Question

Two ships leave port at 4 p.m. One
is headed N38°E and is travelling at
11.5 km/h. The other is travelling at
13 km/h, with heading S47°E. How far
apart are the two ships at 6 p.m.?

i am in grade 11 :), studying ahead of the class :) its 8 pm here, this question is hard for me, i spent 10 minutes on it.

Answer 1

draw the diagram, figure each distance (velocity*time). time is 2 hours for each ship

Now you have a triangle, 2 known legs, and the included angle.

Find the opposite side: I recommend use the law of cosines.

Answer 2

The standard graph in polar coordinates has E at 0' ( o degree ) .
The makes N 38 E would be 90'- 38' = 52' and S 47 E at 2700'+ 47'= 314' , or -46 .
The difference in angle between the two is 53+46= 99'
At 6 PM , two hours have passed . This means the 1st ship went 2(11.5)=23 miles and the 2nd ship went 2(13)= 26 miles . Here , a= 23 , B=26 and A= 99'

This is the only way I can help , I am in 11th grade but I already did some excercise like this , not this one but some like it . I am a Haitian student , ( Haiti) so please maybe my English don't realy make you understand it because I do it in French at school but I translate it in English for you . This is the way they learn me to do it !!! Good luck Karen , be smart !!! Study hard !

Answer 3

Unless otherwise indicated, all angles
are measured CCW from +x-axis.

Ship #1:
d1 = 11.5km/h[52o] * 2h = 22km[52o].

Ship #2:
d2 = 13km/h[ 317o] * 2h = 26km[317o].

d2-d1 = 26[317o] - 22[52o].

X = 26*Cos317 - 22*Cos52) = 5.47 km.
Y = 26*sin317 - 22*sin52 = -35.1 km.

d2-d1 = Sqrt(X^2 + Y^2) = Distance apart.