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Grade: 11
        
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
3 months ago

Answers : (2)

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