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I was wondering is it possible to solve this without assuming that CAD=DAB. As I use the law of sines, trigonometry and have tried to apply law of cosines. However, I cannot see how you can solve this with just using a and $\alpha$.

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    sorry, i didn't see figure properly. I guess the answer is no, as we can change $AD$ freely within $\alpha $ and yet construct a right angled triangle to fit it2012-12-26
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    Extend $CA$ a little, to $CA'$. Extend $CB$ a little, to $CB'$, so that $A'B'$ is parallel to $AB$. Then we can find $D'$ in $AD$ such that $A'D'=AD=a$. The shaded area has increased. Your picture was very helpful.2012-12-26
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    Yes indeed, the question was well expressed.2012-12-26

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Are you asking if $a$ and $\alpha$ alone uniquely define the triangle? If so, the answer is no.

To see this, draw the point $A$ and two lines from $A$ that make an angle $\alpha$. Then draw a line of length $a$ 'inside' this angle, and then draw $CB$ appropriately. This gives a range of triangles from $|AB| = a$, $|AC| = a \cos \alpha$ to $|AC| = a$, $|AB| = \frac{1}{\cos \alpha} a$.

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    I'm asking can you express AC in terms of $a$ and $\alpha$ only.2012-12-26
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    That's what copper hat assumes too, and negates ($A$ in 1st paragraph is a typo, he means $a$.2012-12-26
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    @gnometorule: Thanks for catching that, my typing skills are poor.2012-12-26