Question

在三角形ABC中,角A,B,C所对的边分别为a,b,c,且1+tanA/tanb=2c/b

求∠A

Answer 1

正弦定理，a/sinA=b/sinB=c/sinC=2R,==>sinA/sinB=a/b.
余弦定理，cosA=(b^2+c^2-a^2)/(2bc), cosB=(a^2+c^2-b^2)/(2ac),
1+tanA/tanb=1+sinA/sinB*cosB/cosA=2c^2/(b^2+c^2-a^2)=2c/b,
==>b^2+c^2-a^2=bc,
==>cosA=1/2,
==>∠A=60°.