Obtuse angle triangle

yehoram · 2010-07-28 23:30:37

Abc triangle given:
ac = t
ab = 2t
bc = Kt
Find the values of k for the triangle to be obtuse angle ???

bobbym · 2010-07-28 23:37:38

Hi yehoram;

Do you have a drawing? Capitals are angles and small letters are sides? Your question states triangle Abc, you mean abc?

yehoram · 2010-07-28 23:44:27

No , sorry !

abc triangle given:
let ac = t
and ab = 2t
and bc = Kt

Find the values of k for the triangle to be obtuse angle ???

Last edited by yehoram (2010-07-28 23:44:55)

bobbym · 2010-07-28 23:46:55

Okay those are your sides!

Anakin · 2010-07-28 23:57:03

A guess here, for all k values greater than √5 and less than 3.

That's assuming that t is just a constant (meaning side ac is half as long as side ab). So I just set t=1.

@Bobby, I used the cosine law too and it somehow worked out. I used k^2 = 1^2+2^2 - 2(1*2)cos(x).

(k^2 - 5)/(-4) > -1 and we get -3 < k < 3 but k cannot be negative.
(k^2 - 5)/(-4) > 0 and we get -√5 < k < √5 but once again k cannot be negative.

So I came up with k values greater than √5 and less than 3.

Not entirely sure if that is correct.

Last edited by Anakin (2010-07-29 00:20:00)

bobbym · 2010-07-29 00:08:56

Hi Anakin;

I am making a big mess here with the law of cosines, what are you trying? I am trying to get one angle between 90 and 180 degrees.

Anakin · 2010-07-29 00:24:55

Does that seem correct?

bobbym · 2010-07-29 00:25:17

Hi Anakin;

Looks like you are assuming that k*t is on the left, I am getting some answers with 2 t on the left

Anakin · 2010-07-29 00:31:26

What do you mean on the left? I figured the only angle that could be obtuse in this triangle is the one across from side bc.

This is what I saw it as:

http://i32.tinypic.com/33ljxoy.png

yehoram · 2010-07-29 00:32:13

I think is go like this

Last edited by yehoram (2010-07-29 00:34:42)

bobbym · 2010-07-29 00:40:42

Hi;

I am getting the interval ( - 1 , 3 ) for the second one with ( 2 t )^2 on the left

yehoram · 2010-07-29 00:42:12

The answer is :

Last edited by yehoram (2010-07-29 00:42:43)

Anakin · 2010-07-29 00:43:55

Well, can't we get rid of the t's if they are simply just constants?

So that means
ac = 1
ab = 2
bc = k

Since ONLY the angle across from side bc can be obtuse in this triangle, it would be the reason why we have k^2 on the left side of the equation.

So k^2=1^2+2^2 - 2(1*2) cos (x). For that matter, I think we could simply let t equal a random number like 2 or 4 and it still turns out to be correct.

I still don't see how it could be the second part Yehoram. Because if you say that 1 < k < 3, then that means k can be a number such as 1.1; and that would mean that not a single obtuse angle exists if the three sides are 1.1, 1 and 2.

I believe only the first part of your answer is correct.

Last edited by Anakin (2010-07-29 00:47:21)

bobbym · 2010-07-29 00:47:46

Hi;

For:

( 2 t ) ^2 = 1 + k^2 + 2k * cos(x)

I am getting the open interval ( -√ 3 , 3 ).

Anakin;

I f we are sure it does not equal zero. We can divide through by t.

Anakin · 2010-07-29 00:51:43

But since none of the sides are of zero length, then we can indeed divide by t.

And Bobby, I THINK (I have to be cautious about being wrong lol) the reason why you cannot have 2t^2 on the left side of the equation is because the angle across of it is not the obtuse angle that we are looking for. Only the angle across from (kt)^2 can be obtuse, that is why I put it on the left side.

Last edited by Anakin (2010-07-29 00:52:37)

bobbym · 2010-07-29 00:53:24

Hi Anakin;

Why is that, what is wrong with his diagram that I am missing?

