note 160 = south - 20
North distance = 8 cos 30 - 6 cos 20
East distance = 8 sin 20 + 6 sin 20
distance = sqrt (North^2 + east^2)
2.
call it 132 = south - 48
north distance=-61.1cos 48 +76.5sin 36.5
east distance =61.1sin48 +76.5cos 36.5
tan theta = north/east
1. A party of hikers walks 8 km from camp on 30 degrees, then turns and walks 6 km on course 160 degrees. Find the magnitude of the net displacement from camp.
2. A ship leaves port A and sails 61.1 km course 131 degrees 50' to point B. Then it sails 76.5 km on course 36 degrees 30' to point C. Find the bearing of point C from point A.
3 answers
1. I sketched a triangle, and noticed I could use the cosine law:
R^2 = 8^2 + 6^2 - 2(8)(6)cos 50°
= 38.2923...
R = appr 6.189 km
or
use vectors
first vector = (8cos30 , 8 sin 30) = (6.9282 , 4)
2nd vector = (6cos160, 6sin 160) = (-5.63815.. , 2.05212)
sum = ( 1.29004.... , 6.05212)
magnitude = √( 1.29004..^2 + 5.05212^2)
= appr 6.189
Do the 2nd in the same way.
R^2 = 8^2 + 6^2 - 2(8)(6)cos 50°
= 38.2923...
R = appr 6.189 km
or
use vectors
first vector = (8cos30 , 8 sin 30) = (6.9282 , 4)
2nd vector = (6cos160, 6sin 160) = (-5.63815.. , 2.05212)
sum = ( 1.29004.... , 6.05212)
magnitude = √( 1.29004..^2 + 5.05212^2)
= appr 6.189
Do the 2nd in the same way.
LOL, physicist / mathematician again.