如图,已知在三角新ABC中,D在BC上,边AB=DC,角B=46,角BAD=21,求角DAC的度数?
问题描述:
如图,已知在三角新ABC中,D在BC上,边AB=DC,角B=46,角BAD=21,求角DAC的度数?
急
答
∠ADC=∠B+∠BAD=46+21=67º,则∠ADB=180º-67º设:AB=DC=1;∠DAC=A在ΔADB中:AD/sin46º=1/sin67º==>AD=sin46º/sin67º在ΔADC中:AD/sinC=1/sinA=====>AD=sinC/sinA∴sin46º/si...