In a rt angled triangle ABC, ang B is the right angle and AC = 2rt5 cm. If AB-BC = 2cm, then the value of (cos^2A-cos^2C) is :

• ac=2rt5 ab=4 bc=2 (using a^2+b^2=C^2)
  we get cos(A)=(4/2rt5)
  and cos(c)=2/2rt5
  cos^2A-cos^2c=3/5=.6
• 12 years agoHelpfull: Yes(3) No(0)
• can u explain for me again
  
• 12 years agoHelpfull: Yes(1) No(0)
• ac=2rt5 consider on of the other two sides be bc=x and third side will ab= 2+x(ab-bc=2)
  so x^2+ (2+x)^2=(2rt5)^2
  sloving this we get x=2
  now cos(A)=ab/ac=4/2rt5
  cos(C)=bc/ac=2/2rt5
  then cos^2(a)-cos^2(c)=(4/2rt5)^2-(2/2rt5)^2
  4/5-1/5=3/5=0.6
  
• 12 years agoHelpfull: Yes(1) No(0)
• AB-BC=2....(1)
  sqar it
  (AB-BC)^2=AB^2+BC^2-2AB.BC
  but AB62+BC^2=(2sqt5)^2.... according to Pythagoras Thermo
  put the vale u will get
  AB.BC=8
  now solve with eq(1)
  AB=4
  BC=2
  now you have AB=4 BC=2 CA=2sqt5
  and triangle is right angle
  so find out cosA ,cosC and solve
  ans will be 3/5
• 12 years agoHelpfull: Yes(0) No(0)