perhatikan gambar disamping.Panjang AD =. A. 2 cm B.2√2 cm C.4 cm D.4√2 cm

[tex]sin60^{o}= \frac{AB}{AC} \\ \frac{ \sqrt{3} }{2} = \frac{2 \sqrt{3} cm }{AC} \\ \sqrt{3} \times AC=2 \sqrt{3} cm \times 2 \\ AC= \frac{2 \sqrt{3} cm\times 2}{ \sqrt{3} } \\ AC =4 cm[/tex]
sehingga,
[tex]cos45^{o}= \frac{AD}{AC} \\ \frac{ \sqrt{2} }{2} = \frac{AD}{4cm} \\ \sqrt{2} \times 4 cm = AD \times 2 \\ \frac{4 \sqrt{2} cm}{2} = AD \\ 2 \sqrt{2} cm = AD\\[/tex]
Jadi jawabannya B :)