Triangle problem - Evaluating the given expression

  • Thread starter Saitama
  • Start date
  • #1
Saitama
4,244
93

Homework Statement


In a triangle ABC, with usual notation, if ##a^2b^2c^2 (\sin 2A + \sin 2B + \sin 2C) = λ(∆)^x## where ∆ is the area of the triangle and x ##\in## Q, find (λx).


Homework Equations





The Attempt at a Solution


The usual notation is:
a,b,c are three sides of the triangle opposite to the angles A,B and C respectively.

I remember a formula relating ∆ and the three sides i.e
$$\Delta=\frac{abc}{4R} \Rightarrow abc=4R \Delta$$
Also, from the law of sines,
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
where R is the circumradius of triangle.

The given expression can be written as:
$$2a^2b^2c^2 (\sin A \cos A + \sin B \cos B + \sin C \cos C)$$
I can substitute abc and sines from the above two relations but what should I replace cosines with? Law of cosines doesn't seem to be of much help.

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,915

Homework Statement


In a triangle ABC, with usual notation, if ##a^2b^2c^2 (\sin 2A + \sin 2B + \sin 2C) = λ(∆)^x## where ∆ is the area of the triangle and x ##\in## Q, find (λx).


Homework Equations





The Attempt at a Solution


The usual notation is:
a,b,c are three sides of the triangle opposite to the angles A,B and C respectively.

I remember a formula relating ∆ and the three sides i.e
$$\Delta=\frac{abc}{4R} \Rightarrow abc=4R \Delta$$
Also, from the law of sines,
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
where R is the circumradius of triangle.

The given expression can be written as:
$$2a^2b^2c^2 (\sin A \cos A + \sin B \cos B + \sin C \cos C)$$
I can substitute abc and sines from the above two relations but what should I replace cosines with? Law of cosines doesn't seem to be of much help.

Any help is appreciated. Thanks!

Do you remember that sin(2θ)= 2 sinθ cosθ??

ehild
 
  • #3
Saitama
4,244
93
Hi ehild! :smile:

Do you remember that sin(2θ)= 2 sinθ cosθ??

ehild

Yes, and I have even used it. I wrote ##\sin(2A)=2\sin A\cos A## and similarly the others to get the final expression I wrote.
 
  • #4
ehild
Homework Helper
15,543
1,915
Sorry, it was stupid of me. How can you get the area of the triangle, using R and the central angles?

ehild
 
  • #5
Saitama
4,244
93
Sorry, it was stupid of me. How can you get the area of the triangle, using R and the central angles?

ehild

Only this comes to my mind:
$$\Delta = 2R^2\sin A \sin B \sin C$$

How do I use this?
 
  • #6
ehild
Homework Helper
15,543
1,915
Do you remember that the area is 0.5absin(C)? See the coloured triangles in the picture. How do you get their area in terms of R and the central angles? How are the central angles related to the angles of the big triangle?


ehild
 

Attachments

  • triangleincircle.JPG
    triangleincircle.JPG
    10.2 KB · Views: 353
Last edited:
  • #7
Saitama
4,244
93
Do you remember that the area is 0.5absin(C)? See the coloured triangles in the picture. How do you get their area in terms of R and the central angles? How are the central angles related to the angles of the big triangle?


ehild

Yes, I remember that. I misunderstood what you meant by the central angle, sorry. :redface:

The areas of three triangles formed are 0.5R2sin(2A), 0.5R2sin(2B) and 0.5R2sin(2C). Am I correct?
 
  • #8
ehild
Homework Helper
15,543
1,915
Yes, it is correct. So what is the area of the big triangle?

ehild
 
  • #9
Saitama
4,244
93
Yes, it is correct. So what is the area of the big triangle?

ehild

$$\Delta = \frac{1}{2}R^2\sin 2A +\frac{1}{2}R^2\sin(2B) +\frac{1}{2}R^2\sin(2C)$$

Also,
$$R=\frac{abc}{4\Delta}$$

$$\Rightarrow \Delta=\frac{1}{2}R^2(\sin 2A+\sin 2B+\sin 2C) \Rightarrow \Delta=\frac{1}{2}\frac{a^2b^2c^2}{16\Delta^2}(\sin 2A+\sin 2B+\sin 2C)$$

$$\Rightarrow a^2b^2c^2(\sin 2A+\sin 2B+\sin 2C)=32\Delta^3$$

Hence, ##\lambda=32## and ##x=3## and therefore ##\lambda x=96##.

Thanks a lot ehild! :smile:

How did you think of those triangles?
 
  • #10
ehild
Homework Helper
15,543
1,915
I just drew a picture with triangle and circumcircle. It is obvious, is it not, seeing the double angles in the equation. The central angle is twice the inscribed angle....


ehild
 
  • #11
Saitama
4,244
93
It is obvious, is it not, seeing the double angles in the equation.

It wasn't for me at least. :redface:

Thank you once again ehild! :smile:
 
  • #12
ehild
Homework Helper
15,543
1,915
I forgot to tell you that you gave the idea of using circumcircle... I have never seen the equation ##\Delta=\frac{abc}{4R}##. So thank you, Pranav:smile:

ehild
 

Suggested for: Triangle problem - Evaluating the given expression

Replies
40
Views
990
Replies
7
Views
890
Replies
23
Views
582
Replies
3
Views
56
Replies
1
Views
92
Replies
25
Views
832
  • Last Post
Replies
32
Views
533
  • Last Post
Replies
11
Views
227
  • Last Post
Replies
16
Views
1K
Top