Jump to content

Recommended Posts

I am confused too

It appears from what you are saying that you want to know the Angle a

the calculation for the SIN of an angle is Sin(a) = Opposite/Adjacent (It appears to me that you know Opposite (BC) and you know Adjacent (AB)

So:

From what you have written you simply divide BC/AB and you will get the Sin(a) result

 

If I am correct in my assumptions (probably not) then you know the Opposite and Adjacent

Better off using this cot(a) = Adjacent / Opposite

which is Cot(a) = AB/BC

 

However thats all a complete guess as we dont know what you are doing here

Just one thing if you use MI's then you will not be correct in your calculations (they can only hold whole numbers)

 

(OMG school boy math)

 

 

 

 

Link to comment
Share on other sites

 i trying to find angle  BC

 

Are you using the law of sines?

https://en.wikipedia.org/wiki/Law_of_sines

 

From what I vaguely remember if you get a negative result, the angle is still valid, however if means you're moving in the opposite direction around the unit circle.  To convert to a positive angle, simply add 360 to it.

 

However, you're using a non-standard notation for me, so I'm not sure if you're using the correct numbers, angles and side length, in your equation to spit out good numbers.  Typically, angles are represented with capital letters, and the side opposite the angle is a lower case.  See wiki link above, or post an image of your triangle with everything labeled.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...