- #1

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## Homework Statement

In photo:

https://gyazo.com/6348b5ffe368852ee9868a2ba0dd7e38

## Homework Equations

In image

## The Attempt at a Solution

In the boxes

Where did I go wrong with the angle calculation?

Thanks

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- Thread starter Josh Eiland
- Start date

- #1

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In photo:

https://gyazo.com/6348b5ffe368852ee9868a2ba0dd7e38

In image

In the boxes

Where did I go wrong with the angle calculation?

Thanks

- #2

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I don't know, you didn't show the Calc.

- #3

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https://gyazo.com/efc009de2080feedea420d6b2a5edab4I don't know, you didn't show the Calc.

Triangle on the right shows how I calculated -10° angle by using tan-1(-3.2/17.4), -3.2 being the change in y and 17.4 change in x after I added the vectors.

- #4

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Can't follow your work. Your original post gives the answer as 350. How did you come up with this number?

Last edited:

- #5

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I calculated the angle as -10°, but since the question asked for the angle counter-clockwise from the x axis, I thought it would be correctly expressed as 350.Can't follow your work. Your original post gives the answer as 350. How v did you come up with this number?

- #6

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-10 degrees measured from what?

- #7

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From the positive x axis.-10 degrees measured from what?

- #8

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You can find the coordinate of the player's position through your vectors. After that you can simply use geometry to determine the ccw angle, after that, 360 - positive angle = clockwise angle. The program may not want to accept negative angles, try the positive equivalent between 0 and 360.

- #9

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You can find the coordinate of the player's position through your vectors. After that you can simply use geometry to determine the ccw angle, after that, 360 - positive angle = clockwise angle. The program may not want to accept negative angles, try the positive equivalent between 0 and 360.

https://gyazo.com/efc009de2080feedea420d6b2a5edab4

Triangle on the right shows how I calculated -10° angle by using tan-1(-3.2/17.4), -3.2 being the change in y and 17.4 change in x after I added the vectors.

- #10

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https://gyazo.com/efc009de2080feedea420d6b2a5edab4

Triangle on the right shows how I calculated -10° angle by using tan-1(-3.2/17.4), -3.2 being the change in y and 17.4 change in x after I added the vectors.

And yes, that's how I got 350° counter-clockwise. By doing 360 minus the clockwise angle (360-10).

- #11

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From the positive x axis.

I think if you draw yourself a picture you will see that it can't be -10 degrees from the +ve x-axis.

- #12

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I got -10 from the X-axis counter-clockwise when drawing a picture, as I've explained. Do you have a better answer or suggestion?I think if you draw yourself a picture you will see that it can't be -10 degrees from the +ve x-axis.

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- #14

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From this image, find the vector P.

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That wouldn't make sense, as the angle from initial to final points would be from the top of the red line on the y-axis to the final spot, not from the origin.

From this image, find the vector P.

- #16

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I calculated the final point (17.4, -3.2) and drew a line from the origin (0,0) to this point, creating a triangle with the x and y segments. Using Pythagorean theorem I calculated the magnitude of this vector 17.7

I then used inverse tangent to find the angle -10°

- #17

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- #18

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Tried it, didn't work. I think the teacher probably just put the answer in wrong accidentally. Will ask tomorrow. Thanks for help.

- #19

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I am getting a positive angle of 24.213 degrees counter-clockwise.

Again, if you're not showing your calculations it's impossible for us to tell you where you're going wrong.

Your y-coordinate is wrong. The point is in the first quadrant.

Bx=11.3137

By=11.3137

Cx=6.0622

Cy=3.5

Therefore the resultant vector is

r=[Bx+Cx,By-Cy]

- #20

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I got the same answer as the OP.

- #21

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