Question

三角形三边长是连续的三个自然数,最大角是最小角的二倍,求三边长

Answer 1

设三角形最小角为a,三边长分别为k-1,k,k+1

则根据正弦定理和已知有

（k-1)/sina=(k+1)/sin2a=(k+1)/2sinacosa

∴cosa=(k+1)/(2k-2)

又∵cosa=[k²+(k+1)²-(k-1)²]/[2k(k+1)]

=(k²+k²+2k+1-k²+2k-1)/(2k²+2k)

=(k²+4k)/(2k²+2k)

=(k+4)/(2k+2)

∴(k+1)/(2k-2)=(k+4)/(2k+2)

(k+1)/(k-1)=(k+4)/(k+1)

(k+1)²=(k+4)(k-1)

k²+2k+1=k²+3k-4

∴k=5

∴△ABC的三边长分别为4,5,6.