Triangles and Law of Sines

Date: 03/27/98 at 17:32:36
From: Heathe
Subject: Trigonometry

I am having problems with the Law of Sines. To be specific, I don't 
know how to tell if the information I'm given produces more than one 
triangle. Solving for one is simple. That I can do. But some of the 
problems in my book produce two triangles, and all the info they give 
to solve the problem is like 2 sides and an angle or 2 angles and a 
side.

For example:

     Angle A = 37 degrees
     Side  C = 40
     Side  A = 28

The back of my book gives two solutions for this this problem. How do 
you solve it?

Date: 03/28/98 at 23:02:02
From: Doctor Scott
Subject: Re: Trigonometry

Hi Heathe!

Using the law of sines you get:

       28        40
    -------- = -------
     sin 37     sin C

or 28 sin C = 40 sin 37

             40 sin 37
     sin C = ---------
                28

     sin C = 0.859735747

To solve this equation, we use the inverse sine function.  
InvSin(0.859..) = 59.28 degrees. But recall that the sin of an angle 
and the sin of its supplement are equal. That is, sin x = sin (180-x).  
So, if the sin of 59.28 is 0.8597, then so is sin (180-59.28) or
sin (120.7). 

So another possible measure for C is 120.7. We need to see if it can 
work. Since angle A is 37, keeping in mind that the angles of a 
triangle add up to 180, we can have C = 59.28 and A = 37 (and still 
have some angle measure left over for B) OR C = 120.7 and A = 37 and 
still have some angle measure left over for B. 

That's why there are 2 solutions in the back of your book. Now, we 
have to find all of the other parts of the triangle for BOTH possible 
angle Bs.

-Doctor Scott,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/