1. If m∠ACB=90° and m∠ACD=60°, then m∠BCD=30°. Consider right triangle BCD. BC is its hypotenuse.
The leg that lies opposite to the 30° angle is half of hypotenuse, then BD=3 cm.
By the Pythagorean theorem,
[tex]CD^2+BD^2=BC^2,\\ \\CD^2+3^2=6^2,\\ \\CD^2=36-9=27,\\ \\CD=3\sqrt{3}\ cm.[/tex]
2. Consider right triangle ACD, m∠ACD=60°, then m∠CAD=90°-60°=30°. Thus, the leg CD is opposite to the hypotenuse AC and
[tex]AC=2CD=2\cdot 3\sqrt{3}=6\sqrt{3}\ cm.[/tex]
3. The area of ΔABC is
[tex]A_{ABC}=\dfrac{1}{2}AC\cdot BC=\dfrac{1}{2}\cdot 6\sqrt{3}\cdot 6=18\sqrt{3}\ cm^2.[/tex]
