,在三角形abc中,点d在bc上,ad=33,sin bad=5/13 cos adc=3/5 求sinabd和bd
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,在三角形abc中,点d在bc上,ad=33,sin bad=5/13 cos adc=3/5 求sinabd和bd
,在三角形abc中,点d在bc上,ad=33,sin bad=5/13 cos adc=3/5 求sinabd和bd
,在三角形abc中,点d在bc上,ad=33,sin bad=5/13 cos adc=3/5 求sinabd和bd
因为cosadc=3/5,角adb+角adc=180度,因为cosadc=3/5,所以cosadb=-cosadc=-3/5,sinadc=根号(1-sinadb^2)=4/5,在三角形abc中,由正弦定理得:bd/sinbad=ab/sinadb,因为sinbad=5/13,ad=33,所以ab/4/5=bd/5/13,所以,ab/bd=52/25=k即,ab=52k,bd=25k,在三角形abd中,由余弦定理得:ab^2=ad^2+bd^2-2ad*bd*cosadb,所以(52k)^2=33^2+(25k)^2-2*33*25k*(-3/5),21k^2-10k-11=0..k=1...k=-11/21(不合题意,应舍去),所以bd=25k=25,在三角形abd中,由正弦定理得:,bd/sinbad=ad/shiabd,所以,25/5/13=33/sinabd,即,sinabd=33/65,所以sinabd的值是33/65,bd的值是25
作AE⊥BC,垂足E, cos<ADC=3/5,sin<ADC=4/5, 设AD=x,DE=xcos>ADC=3x/5, AE=AD*sin<ADC=4x/5, sinB=5/13, cos=12/13,(<ADC是锐角,