# Answer to Question #37105 in Mechanics | Relativity for Roshauna

Question #37105

A person walks 38 m East and then walks 33 m at an angle 46° North of East. What is the magnitude of the total displacement? Answer in units of m.

Expert's answer

This is direct application of the cosine theorem.

c^2=a^2+b^2 - 2*a*b*cos(Theta)

in our case:

a=38

b=33

c=?

Theta=180-46=134

cos(Theta)=cos(134)=-0.7

c^2=33^2+38^2+2*33*38*0.7

c^2= 1089+1444+1755.6

c^2=4288.6

c=sqrt(4288.6)=65.48=65.5m

ANSWER:

distance=65.5m

c^2=a^2+b^2 - 2*a*b*cos(Theta)

in our case:

a=38

b=33

c=?

Theta=180-46=134

cos(Theta)=cos(134)=-0.7

c^2=33^2+38^2+2*33*38*0.7

c^2= 1089+1444+1755.6

c^2=4288.6

c=sqrt(4288.6)=65.48=65.5m

ANSWER:

distance=65.5m

## Comments

## Leave a comment