The acute triangle $\displaystyle \triangle ABC$ inscribes a circle. D, E, and F are the tangent points. It is given that: $\displaystyle \angle EFD=\alpha$, $\displaystyle \angle FED=\beta$, and AF=a (see picture).

Using $\displaystyle \alpha ,\beta $ and a, express the following:

(a) the length of EF.

(b) the length of DF.

Well, first thing I thought about is that the angle between a tangent and chord is equal to the subtended angle on the opposite side of the chord, hence $\displaystyle \angle BFD=\beta , $ and $\displaystyle \angle AFE=180-\alpha -\beta $, and since the triangle $\displaystyle \triangle AFE$ is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), $\displaystyle \angle AEF=180-\alpha -\beta $ as well. In addition, $\displaystyle \angle D=180-\alpha -\beta $ and $\displaystyle \angle A=180-2*(180-\alpha- \beta )= -180+2\alpha +2\beta $ (is that correct?

)

Does what I did help in solving the question? And what am I supposed to do next? (I must use the sine/cosine formulas)

Thanks.