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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!

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

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