Question

4 .A boat in calm seas travels in a straight line and ends the trip 22 km west and 53 km north of its original position. To the nearest tenth of a degree, find the direction of the trip.

5. Find the magnitude and direction of the vector. Round the length to the nearest tenth and the degree to the nearest unit.
(There is a picture for number 5, I also need to show my work on both questions. I also need the direction for number 4. Thank you)

Answer 1

4) If you draw the triangle that is associated with the vectors your can use your trig ratios to solve : )
I see Tan theta being useful.
5) The magnitude is the value of the resultant : )

Answer 2

Did you want to see the picture?

Answer 3

I believe the picture shows the boat travelling directly west for 22km then turning and heading North for 53 km. These are the legs of the triangle, and the hypotenuse is the resultant vector.

Answer 4

I meant for number 5, there is no picture for number 4

Answer 5

Well... without a picture for 5) there is no way for me to help you.

Answer 6

#4. Measured from due west, the angle θ has
tanθ = 53/22
so, θ = 67.45°
So, the direction is W67.5°N or N22.5°W or a heading of 337.5°

#5. huh? #4 already gave the direction. The magnitude is just
√(53^2+22^2) = √3293 = 57.38 km

Answer 7

4. All angles are measured CW from +y-axis.
Tan A = x/y = -22/53 = -0.4151, A = -22.5o = 22.5o W of N. = 337.5o CW.

5. D = 22[270] + 53[0],
D = (22*sin270+53*sin0) + (22*cos270+53*cos0)I,
D = -22 + 53i = 57.4[-22.5o] = 57.4[22.5o] W. of N. = 57.4km[337.5o] CW.

Answer 8

I can't believe oobleck because of his #4 work where he has a theta in his work which is a greek letter and it has nothing to do with math and he also has a bunch of other stuff you can see that don't belong in math and henry2, has a lot of crud on his work for #5 making it non-believable.