Year 10 trig

Alan is right!

But I thought I would point out that all these triangles would be similar figures.  

And, since the sides would all be in the same ratio then the length of the sides could all be expressed using just one variable.

I am thinking as I am writing.

Let the sides of one of these similar figures  be 1, a and b and let the given angles be <A, <B and <C

Using the Sine rule.

$$\begin{array}{rll} \frac{a}{SinA}&=&\frac{1}{SinC}\\\\ a&=&\frac{SinA}{SinC}\\\\ \end{array}\\ Also\\\\ \begin{array}{rll} \frac{b}{SinB}&=&\frac{1}{SinC}\\\\ b&=&\frac{SinB}{SinC}\\\\ \end{array}\\\\ \mbox{So the sides are in the ratio } \quad \frac{SinA}{SinC}\;:\;\frac{SinB}{SinC}\;:\;1\\\\ \mbox{So the sides are equal to } \quad \frac{kSinA}{SinC}\;,\;\frac{kSinB}{SinC}\;,\;k\qquad \mbox{Where k}\in Z \\\\$$

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Here is the code for people trying to learn LaTex. 

\begin{array}{rll}

\frac{a}{SinA}&=&\frac{1}{SinC}\\\\

a&=&\frac{SinA}{SinC}\\\\ \

end{array}\\

Also\\\\

\begin{array}{rll}

\frac{b}{SinB}&=&\frac{1}{SinC}\\\\

b&=&\frac{SinB}{SinC}\\\\

\end{array}\\\\

\mbox{So the sides are in the ratio } \quad \frac{SinA}{SinC}\;:\;\frac{SinB}{SinC}\;:\;1\\\\

\mbox{So the sides are equal to } \quad \frac{kSinA}{SinC}\;,\;\frac{kSinB}{SinC}\;,\;k\qquad \mbox{Where k}\in Z \\\\

You can't, unless you have at least one side length!  There are an infinite number of similar triangles with the same three angles - they are just different sizes.