Was the Angle Discovered? Thread Closed Due to Existing Problem

[AzureSekki]

Homework Statement

How did they get all the angle here in this problem
Except for the angle 65° which is given it's not stated on how they got it so im having a hard time to understand the problem

Relevant Equations

Find angle

 

[BvU]

Homework Statement:: How did they get all the angle here in this problem

That's a statement of your problem, not of the homework problem.
Provide a complete problem statement

given:​

knowns:​

unknowns:​

I am having a hard time to understand the problem

This way, so am I !

$C$ doesn't exist ?

 

[Lnewqban]

Didn't you solve this same problem this morning?
Finding all those angles may not be the best way to learn about sum of vectors and to solve the problem.

https://www.physicsforums.com/threads/vector-resolution-onto-axes-not-at-right-angles.984637/

 

[Master1022]

Hi,

So I am slightly confused by some of the lettering, but perhaps I am missing something. If you are told that \angle BON = 65, then do you understand the working to find \angle AOC = 30 degrees (I am assuming point C is where point N is)?

If we can accept that, then I think the easiest way to get \angle BOA is just to consider \angle BON = \angle BOA + \angle AON. Solving this yields the same answer of 35 degrees.

I believe they have labeled the 115 degrees by using the fact that line segments BA and ON are parallel and \angle OBA and \angle BON are therefore supplementary ('internal' angles of parallel lines add up to 180 degrees), so they could get 115 from \angle OBA = 180 - 115

Hope that answers your question. If not, let me know and I will respond appropriately.

Kind regards

 

[BvU]

If not, let me know and I will respond appropriately.

Post the complete problem statement !

 

[AzureSekki]

Hi,

So I am slightly confused by some of the lettering, but perhaps I am missing something. If you are told that \angle BON = 65, then do you understand the working to find \angle AOC = 30 degrees (I am assuming point C is where point N is)?

If we can accept that, then I think the easiest way to get \angle BOA is just to consider \angle BON = \angle BOA + \angle AON. Solving this yields the same answer of 35 degrees.

I believe they have labeled the 115 degrees by using the fact that line segments BA and ON are parallel and \angle OBA and \angle BON are therefore supplementary ('internal' angles of parallel lines add up to 180 degrees), so they could get 115 from \angle OBA = 180 - 115

Hope that answers your question. If not, let me know and I will respond appropriately.

Kind regards

Thanks for the clarification

 

[Mark44]

Thread closed, as this is a new thread about an existing problem.