两道高中数学题

一、cotC/cotA+cotB
=(cosC/sinC)/[(cosA/sinA)+(cosB/sinB)]
=[(cosC·sinA·sinB)/sinC]/(sinA·cosB+cosA·sinB)
=(cosC·sinA·sinB)/[sinC·sin(A+B)]
=(cosC·sinA·sinB)/[sinC·sin(π-C)]
=(cosC·sinA·sinB)/(sinC·sinC)
=cosC·(sinA/sinC)·(sinB/sinC)
根据正弦定理,
a/sinA=b/sinB=c/sinC
→sinA/sinC=a/c;
sinB/sinC=b/c;
则原手伍盯式=cosC·(sinA/sinC)·(sinB/sinC)
=cosC·(a/c)·(b/c)
=(ab·cosC)/c^2
根据余弦定理有
a^2+b^2-c^2 = 2ab·cosC
→ab·cosC=(a^2+b^2-c^2)/2
=（mc^2-c^2)/2
所橘姿以（ab·cosC)/c^2=（mc^2-c^2)/2c^2
=（m-1）毕和/2=999
所以m=1999

二、 A>B>C,且三边长为连续自然数。b=b, a=b+1, c=b-1
a=2·c·cosC b+1=2(b-1)CosC; CosC=(b+1)/[2(b-1)]
c²=a²+b²-2abCosC; (b-1)²=(b+1)²+b²-2(b+1)bCosC, CosC=(b+4)/[2(b+1)]
连立，得b=5,a=6,c=4
a/SinA=b/SinB=c/SinC

sinA:sinB:sinC=a:b:c=6:5:4