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How to find this angle with $tan(alpha)$.
How should I start with the question?

define $beta$ is the $angle FCD$ in rogerl's answer.

$$
left{
begin{aligned}
2 R sinbeta &= 6 \
6 cos beta &= sqrt{14^2 - R^2}
end{aligned} right.
$$

define $x = cos^2beta$

we can get $36x^2 - 232x + 187 = 0 Rightarrow x = frac{17}{18}$

so $sinbeta = frac{1}{3sqrt{2}}$ and $R=9sqrt{2}$

thus $tanalpha = frac{R}{6cosbeta} = frac{9sqrt{2}}{6sqrt{frac{17}{18}}} = frac{9}{sqrt{17}}$

Let $R$ be the radius of the circle. Let $(R,0)$, $(x,y)$ and $(0,R)$ be the coordinates of the points belonging to the circle. Writing Pythagoras' theorem for the three right triangles that can be seen gives :

$$left{begin{eqnarray}
(a) &x^2+(y-R)^2&=&6^2\
(b) &x^2+y^2&=&R^2\
(c) &x^2+R^2&=&14^2
end{eqnarray}right.$$

From (c), one has $x^2=196-R^2$. Plugging this expression of $x^2$ in (a) and (b) gives a system of two equations with two unknowns $y$ and $R$ from which you should conclude that $y=8 sqrt{2}, R=9 sqrt{2}.$

Thus $x^2=196-R^2=196-162=34$; then $x=sqrt{34}$.

Then plug the results into the following formula : $tan alpha = R/x$ whence :

$alpha = mathbb{atan}(R/x)=mathbb{atan}(9 sqrt{2}/sqrt{34})=mathbb{atan}(9/sqrt{17}) approx 1.1412$ (radians).

i.e., approx. $64$ degrees and a half.

In the diagram below, from the given data, we have $AC = AD = r$, $CD=6$, $CE=14$. We want to find $CA$ and $AE$. Call $a=AE$, $b=ED$. Then from Pythagoras,
$$r^2 + a^2 = 196,quad a^2+b^2 = r^2,quad a^2 + (r-b)^2 = 36.$$
Solve these equations to find $r$ and $a$.

The task is to find $tanalpha$, you should favor a trigonometric approach.

Note that $R=14sinalpha$.

Denote the angle between the tangent and chord of 6cm with $beta$.
Also note that if you draw lines from the center of the circle to both ends of the chord of 6cm, the angle between these two lines is $2beta$ (the proof is elementary).

$$14cosalpha=6cosbetatag{1}$$

$$2Rsinfrac{2beta}{2}=2times14sinalphasinbeta=6$$

$$28sinalphasinbeta=6tag{2}$$

Try to solve $alpha,beta$ from (1),(2). For example, express $cosbeta$ from (1) and $sinbeta$ from (2). Square these two expressions and add, $beta$ will disappear. You'll get a simple equation with angle $alpha$ being the only unknown.