Toán cho tam giác ABC có BC=9cm,góc B=60,góc C=40 tìm độ dài AC,AB 01/07/2021 By Hadley cho tam giác ABC có BC=9cm,góc B=60,góc C=40 tìm độ dài AC,AB
Kẻ đường cao AH CÓ HC=AH.cot$40^{0}$ ta lại có HB=AH.cot$60^{0}$ => $\frac{HC}{HB}$ = $\frac{cot40^{0} }{cot60^{0} }$ =>HC=$\frac{cot40^{0} }{cot60^{0} }$ .HB =>BC=($\frac{cot40^{0} }{cot60^{0} }$+1) .HB=>HB=($\frac{cot60^{0}.9}{cot40^{0} +cot60^{0} }$) =>HC=$\frac{cot40^{0} }{cot60^{0} }$.($\frac{cot60^{0}.9}{cot40^{0} +cot60^{0} }$) =($\frac{cot40^{0}.9}{cot40^{0} +cot60^{0} }$) =>AH=($\frac{cot40^{0}.9}{cot40^{0} +cot60^{0} }$).$cot40^{0}$=($\frac{9}{cot40^{0} +cot60^{0} }$) =>AB=$\sqrt[]{AH ^{2}+HB ^{2}}$ =5,87433 cm =>AC=$\sqrt[]{AH ^{2}+HC ^{2}}$ =7,91446 cm Trả lời
Kẻ đường cao AH
CÓ HC=AH.cot$40^{0}$
ta lại có HB=AH.cot$60^{0}$
=> $\frac{HC}{HB}$ = $\frac{cot40^{0} }{cot60^{0} }$
=>HC=$\frac{cot40^{0} }{cot60^{0} }$ .HB
=>BC=($\frac{cot40^{0} }{cot60^{0} }$+1) .HB=>HB=($\frac{cot60^{0}.9}{cot40^{0} +cot60^{0} }$)
=>HC=$\frac{cot40^{0} }{cot60^{0} }$.($\frac{cot60^{0}.9}{cot40^{0} +cot60^{0} }$)
=($\frac{cot40^{0}.9}{cot40^{0} +cot60^{0} }$)
=>AH=($\frac{cot40^{0}.9}{cot40^{0} +cot60^{0} }$).$cot40^{0}$=($\frac{9}{cot40^{0} +cot60^{0} }$)
=>AB=$\sqrt[]{AH ^{2}+HB ^{2}}$ =5,87433 cm
=>AC=$\sqrt[]{AH ^{2}+HC ^{2}}$ =7,91446 cm