[tex]\displaystyle\bf\\Given:\\\Delta AKL\\AK=9~cm\\m\sphericalangle K=90^o\\m\sphericalangle A=60^o\\m\sphericalangle L=30^o\\Find:\\P\Delta AKL~\text{\bf and~the area of }\Delta AKL\\\\Answer:\\\text{\bf AL is hypotenuza}\\\\\frac{AK}{AL}=sin(L)=sin30^o=\frac{1}{2}\\ \\AL=\frac{AK}{sin30^o}= \frac{AK}{\dfrac{1}{2}} =AK\times2=9\times2=18~cm\\\\KL=AL\times sin A=AL\times sin60^o=18\times \frac{\sqrt{3}} {2}=9\sqrt{3}~cm[/tex]
[tex]\displaystyle\bf\\P=AK+AL+KL=9+18+9\sqrt{3}=27+9\sqrt{3}=\boxed{\bf9(3+\sqrt{3})cm}\\\\Area=\frac{AK\times KL}{2}=\frac{9\times9\sqrt{3}}{2}=\boxed{\bf\frac{81\sqrt{3}}{2}~cm^2}[/tex]
