Question

在三角形ABC之中,点D在BC边上,AD=33,sin角BCD=5/13,cos角ADC=3/15.求sin角ABD的值

Answer 1

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