Prove tan C > b/c in Acute Angled Triangle

In summary, in an acute angled triangle, the tangent of the angle C is always greater than the ratio of the length of the side opposite to angle C to the length of the hypotenuse. This can be proven using the Pythagorean theorem and the definition of tangent as the opposite side divided by the adjacent side. Therefore, in an acute angled triangle, the tangent of any angle will always be greater than the ratio of the opposite side to the hypotenuse.
  • #1
ritwik06
580
0
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:
Physics news on Phys.org
  • #2
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.
 
  • #3
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
 
  • #4
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:
 
  • #5


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?
 
  • #6
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:
 
  • #7
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?
 
  • #8
ritwik06 said:
sin A/a= sin B/ b= sin C/c

Oh come on! :rolleyes:

That's not using the sine rule. :frown:
 
  • #9
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
 
Last edited:
  • #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?
 
Last edited:
  • #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).​
 
Last edited by a moderator:
  • #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
 

What is an acute angled triangle?

An acute angled triangle is a triangle where all three angles are less than 90 degrees.

What is the definition of tan?

Tan, or tangent, is a trigonometric function that represents the ratio of the opposite side to the adjacent side in a right triangle.

Why is tan C greater than b/c in an acute angled triangle?

In an acute angled triangle, the opposite side (b) is always shorter than the hypotenuse (c). Since tan C is defined as the ratio of the opposite side to the adjacent side, and the adjacent side (c) is always longer than the opposite side (b), tan C will always be greater than b/c in an acute angled triangle.

How can we prove that tan C > b/c in an acute angled triangle?

We can use the Pythagorean theorem to prove this statement. By rearranging the equation c^2 = a^2 + b^2, we get a^2 = c^2 - b^2. Then, substituting this into the definition of tan C (opposite/adjacent), we get tan C = b/(c^2 - b^2). Since c^2 - b^2 is always positive in an acute angled triangle, we can see that tan C will always be greater than b/c.

Is this statement always true for any acute angled triangle?

Yes, this statement is always true for any acute angled triangle because the definition of tan only depends on the angles of the triangle, not the specific lengths of the sides. As long as the triangle is acute angled, the ratio of the opposite side to the adjacent side will always be greater than the ratio of the opposite side to the hypotenuse.

Similar threads

Finding area of a non right angled triangle
    • Precalculus Mathematics Homework Help
Replies
9
Views
2K
Why is this result different? (calculating the sides of a triangle)
    • Precalculus Mathematics Homework Help
Replies
6
Views
588
Finding Angle C in an ABC Triangle
    • Precalculus Mathematics Homework Help
Replies
19
Views
2K
Solve 2 Triangle Problem: Calculate Exact Value of x
    • Precalculus Mathematics Homework Help
Replies
7
Views
1K
Prove the sums of the angles of any given triangle = 180°
    • Precalculus Mathematics Homework Help
Replies
12
Views
2K
Proving |a|=|-a|: Using Cases and Triangle Inequality"
    • Precalculus Mathematics Homework Help
Replies
6
Views
2K
I Exploring the Uses of Triangle Inequality
    • General Math
Replies
2
Views
1K
MHB Proving Sum of Tangents $\ge$ Sum of Cotangents in Acute Triangle
    • General Math
Replies
1
Views
713
MHB Proving $\dfrac{3}{2}$ Inequality in Acute Triangle $ABC$
    • General Math
Replies
1
Views
582
Combinatorics: Counting quadrilaterals in triangle pattern
    • Precalculus Mathematics Homework Help
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top