Q 52. 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
  
  
• 10 years agoHelpfull: Yes(24) No(2)
• n=22,23,24,25
  ans (d)
• 10 years agoHelpfull: Yes(12) No(3)
• sorry both time answer printed wrong
  ans is (D)
  for acute angle triangle,if c is greatest side then, a^2 + b^2 > c^2
  for 10,24,n
  if n is greatest,then
  10^2+24^2>n^2
  or 26^2>n^2
  or 26>n eqn(1)
  if 24 is greatest
  10^2+n^2>24^2
  n^2>476
  n>21.9 eqn(2)
  cobining eqn 1&2
  21.9
• 10 years agoHelpfull: Yes(7) No(4)
• for acute angle triangle,if c is greatest side then, a^2 + b^2 > c^2
  for 10,24,n
  if n is greatest,then
  10^2+24^2>n^2
  or 26^2>n^2
  or 26>n
  or n24^2
  or n^2>476
  or n>21.9 ----(2)
  combining (1)&(2)we get,
  21.9
• 10 years agoHelpfull: Yes(4) No(3)
• i am correct but prints wrong
  from last answer
  n21.9
  combining two
  21.9
• 10 years agoHelpfull: Yes(3) No(0)
• 1 coz its nt a right triangle
  
• 10 years agoHelpfull: Yes(3) No(1)
• 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
• 8 years agoHelpfull: Yes(3) No(0)
• tan^-1(2.4)=1.17
  so (A)1
  not sure
• 10 years agoHelpfull: Yes(2) No(7)
• For a triangle to be acute the square of each side must be less than the sum of the squares of the other two sides. In this case it is required that
  
  242−102
• 4 years agoHelpfull: Yes(2) No(0)
• 
  for acute angle triangle,if c is greatest side then, a^2 + b^2 > c^2
  for 10,24,n
  if n is greatest,then
  10^2+24^2>n^2
  or 26^2>n^2
  or 26>n
  or n24^2
  or n^2>476
  or n>21.9 -----(2)
  combining eqn 1 & 2
  21.9
• 10 years agoHelpfull: Yes(1) No(1)
• n+10>24.
  n^2+10^2
• 10 years agoHelpfull: Yes(0) No(2)
• c^2>24^2-10^2
  c^2>476(but 14
• 9 years agoHelpfull: Yes(0) No(0)