Question

三角形ABC中,若C=2B,且b+c=2a,判断三角形形状

Answer 1

由于b+c=2a,由正弦定理知sinB+sinC=2sinA.
sinB+sinC=2sin(B+C)/2cos(B-C)/2
所以有sin(B+C)/2cos(B-C)/2=sinA
题意给出C=2B,且三角形有（B+C)/2=90°-A/2
将C=2B,（B+C)/2=90°-A/2代入sin(B+C)/2cos(B-C)/2=cos(A/2)cos(B/2)=2sin（A/2）cos(A/2)
所以cos(B/2)=2sin(A/2)=2cos(B+C)/2=2cos(3B/2)
由cos(B/2)=2cos(3B/2)
解cos(B/2)=2cos(3B/2)方程得sin(B/2)=（根号2）/4.
B=41.4°
C=2B=2*41.4°=82.8°
A=180°-41.4°-82.8°=55.8°
故该三角形是锐角三角形