Anakin · 2010-07-29 01:01:33

Perhaps I was wrong.

I drew it out, I guess you were right. Then I suppose there is a set of answers depending on which angle is obtuse?

Last edited by Anakin (2010-07-29 01:05:22)

bobbym · 2010-07-29 01:04:29

Hi Anakin;

In post #10 it does look scaled correctly. I think you are assuming that k t is larger than 2t. That is not necessarily true. Another idea is that I am a dummy.

Anakin · 2010-07-29 01:06:27

No dummy here. Like always, you were right.

However, you said -√3 < x < 3. It cannot be negative, no?

Last edited by Anakin (2010-07-29 01:10:01)

bobbym · 2010-07-29 01:08:22

Why don't we have the same answers he has in post #12?

Anakin · 2010-07-29 01:10:26

Is it because in your equation ( 2 t ) ^2 = 1 + k^2 + 2k * cos(x), should that not be a negative sign before the 2k?

soroban · 2010-07-29 01:13:17

    A *
      *   *    t
   2t *       *
      *           *
    B *   *   *   *   * C
             kt

    C *
      *   *    kt
    t *       *
      *           *
    A *   *   *   *   * B
             2t

. .

    A *
      *   *    2t
    t *       *
      *           *
    C *   *   *   *   * B
             kt

. .

.

bobbym · 2010-07-29 01:15:21

God! I told you I was a dummy!

The answer I get now is the open interval ( -3 , √ 3 )

Okay so we can't have a side that is minus or a side where that is zero.. So now I see it.

Anakin · 2010-07-29 01:21:28

Working with your equation: I solved the inequality -1<(3-k^2)/(-2k)<0 and I get:

-3 < k < -√3
1 < k < √3

The first one cannot be accepted so therefore, the second is correct.

bobbym · 2010-07-29 01:25:06

Yes, I see that no negative or zero sides of course. Forgot to prune out the impossible cases, Like you and soroban did.

Also hurriedly solved the inequality by plugging in -1 and 0, to get a fast answer but that produced extra solutions. So all in all a pretty bad performance.

Math Is Fun Forum

#1 2010-07-28 23:30:37

Obtuse angle triangle

#2 2010-07-28 23:37:38

Re: Obtuse angle triangle

#3 2010-07-28 23:44:27

Re: Obtuse angle triangle

#4 2010-07-28 23:46:55

Re: Obtuse angle triangle

#5 2010-07-28 23:57:03

Re: Obtuse angle triangle

#6 2010-07-29 00:08:56

Re: Obtuse angle triangle

#7 2010-07-29 00:24:55

Re: Obtuse angle triangle

#8 2010-07-29 00:25:17

Re: Obtuse angle triangle

#9 2010-07-29 00:31:26

Re: Obtuse angle triangle

#10 2010-07-29 00:32:13

Re: Obtuse angle triangle

#11 2010-07-29 00:40:42

Re: Obtuse angle triangle

#12 2010-07-29 00:42:12

Re: Obtuse angle triangle

#13 2010-07-29 00:43:55

Re: Obtuse angle triangle

#14 2010-07-29 00:47:46

Re: Obtuse angle triangle

#15 2010-07-29 00:51:43

Re: Obtuse angle triangle

#16 2010-07-29 00:53:24

Re: Obtuse angle triangle

#17 2010-07-29 01:01:33

Re: Obtuse angle triangle

#18 2010-07-29 01:04:29

Re: Obtuse angle triangle

#19 2010-07-29 01:06:27

Re: Obtuse angle triangle

#20 2010-07-29 01:08:22

Re: Obtuse angle triangle

#21 2010-07-29 01:10:26

Re: Obtuse angle triangle

#22 2010-07-29 01:13:17

Re: Obtuse angle triangle

#23 2010-07-29 01:15:21

Re: Obtuse angle triangle

#24 2010-07-29 01:21:28

Re: Obtuse angle triangle

#25 2010-07-29 01:25:06

Re: Obtuse angle triangle

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