# The sides of a triangle have lengths 4x+1, 2x+1, and 6x-1. If the length of the longest side is 6x-1, what values of x make the triangle obtuse?

meagan18 2011/01/07 00:58:09
 You d I have no idea.
You!
I NEEED HELP

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• Doc Frank stein 2011/01/07 10:19:35 (edited)
You d
As a maths tutor in OZ [though to be honest with you, here in this country, this type of problem is not part of the curriculum, but I've given it my best shot for you ... so I hope it's right], I've had a go at the problem.
The only way I could see how to gain the result is by using Pythagoras' principles of right angled triangles:

HAVE NOW WORKED OUT THE CORRECT ANSWER. USING 2 METHODS, I ARRIVED AT THE SAME - SO SHOULD BE CORRECT => X>1.54
[The 2 Methods: Completing the Square AND the Quadratic Formula]

If you begin by using Pythagoras formula [below] it will set you on the right track.

Pythagoras states that a^2 b^2 = c^2 in a right-angled Tri...
But really, we want the largest angle > than 90 deg, because obtuse ang's > than 90, by def'n.
To achieve this in our equation, the c^2 must be Greater than the SUM of the squares of the other 2 sides, not just EQUAL to them. In diagram, the 6x-1 side has to be > than in the regular equation ... which would mathematically alter the Pythag formula thus:
(6x-1)^2 > (4x 1)^2 (2x 1)^2
[Please Note: Ot is hard to explain this w/o Diagram, adn even my Mum always said I was a Diagram manLOL!...
Basically, we are beginning with a 90 deg tri - using Pythag ... but THEN, if we make the longest side even longer (meaning "greater than" ...

>
>
>>

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>
>

>
As a maths tutor in OZ [though to be honest with you, here in this country, this type of problem is not part of the curriculum, but I've given it my best shot for you ... so I hope it's right], I've had a go at the problem.
The only way I could see how to gain the result is by using Pythagoras' principles of right angled triangles:

HAVE NOW WORKED OUT THE CORRECT ANSWER. USING 2 METHODS, I ARRIVED AT THE SAME - SO SHOULD BE CORRECT => X>1.54
[The 2 Methods: Completing the Square AND the Quadratic Formula]

If you begin by using Pythagoras formula [below] it will set you on the right track.

Pythagoras states that a^2 b^2 = c^2 in a right-angled Tri...
But really, we want the largest angle > than 90 deg, because obtuse ang's > than 90, by def'n.
To achieve this in our equation, the c^2 must be Greater than the SUM of the squares of the other 2 sides, not just EQUAL to them. In diagram, the 6x-1 side has to be > than in the regular equation ... which would mathematically alter the Pythag formula thus:
(6x-1)^2 > (4x 1)^2 (2x 1)^2
[Please Note: Ot is hard to explain this w/o Diagram, adn even my Mum always said I was a Diagram manLOL!...
Basically, we are beginning with a 90 deg tri - using Pythag ... but THEN, if we make the longest side even longer (meaning "greater than" in our equation), we will automatically open up the 90 deg angle to make it 'wider' - this will therefore make this angle "Obtuse" - which is the point of the exercise]

[You may figure it from here but I'll keep working. It's at this point, you may expand and simplify, THEN you may solve by one of the 2 methods mentioned]
Begin by expanding the 3 perfect squares => then it simplifies to
36x^2-12xPLUS1 > 20x^2 PLUS12x PLUS2
=> 16x^2-24x-1 > 0 [a Quadratic equation]

So, using "Quadratic formula": x = [-b PLUS/- sq.rt.(b^2-4ac)] / 2a
FOR SOME REASON, THE SH PROGRAM IS NOT INCLUDING THE "PLUS" SINGS i AM ENTERING ... SO I WILL HAVE TO WORD IN THE PLUS SIGNS - Sorry]

=> x = {24 PLUSsq.rt.(576 PLUS64)} / 32
=> x = 3/4 PLUSsq.rt.(5/8)
=> x = 1.54...approx

