Prove tan C > b/c in Acute Angled Triangle

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Homework Help Overview

The discussion centers around proving the inequality tan C > b/c in the context of an acute angled triangle. Participants explore various approaches and interpretations related to trigonometric relationships within the triangle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of different expressions involving sine and tangent, questioning the assumptions behind the inequalities. Some attempt to use the sine rule and relate different angles and sides, while others express confusion about the implications of their findings.

Discussion Status

The conversation reflects a mix of attempts to clarify the problem and explore various mathematical relationships. Some participants have provided hints and suggestions, while others express uncertainty about the direction of their reasoning. There is no explicit consensus on the approach to take.

Contextual Notes

Participants note the importance of the triangle being acute angled and discuss specific angle relationships. There are references to previous misunderstandings regarding the full problem statement and the necessity of providing complete information for effective assistance.

ritwik06
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Hi guys!
There is one question after which I have been for hours. I need to prove that:
tan C > b/c
in an acute angled triangle.
I cannot get started.
I tried to convert b/c to sin B/sin C but no useful results. .. and so on.
Please help me as soon as possible!
 
Last edited:
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ritwik06 said:
There is one question after which I have been for hours. I need to prove that:
sin C/ cos B > b/c
in an acute angled triangle.

Hi ritwik06! :smile:

That's obviously not true :frown: …

try it for B = C < 45º.
 
tiny-tim said:
Hi ritwik06! :smile:

That's obviously not true :frown: …

try it for B = C < 45º.

Hey tim,
HI!
Actually, if you read the question carefully, its said that the triangle is acute angled. If B=C<45
=> A>90
 
oops!

ritwik06 said:
Hey tim,
HI!
Actually, if you read the question carefully, its said that the triangle is acute angled. If B=C<45
=> A>90

oops! :redface:

let me try again …

try it for C < 45º and B > 90º - C.

have i got it right this time? :rolleyes:
 


tiny-tim said:
oops! :redface:

let me try again …

try it for C < 45º and B > 90º - C.

have i got it right this time? :rolleyes:

Hey tim,
thanks a lot for trying this. :smile:
The actual thing is that I have to prove that
c tanC / b > 1
can you help me?
 
ritwik06 said:
The actual thing is that I have to prove that
c tanC / b > 1

Use the sine rule! :rolleyes:

What do you get? :smile:
 
tiny-tim said:
Use the sine rule! :rolleyes:

What do you get? :smile:
sin A/a= sin B/ b= sin C/c
But how does this help?
 
ritwik06 said:
sin A/a= sin B/ b= sin C/c

Oh come on! :rolleyes:

That's not using the sine rule. :frown:
 
ritwik06 said:
sin A/a= sin B/ b= sin C/c
But how does this help?

From sine rule I get ;
c tan C/b = (c*c sin B)/ (b*b cos C)

What next??
Did you get the answer sir? I mean were you able to prove it? :smile:
Thanks a lot. But I think my brain's week- I still can't see through it. Please help me further tim.
Ritwik
 
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  • #10
I have noticed that:
B+C>90
B>90-C
as both angles are less than 90
sin B> cos C
But still, I don't get what I want to prove.
 
  • #11
ritwik06 said:
I have noticed that:
B+C>90
B>90-C
as both angles are less than 90
sin B> cos C

Yes, that's good! :smile:

Are you sure the question isn't to prove b tanC / c > 1 ?
 
  • #12
tiny-tim said:
Yes, that's good! :smile:

Are you sure the question isn't to prove b tanC / c > 1 ?

the question is to prove that (c tan C)/b > 1 !
Is it difficult?
 
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  • #13
ritwik06 said:
the question is to prove that (c tan C)/b > 1 !
Is it difficult?

This is the PM you sent me while the PF server was offline:
*_Here is the actual question:_*

http://img510.imageshack.us/img510/8586/diagqp4.jpg

The triangle ABC is a acute angled triangle. The object of the problem is to prove that angle CFD is greater than 45 degree.

*_Here is what I have done:_*

If angle CFD >45
=> tan CFD>1

What I did was to drop a perpendicular from D on CF(say DR).
*The length of this perpendicular is (ac sin C)/ (b+c)*
as BD:DC = b:c

DPC and BFC are similar.
FC=a cos C
FR:FC=b:(b+c)
*FR=(ab cos C)/ (b+c)*

now *tan angle CFD=(c tan C)/(b)*
Thats what I asked you to help me on. That is to prove that *(c tan C)/b >1*

But I wonder, if it can be proved. Say for example: C=30, B=65, A=85 ctanC/b<1 Even though the triangle is perfectly valid.

How is this possible?


I also know one inequality that holds for an acute angled triangle. A+B>90 a>90-B sin A>cos B This is true for all the pairs of angles A,B,C.


Please help me prove this. Its taking me hours. I wonder if it would be helpful but how can I ever prove that tan C> b/c for a acute angled triangle??

Is there any other simple way out to prove that angle CFD>45??

thanks a lot,
Ritwik

Ritwik, in this forum, you should always state the full question. :frown:

The picture, http://img510.imageshack.us/img510/8586/diagqp4.jpg gives information that was not in any of your previous posts.

You have wasted your time and mine by not giving the full question.

Do not ever do that again. :mad:

Hint: draw the line through AG perpendicular to AC and meeting BC at G (so AG is parallel to FB).​
 
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  • #14
tiny-tim said:
This is the PM you sent me while the PF server was offline:

Ritwik, in this forum, you should always state the full question. :frown:

The picture, http://img510.imageshack.us/img510/8586/diagqp4.jpg gives information that was not in any of your previous posts.

You have wasted your time and mine by not giving the full question.

Do not ever do that again. :mad:

Hint: draw the line through AG perpendicular to AC and meeting BC at G (so AG is parallel to FB).​

I am really very very sorry for that. I thought I had reached close enough to the answer, that's why I didn't type the whole question.

Sir, Please be specific about which line are you talking about?
I don't understand this:
Hint: draw the line through AG perpendicular to AC and meeting BC at G (so AG is parallel to FB).

Please re-read your statement. Are you talking about a perpendicular from point D on AC?? If, yes I have done it already. If no, Please specify clearly what constructions do you hint at.

Sorry once again for that!
Ritwik
 
Last edited by a moderator:
  • #15
ritwik06 said:
I am really very very sorry for that. I thought I had reached close enough to the answer, that's why I didn't type the whole question.

Sir, Please be specific about which line are you talking about?
I don't understand this:
Hint: draw the line through AG perpendicular to AC and meeting BC at G (so AG is parallel to FB).

Please re-read your statement. Are you talking about a perpendicular from point D on AC?? If, yes I have done it already. If no, Please specify clearly what constructions do you hint at.

Sorry once again for that!
Ritwik


Hey Tim,
Were you talking about this construction??
http://img147.imageshack.us/img147/1702/diagqp4pq9.jpg
Please help me yar!
 
Last edited by a moderator:
  • #16
ritwik06 said:
Hey Tim,
Were you talking about this construction??
http://img147.imageshack.us/img147/1702/diagqp4pq9.jpg
Please help me yar!

No, that's the line through D parallel to FB.

(it probably works in the end, but it would be complicated)

It is much simpler to draw the line through A parallel to FB.
 
Last edited by a moderator:
  • #17
tiny-tim said:
No, that's the line through D parallel to FB.

(it probably works in the end, but it would be complicated)

It is much simpler to draw the line through A parallel to FB.
SOLVED
Thanks a lot tim for your efforts. I have solved this . Thanks a lot for the help!
regards,
Ritwik
 

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