If in a triangle ABC,
cosAcosB+sinAsinBsinc=1, then a:b:c=?

• answer:1 : 1 : √ 2
  
  cosA.cosB +sinA.sinB.sinC = 1;
  
  multiply by 2 both side we get
  
  2 cosA.cosB + 2 sinA.sinB.sinC = 2
  
  then do some manipulation because its IIT JEE question :)
  we get
  
  ( cos^2 A + cos^2 B – 2.cosA.cosB )+ ( sin^2 A + sin^2 B - 2.sinA.sinB ) + 2.sinA.sinB ( 1 – sinC) = 0
  
  simplify it
  
  ( cosA – cosB )^2 + ( sinA – sinB )^2 + 2.sinA.sinB ( 1 – sinC) = 0 eqn(1)
  
  By triangle property
  cosA – cosB = 0 , sinA - sinB = 0 and (1 – sinC) = 0
  
  so A=B and C=pi/2 ;
  
  and
  A = B = pi/4 ;
• 10 years agoHelpfull: Yes(0) No(0)