Here's another algebra problem

GurraTedden · 2005-07-20 05:01:43

Hi, thought that i might just put up all my problems in here, to hope
for the best. Well, i have other problems too, but to get rid of one or
two of them would make my day!
Just check this link out, for the problem:

http://engman.bravehost.com/Jobb/algebra2.JPG

/Gustav

mathsyperson · 2005-07-20 05:35:07

I've got an answer and it should take around 5 minutes to type up, so don't go away!

Last edited by mathsyperson (2005-07-20 05:55:34)

GurraTedden · 2005-07-20 05:46:24

Hi! Thanks so much for the fast intrest. I mean, this is so great.
Hate having to sit by my self with this kind of lousy problems.
Actually this is a part of a quite simple mechanical problem, and
I'm almost 100 percent sure of that the way they mean, is to be
able to calculate that angle. But, it doesn't seem like that should
be a big problem. I mean, calculating that angle shouldn't be the
head task of the exercise. So, I hope that your answer is not too
advanced, even thought I will appreciate it equally much, my friend!
/Gustav

mathsyperson · 2005-07-20 05:57:30

I had a much more complicated way of working this out, involving advanced trigonometry and lots of other horrible stuff, but while I was typing it up I suddenly realised there was a far easier way, so that's why I'm a bit delayed.

Start by extending the bottom of the triangle towards the red line to make a bigger, right-angled triangle. Then take away the triangle that contains the angle that you want, leaving you a smaller right-angled triangle. We know that the left side of this is 210m, because we are told that part of this is 120m and the rest of it is 90m, and we know that its bottom side is 70m, because looking at the measurements of 30 and 90 tell you that the hypotenuse goes up 3 times as much as it goes across.

Using trigonometry, we can work out that the angle next to your wanted angle is tan^-1(210/70), which is ∼71.6°. As angles on a straight line add up to 180°, the final step is to take 71.6° from 180° to give your answer: ∼108.4°.
If you want the exact answer, it's 180- tan^-1(3).

I'm interested in your other problem too, but I'm a bit exhausted after doing this one so I'll do it later.

P.S. To MathsIsFun, can you add little -1 to the list of symbols?

GurraTedden · 2005-07-20 06:39:59

Ahh! An easy solution, just what i was hoping to be buried somewhere
under. And the mechanical excercise is now solved, thank
to your algebraic consultancy. A big thanks to you!

MathsIsFun · 2005-07-20 09:47:40

THIS is way cool !

WELL done, mathsy, and I hope you are right, too!

(Regarding symbols, I don't think there is a special "-1" defined, but there *may* be some kludge to make a little "-")

Math Is Fun Forum

#1 2005-07-20 05:01:43

Here's another algebra problem

#2 2005-07-20 05:35:07

Re: Here's another algebra problem

#3 2005-07-20 05:46:24

Re: Here's another algebra problem

#4 2005-07-20 05:57:30

Re: Here's another algebra problem

#5 2005-07-20 06:39:59

Re: Here's another algebra problem

#6 2005-07-20 09:47:40

Re: Here's another algebra problem

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