2. A triangle has sides of lengths 10, 24 and n, where n is a positive n is a positive integer. The number of values of n for which this triangle has three acute angles is
A. 1
B. 2
C. 3
D. 4

• triangle with acute angle:a^2+b^2>c^2
  i)let a=10,b=n,c=24
  so 10^2+n^2>24^2
  n^2>(24^2-10^2)
  n^2>576-100
  n^2>476
  
  ii)let a=10,b=24,c=n
  10^2+24^2>n^2
  100+576>n^2
  676>n^2
  
  so from i and ii
  476>n^2>676 hence root476>n>root676
  ans=22,23,24,25
  sol:4
  
  
• 8 years agoHelpfull: Yes(17) No(0)
• In an acute angled triangle; the sum of the squares of any two sides of a triangle have to be greater than the third side
  i.e. 10^2+24^2>x^2 ; x^2+10^2>24^2 ;and x^2+>24^2>10^2 computing the minimum and maximum “x”, we get the range as 26>x>√476. the values of “x” which satisfy the above conditions are 22,23,24 and 25. Hence 4
• 8 years agoHelpfull: Yes(16) No(4)
• a^2+b^2>c^2
  10^2+24^2>c^2
  c^224^2
  b^2>476
  so 476
• 8 years agoHelpfull: Yes(4) No(11)
• I think ans:3
  n should be 22 or 23 or 25 to satisfy condition
  a^2+b^2>c^2
• 8 years agoHelpfull: Yes(3) No(5)
• a^2+b^2>c^2=>=>10^2+24^2>c^2
  c^224^2=>=>b^2>476
  so 476>>>so value is 4
• 8 years agoHelpfull: Yes(2) No(7)
• triangle with acute angle:a^2+b^2>c^2
  i)let a=10,b=n,c=24
  so 10^2+n^2>24^2
  n^2>(24^2-10^2)
  n^2>576-100
  n^2>476
  
  ii)let a=10,b=24,c=n
  10^2+24^2>n^2
  100+576>n^2
  676>n^2
  
  so from i and ii
  476>n^2>676 hence sqrt(476)>n>sqrt(676)
  ans=22,23,24,25
  sol:4
• 7 years agoHelpfull: Yes(1) No(0)
• plzzz explain this properly
  
• 8 years agoHelpfull: Yes(0) No(3)