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Inverse Sine of a Value Greater than One

Date: 12/22/2003 at 07:28:25
From: Kaylie
Subject: Sine Rule Problem

Why is it not possible to calculate an angle when I have the formula 
sin(B) = 1.732?  Using my scientific calculator I do (shift)sin 1.732.  
This should give me an angle size, but my calculator says 'error'.  Is 
this something to do with the 'sine wave' or is there a different 
reason?  I'm trying to find angle B in this problem:

  80       40                 80 x sin 60   
------ = -------  so sin(B) = ------------ = 1.732     
sin(B)   sin(60)                   40 

But shift-sin(1.732) = not possible.  Why?



Date: 12/22/2003 at 08:23:29
From: Doctor Peterson
Subject: Re: Sine Rule Problem

Hi, Kaylie.

Thanks for writing.  You have partially answered your own question.   
Think about what the sine wave or curve looks like.  As the angles 
change, the sine of the angle goes up and down, but it never goes
above 1 or below -1.  In other words, there is no angle with a sine
that is greater than 1.
 
When you use the sine key, you put in an angle and get out the sine of 
that angle.  For instance, when you do sin(30) you are finding the
sine of 30 degrees and the calculator says it's 0.5.  When you use the 
inverse sine (shift-sine) you put in the value of the sine and the 
calculator tells you the angle.  So, the inverse sine of 0.5 is 30 
since a 30 degree angle has a sine of 0.5.

So when you ask the calculator to do the inverse sine of 1.732, you
are asking it what angle has a sine of 1.732.  But as we said above,
there is no angle that has a sine greater than 1.  The inverse sine or
arcsin of 1.732 does not exist.  That's what the calculator is saying.  

It sounds like your problem is asking you to try to find angle B in a 
triangle with a = 40, b = 80, and A = 60 degrees.  Try constructing 
such a triangle, and see what happens.  We make angle A and mark 80 
units on one of its rays for side b, then swing an arc around the 
resulting point C with a radius or length of 40, so that its 
intersection with the other ray will give point B.  What happens?

                    B
                  /
                 /
                /
               /
              /
         c   /                    ooooooooo
            /                 oooo         oooo
           /               ooo \               ooo
          /              oo     \
         /              o        \ a=40
        /              o          \
       /              o            \
      /               o             \
     /60             o               \
    A----------------o----------------C-----------
                     b=80

So the calculator is correct: there is no such triangle!  Obviously we 
can't measure angle B at the top if side 'a' is not long enough to 
complete the triangle and form that angle.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Triangles and Other Polygons
High School Trigonometry

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