north distance = N = 7 cos 30
east distance = E = 1o (? assume 10) + 7 sin 30
magnitude of displacement = sqrt (N^2 + E^2)
angle east of north = tan^-1 (E/N)
Use scale drawing method:
If a man travels 7km 30degree east of north,then 1okm east,find the resultant displacement
2 answers
Did you make your sketch?
I see a triangle where I can use the cosine law, and
R^2 = 7^2 + 1^1 - 2(7)(1)cos120°
= 50 - 14(-.5)
= 50+7 = 57
R = √57
by sine law:
sinØ/1 = sin120/√57
Ø = 6.586
direction = 6.586+30 = N 36.586°E
or
using vectors:
vector r = (7cos60,7sin60) + (cos0, sin0)
= (7/2 , 7√3/2) + (1,0)
= (9/2 , 7√3/2)
|r| = √(81/4 + 147/4) = √57
direction : tan^-1 β = (7√3/2) / (9/2)
β = 53.413° or N 36.586°E
I see a triangle where I can use the cosine law, and
R^2 = 7^2 + 1^1 - 2(7)(1)cos120°
= 50 - 14(-.5)
= 50+7 = 57
R = √57
by sine law:
sinØ/1 = sin120/√57
Ø = 6.586
direction = 6.586+30 = N 36.586°E
or
using vectors:
vector r = (7cos60,7sin60) + (cos0, sin0)
= (7/2 , 7√3/2) + (1,0)
= (9/2 , 7√3/2)
|r| = √(81/4 + 147/4) = √57
direction : tan^-1 β = (7√3/2) / (9/2)
β = 53.413° or N 36.586°E
3 answers