IIT JEE Entrance Exam> If sides of a triangle are in gp and its ...

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

kartik , 5 Years ago

Grade 11

2 Answers

Samyak Jain

Last Activity: 5 Years ago

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 $\theta$ cos $\theta$ .

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. Thus, sum of two angles should be less than 180.

i.e. $\theta$ + 2 $\theta$ $\Rightarrow$ $\theta$ or more specifically 0 $\theta$ .

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 and 60 , 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.

Samyak Jain

Last Activity: 5 Years 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 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 $\theta$ cos $\theta$ .

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. Thus, sum of two angles should be less than 180.

i.e. $\theta$ + 2 $\theta$ $\Rightarrow$ $\theta$ or more specifically 0 $\theta$ .

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 and 60 , 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.

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IIT JEE Entrance Exam> If sides of a triangle are in gp and its ...

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

kartik , 5 Years ago

Samyak Jain

Samyak Jain

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