在一个三角形里,AB=6 AC=4 BC=5 1.求证:∠C=2∠B2.求中线AD3.求角平分线AE4.求高AF快来回答把谢谢!
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余弦定理(1)cosC=(25+16-36)/(2*5*4)=1/8,cosB=(25+36-16)/(2*5*6)=3/4,2(cosB)^2-1=1/8,所以∠C=2∠B(2)BD=2.5,AD^2=36+6.25-2*6*2.5*cosB=79/4,AD=(根号79)/2(3)AB/AC=BE/EC,所以BE=3,AE^2=36+9-2*6*3*cosB=18,AE=3根号2(4)sinB=(根号7)/4,S=0.5*BC*AF=0.5*AB*BC*sinB,AF=(3根号7)/2