Question 193994
the lengths of the longest and shortest sides of an acute 
scalene triangle are 9 m and 41 m, what could be the length 
of third side?
<pre><font size = 4 color = "indigo"><b>
Let a = 9m be the one side of the triangle
Let b = 41m be the second side of the triangle
Let c = the unknown third side.

Since c is the middle-size side, and b=41m is the
longest side, we know that c < 41m

Let the angle between sides a and c be angle B, which is also
the angle opposite side b.
By the law of cosines:
b<sup>2</sup> = a<sup>2</sup> + c<sup>2</sup> - 2ac*cos(B)

Solve for cos(B)

              a<sup>2</sup> + c<sup>2</sup> - b<sup>2</sup>
    cos(B) =  ------------
                  2ac
              9<sup>2</sup> + c<sup>2</sup> - 41<sup>2</sup> 
    cos(B) =  ------------
                 2(9)c

              81 + c<sup>2</sup> - 1681 
    cos(B) =  ------------
                 2(9)c

              c<sup>2</sup> - 1600 
    cos(B) =  ------------
                 2(9)c

              c<sup>2</sup> - 1600 
    cos(B) =  ------------
                  18c

In order for B to be less than 90°, cos(B) must be greater than 0.

So            cos(B) > 0

or

       c<sup>2</sup> - 1600 
       ------------  > 0
            18c

Multiplying through by 18c will not reverse
the inequality since c is positive.

       c<sup>2</sup> - 1600 > 0       

Factor the left side:

    (c - 40)(c + 40) > 0   

This is only true when c > 40

So the answer is that c must be between 40m and 41m,
exclusive of both 40m and 41m. That can be written:

   40 < c < 41

Edwin</pre>