用正弦定理判断三角形形状sinA=2sinBcosc 且sin^2A=sin^2B+sin^C请判断三角形形状应是sin^2C
问题描述:
用正弦定理判断三角形形状
sinA=2sinBcosc 且sin^2A=sin^2B+sin^C
请判断三角形形状
应是sin^2C
答
∵sinA/a=sinB/b=sinC/c=1/(2R)∴sinA=2sinBcosC===>a=2bcosC===>a/2=bcosC(AD⊥BC,BD=CD)∴b=c(等腰三角形三线合一)sin^2A=sin^2B+sin^2C====>a²=b²+c²====>∠A=90º∴△ABC为等腰直角三角形...