + 4 ) / √2 j

= - 4  / √2 i + 2 / √2 j

D = - 2.8288 i + 1.414 j

Magnitude of D

= √ ( 2.8288² + 1.414² )

= 3.16 miles

For direction we calculate angle with X axis

Tanθ = 1.414 / 2.8288

θ = 26 °

As x is negative and Y is positive ,

the direction will be north of west .

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)