如图,已知△ABC中,∠B=45°,∠C=30°,BC=4+2√3,求AB、AC的长
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解答:
∠A=180°-∠B-∠C=105°
∴ sinA=sin105°=(√6+√2)/4
利用正弦定理:AB/sinC=AC/sinB=BC/sinA
∴ AB=BCsinC/sinA
=(4+2√3)*(1/2)/[(√6+√2)/4]
=(8+4√3)/(√6+√2)
=(6+4√3+2)/(√6+√2)
=(√6+√2)²/(√6+√2)
=√6+√2
AC=ABsinB/sinC
=(√6+√2)*sin45°/sin30°
=(√6+√2)√2
=2√3+2
∠A=180°-∠B-∠C=105°
∴ sinA=sin105°=(√6+√2)/4
利用正弦定理:AB/sinC=AC/sinB=BC/sinA
∴ AB=BCsinC/sinA
=(4+2√3)*(1/2)/[(√6+√2)/4]
=(8+4√3)/(√6+√2)
=(6+4√3+2)/(√6+√2)
=(√6+√2)²/(√6+√2)
=√6+√2
AC=ABsinB/sinC
=(√6+√2)*sin45°/sin30°
=(√6+√2)√2
=2√3+2
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