Question

若△ABC中,∠C＝90°,∠A,∠B,∠C的对边分别是a,b,c,且c-4ac+4a=0,则sinA+cosA的值为

Answer 1

/2

所以

a*cos^2(C/2)+c*cos^2(A/2)=(3/2)b

a*(1+cosC)/2+c*(1+cosA)/2=(3/2)b

acosC+ccosA=3b-a-c

b=3b-a-c

2b=a+c

证毕.

2.

b^2=a^2+c^2-2accosB

(a+c)^2=4a^2+4c^2-8accosB

2ac=3a^2+3c^2-8accosB

8accosB=3a^2+3c^2-2ac

3a^2+3c^2≥6ac

3a^2+3c^2-2ac≥4ac

8accosB≥4ac

cosB≥1/2

B≤π/3