Question

PLEAASEEE HELLLPPPPP I DON'T GET IT ;((

Given: ∆AKL, AK = 9

m∠K = 90°

m∠A = 60°

m∠L = 30°

Find: P∆AKL, The area of ∆AKL

Answer 1

`\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}\\\\(AK)/(AL)=sin(L)=sin30^o=(1)/(2)\\ \\AL=(AK)/(sin30^o)= (AK)/((1)/(2)) =AK*2=9*2=18~cm\\\\KL=AL* sin A=AL* sin60^o=18* \frac{√(3)} {2}=9√(3)~cm`

`\displaystyle\bf\\P=AK+AL+KL=9+18+9√(3)=27+9√(3)=\boxed{\bf9(3+√(3))cm}\\\\Area=(AK* KL)/(2)=(9*9√(3))/(2)=\boxed{\bf(81√(3))/(2)~cm^2}`