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Because AB is a diameter, by Thale's theorem,

\(\angle ACB = \angle ADB = 90^\circ\)

 

That means \(\angle CAB = \arccos\left(\dfrac24\right) = 60^\circ\) by simple trigonometry.

 

Also, because \(\angle ADB = 90^\circ\), considering the interior angle sum of \(\triangle ADB\),

\(\angle DBA = 45^\circ\). This means \(\triangle ADB\) is a right isosceles triangle, which means AD = DB.

 

By Pythagorean theorem, we get \(AD = DB = 2\sqrt 2\).

 

Now consider \(\triangle ADC\) and use Law of Cosine. I believe you can finish it from here on your own.