Atr=[tex] \frac{a.b.sinC}{2} [/tex]
a)162=[tex] \frac{18.18.sin\ \textless \ O}{2} \\ sin\ \textless \ O=1 \\ m\ \textless \ O=90[/tex] m(arcmicAB)=90
b)[tex] \frac{18.18.sin\ \textless \ O}{2} =81 \\ sin \ \textless \ O= \frac{1}{2} \\ m\ \textless \ O=30 [/tex] m(arcmicAB)=30