# Math Help - circles again

1. ## circles again

CD is tangent to the circle and AE is perpendicular to AC. If a radius of the circle measures 12 in., which is the vale of DF + AF?

2. I can't see the image

3. Let $R=12,CF=x,FA=y\Rightarrow x+y=12$.
In $\triangle DCF \ \sin 30=\frac{x}{DF}\Rightarrow DF=2x$.
In $\triangle EAF \ \tan 30=\frac{y}{R}\Rightarrow y=4\sqrt{3}\Rightarrow x=12-4\sqrt{3}$.
Then $DF+FA=2x+y=24-8\sqrt{3}+4\sqrt{3}=24-4\sqrt{3}$

4. Hello, sanee66!

In right triangle $EAF\!:\;\tan30^o \,=\,\frac{AF}{12}\quad\Rightarrow\quad AF \:=\:12\tan30^o \:=\:4\sqrt{3}$

Then: . $CF \:=\:12 - AF \:=\:12 - 4\sqrt{3}\quad\Rightarrow\quad CF\:=\:4(3-\sqrt{3})$ .[1]

In right triangle $DCF\!:\;\sin30^o \:=\:\frac{CF}{DF}\quad\Rightarrow\quad DF \:=\:\frac{CF}{\sin30^o} \:=\:2\!\cdot\!CF$ .[2]

Substitute [1] into [2]: . $DF \:=\:8(3-\sqrt{3})$

Therefore: . $DF + AF \;=\;8(3-\sqrt{3}) + 4\sqrt{3} \;=\;24 - 4\sqrt{3}$

[Darn, this is identical to red_dog's solution!]

5. Using trigo. for this problem is like kill a fly with a Grand Cannon.
• $\measuredangle~AEF=30^\circ~\because~\overline{CD} \parallel\overline{AE}$
• We have two triangles $30^\circ,\,60^\circ,\,90^\circ$
• Finally $\overline{AF}=4\sqrt3\implies\overline{DF}=24-8\sqrt3~\therefore~\overline{DF}+\overline{AF}=24-4\sqrt3$
Properties of triangle $30^\circ,\,60^\circ,\,90^\circ$ are easily proved applying the Pythagorean Theorem.