Guest

(c2+a2)/(b2+c2)={1+cosBcos(C-A)}/{1+cosAcos(B-c)} (a+b+c)[tanA/2+tanB/2]=2c cotC/2 sin(A-B)/2=(a-b)/c.cosC/2 b.cosb + c.cosc = a.cos(b-c) a cosA+b cosB=c cos(A-B) b2=(c-a)2.cos2(B/2)+(c+a)2.sin2(B/2)

  1. (c2+a2)/(b2+c2)={1+cosBcos(C-A)}/{1+cosAcos(B-c)}
  2. (a+b+c)[tanA/2+tanB/2]=2c cotC/2
  3. sin(A-B)/2=(a-b)/c.cosC/2
  4. b.cosb + c.cosc = a.cos(b-c)
  5. a cosA+b cosB=c cos(A-B)
  6. b2=(c-a)2.cos2(B/2)+(c+a)2.sin2(B/2)

Grade:

0 Answers

No Answer Yet!

ASK QUESTION

Get your questions answered by the expert for free