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# Im trying to make something... Help?

I'm trying to make a code that will solve for a Mathematical reference angle in python. I dont know how to make the code subtract from user input to solve for the reference angle. Help?

17 ответов

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Well I want the user to input an angle say 150 (which is 30 away from the x axis) and make it subtract and set as new value. Should I set user input as a variable like:
input ("Enter an angle.")
input == x
Then just subtract like:
x -= 180
x ** 2
x // 2 (I think that's root.)
>>>
30
>>>
The reason I put the square and then root is to get rid of any negatives just in case it is. I could of also done:
If x < 0:
x *= -1
else:
x += 0
Thoughts?

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I guess I'm talking for most people here if i say: What angle are you talking about?

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A reference angle is just an angle between the terminal side and the x axis. But the thing I'm asking about is how to subtract from the user input say if it's greater than 180, to subtract and set new value. If I need to teach a lesson on reference angles I can but I don't need someone to make the whole code for me, just what I specified.

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You probably talking about angle functions like sin cos and tan
And why the subtraction, a simple if statement should do the trick to set the imput to 180.

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I still don't see a problem(and there is no need for else)

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True, but it says invalid syntax when I do x -= 180
^

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Same problem with x = x - 180?

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I'll try that.

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How should I set input to equal a variable like x? It says it's not defined.

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Define it with x = 0

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Now it's saying there is a problem with:
if x > 180
^

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With or without the :

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With

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OK send the whole statement

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#terminal angle aka Reference angle...
#solver
input ("Enter an angle!")
x = 0
x = input
if x >= 180:
x = x - 180
else:
print("Working...")
print ("This is your Reference Angle: " + x)

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Kinda the same trick if x > 179:

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I'll work on this later, if you want to help, see what you can do. Goodnight!