that is not the right answer.

first you draw the side that is given (say BC)
then you draw one given angle on one endpt.say angle PBC,
such that P is random point on the ray.
draw the remaining angle on that point, say angle BPQ,
such that Q lies on line BC.
Draw line CA parallel to PQ such that A lies on ray BP.
Then pts. A, B, C form the required triangle.

What do you mean by that? Can you point out what is wrong with my answer?

What you did works as well. This does not mean that my answer is wrong, even though it is different from yours.

I would like it very much if you thought of my construction and either point a mistake or admit that the construction is correct.

i meant the construction to be done using
only compass and straight edge.
you're not supposed to find the third angle first.
using only the given info, using no algebra,
is the way its supposed to be done.

yours is correct too, but it is not the ruler-and-compass style.

again, i'm really sorry for offending you. :)

Of course it is a ruler-and-compass construction. Much in the spirit of what you've done. Pick a line and two arbitrary points on that line. At one point draw one angle, at the other the other angle. At the intersection, you get the third angle. You can proceed from there.

This is a mathematical tactics known as the reduction of the problem to a previously solved one. Of course you can combine the two steps.

Instead of going to the ASA construction, place on one of the constructed angles the given segment and draw a parallel, as you've done.

However, it is now quite clear that, in general, there are two solution. You can place your segment between angles 1 and 3 or between the angles 2 and 3. (1 and 2 are given and 3 is the one you just found.) Unless the given angles (1 and 2) are equal, there are two solutions. The SAA do not determine a unique triangle.