# Homework Help: Triangle Inequality

1. Oct 20, 2008

### ritwik06

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.

Last edited: Oct 20, 2008
2. Oct 20, 2008

### tiny-tim

Hi ritwik06!

That's obviously not true

try it for B = C < 45º.

3. Oct 20, 2008

### ritwik06

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

4. Oct 20, 2008

### tiny-tim

oops!

oops!

let me try again …

try it for C < 45º and B > 90º - C.

have i got it right this time?

5. Oct 20, 2008

### ritwik06

Re: oops!

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

6. Oct 21, 2008

### tiny-tim

Use the sine rule!

What do you get?

7. Oct 21, 2008

### ritwik06

sin A/a= sin B/ b= sin C/c
But how does this help????

8. Oct 21, 2008

### tiny-tim

Oh come on!

That's not using the sine rule.

9. Oct 21, 2008

### ritwik06

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?
Thanks a lot. But I think my brain's week- I still cant see through it. Please help me further tim.
Ritwik

Last edited: Oct 21, 2008
10. Oct 21, 2008

### ritwik06

I have noticed that:
B+C>90
B>90-C
as both angles are less than 90
sin B> cos C
But still, I dont get what I want to prove.

11. Oct 21, 2008

### tiny-tim

Yes, that's good!

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

12. Oct 21, 2008

### ritwik06

the question is to prove that (c tan C)/b > 1 !!!!
Is it difficult?

Last edited: Oct 21, 2008
13. Oct 21, 2008

### tiny-tim

This is the PM you sent me while the PF server was offline:
Ritwik, in this forum, you should always state the full question.

The picture, http://img510.imageshack.us/img510/8586/diagqp4.jpg [Broken] 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.

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

Last edited by a moderator: May 3, 2017
14. Oct 21, 2008

### ritwik06

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

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

Sorry once again for that!
Ritwik

Last edited by a moderator: May 3, 2017
15. Oct 22, 2008

### ritwik06

Hey Tim,
http://img147.imageshack.us/img147/1702/diagqp4pq9.jpg [Broken]

Last edited by a moderator: May 3, 2017
16. Oct 22, 2008

### tiny-tim

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: May 3, 2017
17. Oct 22, 2008

### ritwik06

SOLVED
Thanks a lot tim for your efforts. I have solved this . Thanks a lot for the help!
regards,
Ritwik