(1)根据正弦定理,得
b/sinB=c/sinC
3√2/sin45°=3√3/sinC
解得
sinC=√3 /2
∵c=3√3>b=3√2
∴∠C>∠B
∴∠ C=60° ∠ C=120°(舍去)
(2)
A=180º-∠B-∠C=105°
S△ABC=½bcsin105°
=½bcsin(60º+45º)
=½x9√6x(√6+√2)/4
=9(√3+3)/4
(1)根据正弦定理,得
b/sinB=c/sinC
3√2/sin45°=3√3/sinC
解得
sinC=√3 /2
∵c=3√3>b=3√2
∴∠C>∠B
∴∠ C=60° ∠ C=120°(舍去)
(2)
A=180º-∠B-∠C=105°
S△ABC=½bcsin105°
=½bcsin(60º+45º)
=½x9√6x(√6+√2)/4
=9(√3+3)/4