Therefore, x > 1.54...
(more)
• FD-Fire... Doc Fra... 2011/01/07 11:04:00
an obtuse triangle cannot be a right one
• Doc Fra... FD-Fire... 2011/01/07 12:29:09
Try my new calculations and see how it goes.
Perseverence, my Friend ;)
• Gun665 2011/01/07 01:44:20
• meagan18 Gun665 2011/01/07 02:10:07
haha tried that
• keeper 2011/01/07 01:34:50
I have no idea.
good luck... but I think "1" works....
• FD-Fire... keeper 2011/01/07 01:37:20
I thought that too actually...
• keeper FD-Fire... 2011/01/07 02:01:07
If you plug in "1", the sides are 5, 5, and 3... that is Obtuse to me...
• FD-Fire... keeper 2011/01/07 11:03:11
but that would be isosceles, not obtuse
• Doc Fra... FD-Fire... 2011/01/07 12:31:13
Yup ... that's right.
I worked it out at top.
• keeper FD-Fire... 2011/01/07 19:05:59
you are right~~
• FD-Firefighter 2011/01/07 01:22:34
You d
have any angle measurements? you can use law of cosines to figure this one out if you have one angle measurement
• Doc Fra... FD-Fire... 2011/01/07 12:32:24
Yeah ,,, maybe ... but we don't have one angle given.
Try Pythagoras
• FD-Fire... Doc Fra... 2011/01/07 23:58:16
actually, you could do it with LOC the way it is
• Doc Fra... FD-Fire... 2011/01/08 04:43:52
Being from OZ, we might use diff terms out here: what is LOC??
Sorry.
• FD-Fire... Doc Fra... 2011/01/08 12:30:52
Law of cosines
• Doc Fra... FD-Fire... 2011/01/08 19:54:45
Thanks very much ... I'll have to give it a go sometime. I think they just call it 'cosine rule' out here. But I believe you still need an angle size somewhere ... so give us a starter of what you had in mind, and we can work from there. There are also no actual dimensions given = so what was your plan?
• FD-Fire... Doc Fra... 2011/01/08 20:01:50
plug in the side values into the loc formula.
c^2 = a^2 + b^2 - 2abcosA , where A is an angle
so set that up with the the sides a,b,c as 2x+1 ect
then solve for the angle in terms of x
that formula can be rewritten
a^2 = b^2 + c^2 - 2bacosB,
b^2 = a^2 + c^2 - 2accosC,
and do the same thing, in each case, solving for the angle in terms of X

we know that A+B+C has to = 180, so plug in the values you got from the above formulas into that, and solve for x
• Doc Fra... FD-Fire... 2011/01/08 21:35:12
Thanks very much for this.
I've got to prepare a Sunday School lesson for today [only partly done] ... so when I get a chance, I'll give it a go => Did u work it all out down to a value answer yourself??
• FD-Fire... Doc Fra... 2011/01/08 22:19:15
no, I haven't, and When I think about it more, I am not so sure that will work.
• Doc Fra... FD-Fire... 2011/01/08 23:58:12
LOL!
Neither am ILOL!
We have no values to work with!HaHa!
Do you think the idea is workable at all??
I think you'd have to really work it carefully through in the mind, because you might be able to use an 'inequality' sign instead of an = somewhere.
• FD-Fire... Doc Fra... 2011/01/09 00:01:23
yeah... hmmmmmmm
• gary-da fuz 2011/01/07 01:20:16
You d
2x+1
• FD-Fire... gary-da... 2011/01/07 01:23:08
I think she wants a value for x
ie
x=2
• gary-da... FD-Fire... 2011/01/07 01:36:33
• BONNIE 2011/01/07 01:02:45
I have no idea.
• Doc Fra... BONNIE 2011/01/07 10:21:29
Now come on, Kitty Kitty ... you're a smart catLOL!
OK ... Ok ... Well, if u can't do it, hand it over to LB and that should make you jealousLOL!
• BONNIE Doc Fra... 2011/01/07 18:04:10
Kitty Kitty was not a math major...LOL
• Doc Fra... BONNIE 2011/01/08 04:45:50
Then she will have to duck her head and hand it over to LB, and hope he doesn't Bark, "I told u so!" ;)
• BONNIE Doc Fra... 2011/01/08 20:12:46
((Smiling))
• Doc Fra... BONNIE 2011/01/08 21:36:43 (edited)
love ur Gorgeous face :)
• BONNIE Doc Fra... 2011/01/08 21:39:53
• Doc Fra... BONNIE 2011/01/08 21:44:14
LOL
Well sometimes we experince great Grief too :(
• BONNIE Doc Fra... 2011/01/08 21:46:37
Now, now, don't get maudlin...it is too early in the day....