# How to find length AF in this right triangle?

In the given figure, $$AD =1$$, $$DC = 6$$, $$\angle AFC = 90 ^\circ$$, $$\angle ADF = 60 ^\circ$$, how to solve for the length of $$AF$$?

What I got is $$AF^2 + 36\tan^2 \angle AFD\cdot AF^2 = 7$$, using sine law and pythagoras theorem. I didn't know what I was missing, any help will be appreciated. Thanks in advance!

Let $$FD=x$$

By cosine-rule in $$\triangle ADF$$,

$$AF^2 = x^2 + 1^2 - 2\cdot x\cdot 1\cdot \cos 60$$

Cosine-rule in $$\triangle CDF$$,

$$CF^2 = x^2 + 6^2 - 2\cdot x\cdot6\cdot \cos 120$$

Now use $$AC^2 = AF^2 + CF^2$$ where $$AC=7$$.

I got $$\boxed{AF=\dfrac{\sqrt{7}}{2}}$$