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 it – S L Dec 26 '12 at 20:37
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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. – André Nicolas Dec 26 '12 at 20:37
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Yes indeed, the question was well expressed. – copper.hat Dec 26 '12 at 20:47
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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 $|AC| = a$, $|AB| = \frac{1}{\cos \alpha} a$.
copper.hat
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That's what copper hat assumes too, and negates ($A$ in 1st paragraph is a typo, he means $a$. – gnometorule Dec 26 '12 at 20:39
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