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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

Samyak Jain
333 Points
							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 $\dpi{100} \theta$ denote the smallest angle of the triangle. Then largest angle is 2$\dpi{100} \theta$. Case 1.  r > 1$\dpi{100} \Rightarrow$ 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 $\dpi{100} \theta$ and that opposite to ar is 2$\dpi{100} \theta$.By sine rule,  (a/r) / sin$\dpi{100} \theta$  =  ar / sin2$\dpi{100} \theta$.  Cancel a on both sides and replace sin2$\dpi{100} \theta$ by 2sin$\dpi{100} \theta$cos$\dpi{100} \theta$.1/r sin$\dpi{100} \theta$  =  r / 2sin$\dpi{100} \theta$ cos$\dpi{100} \theta$$\dpi{100} \Rightarrow$  r2  =  2 cos$\dpi{100} \theta$     (Observe that r2 = 2 cos$\dpi{100} \theta$ is positive i.e. $\dpi{100} \theta$ is acute.)Maximum value of cos$\dpi{100} \theta$ is 1, so maximum value of r2 is 2 i.e. maximum value of r is $\dpi{80} \sqrt{2}$ .1 $\dpi{80} \sqrt{2}$       ...(1) We know that sum of all angles of a triangle is 180$\dpi{80} \degree$. Thus, sum of two angles should be less than 180$\dpi{80} \degree$.i.e. $\dpi{100} \theta$ + 2$\dpi{100} \theta$ $\dpi{80} \degree$  $\dpi{100} \Rightarrow$  $\dpi{100} \theta$ $\dpi{80} \degree$  or  more specifically 0$\dpi{80} \degree$ $\dpi{100} \theta$ $\dpi{80} \degree$. Case 2.  0 $\dpi{100} \Rightarrow$ a/r and ar are the largest and smallest sides respectively and angles opposite to them are 2$\dpi{100} \theta$ and $\dpi{100} \theta$.By sine rule,   (a/r) / sin2$\dpi{100} \theta$  =  ar / sin$\dpi{100} \theta$.Above equation simplifies to r2 = 1 / 2cos$\dpi{100} \theta$.$\dpi{100} \because$ we have found that $\dpi{100} \theta$ lies between 0$\dpi{80} \degree$ and 60$\dpi{80} \degree$ , cos$\dpi{100} \theta$ lies between 1/2 and 1.So, r2 lies between 1/2 and 1  i.e.  r lies in the interval (1/$\dpi{80} \sqrt{2}$ , 1). $\dpi{100} \therefore$ Complete range of common ratio r is (1/$\dpi{80} \sqrt{2}$ , 1)  $\dpi{100} \cup$  (1 , $\dpi{80} \sqrt{2}$).b is correct option.

7 months ago
Samyak Jain
333 Points
							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 $\dpi{100} \theta$ denote the smallest angle of the triangle. Then largest angle is 2$\dpi{100} \theta$. Case 1.  r > 1$\dpi{100} \Rightarrow$ 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 $\dpi{100} \theta$ and that opposite to ar is 2$\dpi{100} \theta$.By sine rule,  (a/r) / sin$\dpi{100} \theta$  =  ar / sin2$\dpi{100} \theta$.  Cancel a on both sides and replace sin2$\dpi{100} \theta$ by 2sin$\dpi{100} \theta$cos$\dpi{100} \theta$.1/r sin$\dpi{100} \theta$  =  r / 2sin$\dpi{100} \theta$ cos$\dpi{100} \theta$$\dpi{100} \Rightarrow$  r2  =  2 cos$\dpi{100} \theta$     (Observe that r2 = 2 cos$\dpi{100} \theta$ is positive i.e. $\dpi{100} \theta$ is acute.)Maximum value of cos$\dpi{100} \theta$ is 1, so maximum value of r2 is 2 i.e. maximum value of r is $\dpi{80} \sqrt{2}$ .1 $\dpi{80} \sqrt{2}$       ...(1) We know that sum of all angles of a triangle is 180$\dpi{80} \degree$. Thus, sum of two angles should be less than 180$\dpi{80} \degree$.i.e. $\dpi{100} \theta$ + 2$\dpi{100} \theta$ $\dpi{80} \degree$  $\dpi{100} \Rightarrow$  $\dpi{100} \theta$ $\dpi{80} \degree$  or  more specifically 0$\dpi{80} \degree$ $\dpi{100} \theta$ $\dpi{80} \degree$. Case 2.  0 $\dpi{100} \Rightarrow$ a/r and ar are the largest and smallest sides respectively and angles opposite to them are 2$\dpi{100} \theta$ and $\dpi{100} \theta$.By sine rule,   (a/r) / sin2$\dpi{100} \theta$  =  ar / sin$\dpi{100} \theta$.Above equation simplifies to r2 = 1 / 2cos$\dpi{100} \theta$.$\dpi{100} \because$ we have found that $\dpi{100} \theta$ lies between 0$\dpi{80} \degree$ and 60$\dpi{80} \degree$ , cos$\dpi{100} \theta$ lies between 1/2 and 1.So, r2 lies between 1/2 and 1  i.e.  r lies in the interval (1/$\dpi{80} \sqrt{2}$ , 1). $\dpi{100} \therefore$ Complete range of common ratio r is (1/$\dpi{80} \sqrt{2}$ , 1)  $\dpi{100} \cup$  (1 , $\dpi{80} \sqrt{2}$).b is correct option.

7 months ago
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