In a eucldian triangle one side is 20 and one of its others sides is 10... given the area is 80 what is the third side?

• Fisrtly, Draw the the triagle. Now
  Using Relation A= 1/2*a *b* SinC put (A=80,a=20,b=10)
  We can calculate the value of SinC which is equal to 4/5
  Since, SinC=4/5 therefore, CosC= 3/5
  Now using Cosine rule
  c^2=a^2+b^2-2abCosC..
  Now put the values
  C=2sqrt65
  ie.16.12
• 11 years agoHelpfull: Yes(20) No(0)
• Area of triangle A=sqrt of s(s-a)(s-b)(s-c).
  where s=(a+b+c)/2.
  here a=20,b=10,A=80.
  therefore s=15+c/2.
  80^2=(15+c/2)(c/2-5)(c/2+5)(15-c/2)
  c=16.12.
• 11 years agoHelpfull: Yes(15) No(1)
• let 20 be the base,and other side of the triangle be 10,draw a perpendicular to the base which height will be h=8 as(1/2*20*h=80 given),
  the base is two parts now x & 20-x,putting pythagoras theorem we can get x 10^2=8^2+x^2 so x=6,
  other part of base is 20-6=14,
  now we can get the other part of the triangle 14^2+8^2=64 i.e 8
  hence the other side is 8.
• 11 years agoHelpfull: Yes(4) No(6)
• let 10 be the base and draw a perpendicual on it so height will come 16.
  now we get a right triangle with height 16 and one side 20 so we can get base which comes out 12.
  10 is one side so 2 will be an extension for the perpendicular drawn.height=16 perpendicular =16 so third side =root(16*16+2*2)=root260
• 11 years agoHelpfull: Yes(3) No(4)