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Vectors - Physics

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This is someone else's solution, I don't understand why we took the angle as 120° in both the questions.
For Question 17, I don't understand the equation used. To find the angle we use θ=tan ^-1(vertical components / horizontal components).
If B is the first distance covered (25km) and A is the distance travelled after that (30km), Why did we split A into 2 components and take B in the denominator and add it to the horizontal component of A?
I have a feeling that the reason I'm confused is because my vector diagram drawn on the bottom right is wrong😬.
(edited 2 years ago)
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mqb2766
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Theyve just done a slightly different diagram (to the right of the 17 answers). So 120 is the angle between the two vectors.
Then the tan() calculation to get the angle is y/x, so the ratio of the components.
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Original post by mqb2766
Theyve just done a slightly different diagram (to the right of the 17 answers). So 120 is the angle between the two vectors.
Then the tan() calculation to get the angle is y/x, so the ratio of the components.

If I choose to answer the question with the vector diagram I drew on the bottom right, I will use 60° as the angle between the two vectors? Distance equation in Q.16 is √(900+625+2(30)(25) cos(60)) ?

and for question 17 I should use θ=tan ^-1(30 / 25cos(30))?
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mqb2766
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Original post by : ).
If I choose to answer the question with the vector diagram I drew on the bottom right, I will use 60° as the angle between the two vectors? Distance equation in Q.16 is √(900+625+2(30)(25) cos(60)) ?

and for question 17 I should use θ=tan ^-1(30 / 25cos(30))?

In a sense you're should be doing exactly the same for 16 as cos(60) = -cos(120). Is there a - sign different? Or you could do it component wise from the first oasis by calcuating the x-y components then using pythagoras. It will be the same.

For 17, there are a few ways to do it. You could use the cos rule again to find the angle as you now know all three sides. Or you can do it component wise as per the solution here. Does your calc give the right solution? If not, how did you come up with those numbers (a sketch?).
(edited 2 years ago)
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Original post by mqb2766
In a sense you're should be doing exactly the same for 16 as cos(60) = -cos(120). Is there a - sign different? Or you could do it component wise from the first oasis by calcuating the x-y components then using pythagoras. It will be the same.

For 17, there are a few ways to do it. You could use the cos rule again to find the angle as you now know all three sides. Or you can do it component wise as per the solution here. Does your calc give the right solution? If not, how did you come up with those numbers (a sketch?).

Thank you so much!
For question 17, this is what I'm trying to do. I'm getting a different answer. I will use the cos rule and the Pythagoras theorem if I still don't understand. But I can't figure out why I'm getting a different answer for the angle.
Mistake* it's tan inverse not θ inverse sorry!
er.png
(edited 2 years ago)
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mqb2766
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Original post by : ).
Thank you so much!
For question 17, this is what I'm trying to do. I'm getting a different answer. I will use the cos rule and the Pythagoras theorem if I still don't understand. But I can't figure out why I'm getting a different answer for the angle.
er.png

The y component is not 30 on the right. You'll have to subtract a component.
(edited 2 years ago)
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Original post by mqb2766
The y component is not 30 on the right. You'll have to subtract a component.

I think I tried that. Is this what you mean? haw.png
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mqb2766
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Original post by : ).
I think I tried that. Is this what you mean? haw.png

And what is that relative to North?
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Original post by mqb2766
And what is that relative to North?

💡 Thank you so much. I understand!
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mqb2766
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Original post by : ).
💡 Thank you so much. I understand!

Note they calculated the complementary angle directly by doing tan() = x/y, rather than y/x.which you did (then got the complementary angle)
Either is fine as long as you know what they represent.
(edited 2 years ago)
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No wonder, That makes sense.

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