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        If sides of a triangle are in gp and its larger angle is 2 times the smallest then the range of the  common ratio is
7 months ago

							The sides of a triangle are in GP. Let the lengths of the sides bea/r , a , ar. r is the common ratio of the GP.Clearly, a > 0 and r > 0 as they represent length.Let  denote the smallest angle of the triangle. Then largest angle is 2. Case 1.  r > 1 a/r and ar are the smallest and largest sides respectively.We know that side opposite to larger angle of a triangle is greater and vice versa.So, angle opposite to a/r is  and that opposite to ar is 2.By sine rule,  (a/r) / sin  =  ar / sin2.  Cancel a on both sides and replace sin2 by 2sincos.1/r sin  =  r / 2sin cos  r2  =  2 cos     (Observe that r2 = 2 cos is positive i.e.  is acute.)Maximum value of cos is 1, so maximum value of r2 is 2 i.e. maximum value of r is  .1        ...(1) We know that sum of all angles of a triangle is 180. Thus, sum of two angles should be less than 180.i.e.  + 2        or  more specifically 0  . Case 2.  0  a/r and ar are the largest and smallest sides respectively and angles opposite to them are 2 and .By sine rule,   (a/r) / sin2  =  ar / sin.Above equation simplifies to r2 = 1 / 2cos. we have found that  lies between 0 and 60 , cos lies between 1/2 and 1.So, r2 lies between 1/2 and 1  i.e.  r lies in the interval (1/ , 1).  Complete range of common ratio r is (1/ , 1)    (1 , ).b is correct option.

7 months ago
							Due to some error, some characters are missing in above answer. So I’m reposting the answer.The sides of a triangle are in GP. Let the lengths of the sides bea/r , a , ar. r is the common ratio of the GP.Clearly, a > 0 and r > 0 as they represent length.Let  denote the smallest angle of the triangle. Then largest angle is 2. Case 1.  r > 1 a/r and ar are the smallest and largest sides respectively.We know that side opposite to larger angle of a triangle is greater and vice versa.So, angle opposite to a/r is  and that opposite to ar is 2.By sine rule,  (a/r) / sin  =  ar / sin2.  Cancel a on both sides and replace sin2 by 2sincos.1/r sin  =  r / 2sin cos  r2  =  2 cos     (Observe that r2 = 2 cos is positive i.e.  is acute.)Maximum value of cos is 1, so maximum value of r2 is 2 i.e. maximum value of r is  .1        ...(1) We know that sum of all angles of a triangle is 180. Thus, sum of two angles should be less than 180.i.e.  + 2        or  more specifically 0  . Case 2.  0  a/r and ar are the largest and smallest sides respectively and angles opposite to them are 2 and .By sine rule,   (a/r) / sin2  =  ar / sin.Above equation simplifies to r2 = 1 / 2cos. we have found that  lies between 0 and 60 , cos lies between 1/2 and 1.So, r2 lies between 1/2 and 1  i.e.  r lies in the interval (1/ , 1).  Complete range of common ratio r is (1/ , 1)    (1 , ).b is correct option.

7 months ago
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