Question

在三角形ABC中.三内角A,B,C,所对的边分别为a,b,c,若满足a=(√3-1）才，tanB/tanc=2a-c/c,求A,B,C的值

Answer 1

由正弦定理
a/c=sinA/sinC=√3-1

(2a-c)/c=2a/c-c/c=2sinA/sinC-1
所以tanB/tanC=(sinB/cosB)/(sinC/cosC)=sinBcosC/sinCcosB=2sinA/sinC-1
(sinBcosC+sinCcosB)/sinCcosB=2sinA/sinC
sin(B+C)/sinCcosB=2sinA/sinC
sin(180-A)/sinCcosB=2sinA/sinC
sinA/sinCcosB=2sinA/sinC
0<A<180则sinA不会等于0
同理，sinC不等于0
约分
1/cosB=2
cosB=1/2
B=60度

代入tanB/tanC=2sinA/sinC-1
且sinA/sinC=√3-1
√3/tanC=2√3-2-1
tanC=2+√3
C=75度

A=180-B-C

所以
A=45,B=60,C=75