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)
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.
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 : )
Did you want to see the picture?
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.
I meant for number 5, there is no picture for number 4
Well... without a picture for 5) there is no way for me to help you.
#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
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.
To solve question 4, we can use trigonometry to find the direction of the boat's trip. Here's how:
Step 1: Draw a diagram: Draw a coordinate plane and mark the boat's original position, the direction it traveled (towards the west), and the final position (22 km west and 53 km north).
Step 2: Find the displacement: The displacement is the straight-line distance between the original and final positions. In this case, the displacement is the hypotenuse of a right triangle formed by the 22 km and 53 km sides.
We can use the Pythagorean theorem to find the displacement:
Displacement = √((22 km)^2 + (53 km)^2)
Step 3: Find the angle: To find the angle of the trip, we need to use the inverse tangent function (tan^(-1)). We can use the formula:
Angle = tan^(-1)(opposite/adjacent)
Angle = tan^(-1)(53 km/22 km)
Step 4: Calculate the angle: Use a calculator or trigonometric table to find the inverse tangent. The approximate value to the nearest tenth of a degree can be determined:
Angle ≈ 68.3 degrees (rounded to the nearest tenth of a degree)
So, the direction of the boat's trip is approximately 68.3 degrees.
Now, let's move on to question 5, where we need to find the magnitude and direction of a vector represented in a picture.
To solve this, we need the picture you mentioned to perform necessary calculations. Please upload or provide a description of the picture, and I can guide you through finding the magnitude and direction of the vector.