In the diagram below, angle ACB is 45°, angle ADB = 60° and AD = 2*DC. Find the angle at A.

Angle BDC = 120°, so angle DBC = 15°. We'll start by establishing a fact about 15-75-90 triangles.

Let ABCD be a rectangle with AB = 1 and BC = 2.

Choose E on BC so that AD = AE, making triangle ADE isoceles. Then AE = 2 and AB = 1, so triangle ABE is half an equilateral triangle, and hence it's a 30-60-90 triangle. Furthermore, AE = 2, so BE = √3 and EC = 2 - √3.

Since angle BAE = 60°, angle DAE = 30°, angle ADE = 75° and hence angle EDC = 15°. As a result, in a 15-75-90 triangle, the ratio of the side opposite the 15° angle to the perpendicular side ajacent to the 15° angle is 2 - √3.

Returning to the original diagram, drop a perpendicular from B to AC, meeting AC at E.

Now choose units so that BC = √6, just to make things easy. Triangle BEC is a 45-45-90 triangle, so BE = EC = √6/√2 = √3. And since angle ADB = 60° triangle BED is a 30-60-90 triangle, so ED = 1. Hence DC = √3 - 1.

AD = 2DC, so AE = 3DC - EC = 3(√3 - 1) - √3 = 2√3 - 3. Hence AE/BE = (2√3 - 3)/√3 = 2 - √3. So we finally obtain that triangle ABE is a 15-75-90 triangle, and so the angle at A is 75°.