(5分)在$\Delta ABC$中,内角$A$,$B$,$C$所对边分别为$a$,$b$,$c$,若$B=\dfrac{\pi }{3}$,$b^2=\dfrac{9}{4}ac$,则$\sin A+\sin C=$( ) A.$\dfrac{3}{2}$ B.$\sqrt{2}$ C.$\dfrac{\sqrt{7}}{2}$ D.$\dfrac{\sqrt{3}}{2}$ 答案:$C$ 分析:根据已知条件,结合正弦定理,以及余弦定理,即可求解. 解:因为$B=\dfrac{\pi }{3}$,$b^2=\dfrac{9}{4}ac$, 所以由正弦定理可得,$\sin A\sin C=\dfrac{4}{9}\sin ^{2}B=\dfrac{1}{3}$, 由余弦定理可得:$b^{2}=a^{2}+c^{2}-2ac\cdot \cos B=a^{2}+c^{2}-ac=\dfrac{9}{4}ac$,即$a^2+c^2=\dfrac{13}{4}ac$, $\sin ^2A+\sin ^2C=\dfrac{13}{4}\sin A\sin C=\dfrac{13}{12}$, 所以$(\sin A+\sin C)^{2}=\sin ^{2}A+\sin ^{2}C+2\sin A\sin C=\dfrac{7}{4}$,$\sin A+\sin C=\dfrac{\sqrt{7}}{2}$. 故选:$C$. 点评:本题主要考查正弦定理,以及余弦定理,属于基础题.
|