I don't understand why the answer is as it says in the book. Let me write out the task first.

• A If the cosine-rule can be used to find an angle in a triangle, it is only one angle that fits this description. (true, and I agree on that).

• (help)B If cosine-rule can be used to find one side in a triangle, it is only one side that fits this description. (false the book says.).
I don't agree, or understand. $a^2 = b^2 + c^2 - 2 \cdot b \cdot c \cdot \cos (A).$ Let's say we want to find the side a, then we need b, c and the angle A. Hence, we have all the angles, and two sides. How can this one side we discover using this rule not be unique?

• (help)C If the sine-rule is used to find a side in a triangle, there can be two sides that fit's this description. (false the book says).
Again, I don't understand. $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$ If we have two angles, the third is given. And we have one side as well. How can this side not be unique?

• D If sine-rule can be used to find an angle in a triangle, there is always two angles that fits the description. (False, I agree.) Sine could be 1, 90 degrees. And say if we were given two sides, and the opposite angle of the longest of the two there would only be one angle that fits.

I included the things I understand so you can get a grip on what I do actually understand.

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On B I meant that since b and c is fixed, and we have one angle, the other angles must be unique. –  Algific Aug 25 '11 at 7:57
Please do not use [homework] as a single tag. –  Asaf Karagila Aug 25 '11 at 7:59
Ok, fixed it now. –  Algific Aug 25 '11 at 8:00
The best way of seeing all these things is to get out a ruler and a pair of compasses and a protractor and try constructing triangles from the data given. If you try the case of two sides and an acute angle which is not the angle between the given sides you'll understand what B is getting at, and you should be able to unpick C as well (you should find two triangles which fit the data, except in an exceptional case and subject to conditions which ensure that a triangle exists). –  Mark Bennet Aug 25 '11 at 8:26
I assumed at B that the angle given must be the one between the two sides, otherwise the cosine-rule can't be used. –  Algific Aug 25 '11 at 8:31