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.

Provide a better Answer & Earn Cool Goodies

Enter text here...

LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

Full Live Access

Study Material

Live Doubts Solving

Daily Class Assignments

Ask a Doubt

Get your questions answered by the expert for free

Enter text here...

My rank is 29k, General, should I opt for any state college cse branch or some Nits in which in which I am getting ECE , chemical, mechanics etc....

iit jee entrance exam

0 Answer Available

Last Activity: 2 Years ago

How is Ranquer Eduzone,comparing, Akash,Fitjee,Resonance,Naryana

iit jee entrance exam

0 Answer Available

Last Activity: 2 Years ago

Can class 9th students study syllabus classof 11th

iit jee entrance exam

3 Answers Available

Last Activity: 2 Years ago

Sum of series 2[n/2] ! [n/2]!/ n! [2Co-2C2+3C2....+[-1]^n[n+1]2Cn] where n is an even positive integer. Kindly provide solution of this question and how to generalise this. How do we know that which expansion will give this?

iit jee entrance exam

1 Answer Available

Last Activity: 2 Years ago

I m foreigner,How much mark i need to get into IIT??

iit jee entrance exam

1 Answer Available

Last Activity: 3 Years ago

View all Questions

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

Provide a better Answer & Earn Cool Goodies

LIVE ONLINE CLASSES

Ask a Doubt

Other Related Questions on iit jee entrance exam

My rank is 29k, General, should I opt for any state college cse branch or some Nits in which in which I am getting ECE , chemical, mechanics etc....

iit jee entrance exam

How is Ranquer Eduzone,comparing, Akash,Fitjee,Resonance,Naryana

iit jee entrance exam

Can class 9th students study syllabus classof 11th

iit jee entrance exam

Sum of series 2[n/2] ! [n/2]!/ n! [2Co-2C2+3C2....+[-1]^n[n+1]2Cn] where n is an even positive integer. Kindly provide solution of this question and how to generalise this. How do we know that which expansion will give this?

iit jee entrance exam

I m foreigner,How much mark i need to get into IIT??

iit jee entrance exam