Need help...lots of help.....to tackle the Evil Maths Problems >.<!

Heldensheld · 2006-09-06 02:30:36

I'm back at this forum...not to have a chat but to ask for help!

Here are the problems.

(1) Solve the system where a, b and c are positive integers.
ab + bc = 44
ac + bc = 23

(2) A rocket is fired vertically above the North Pole. How high must the rocket rise if the horizon to be seen from it is
a) The latitudinal circle of 30o
b) The latitudinal circle of ao o=degrees

(3) In triangle ABC, if angle A has size $ and angle B has size 2$, and the sides BC, CA and AB have lengths a, b and c respectively, show that b^2 = a(a+c).

Solutions would gladly be welcomed !

Kisses for the girls and beers for the guys!

Dross · 2006-09-06 06:03:58

Am at work, so will only answer question 2 for now. First, I direct you to my very crude drawing: .

Now, the north pole lies along the y-axis, and the distance that the rocket has to travel is x. Using basic trig:

The rest is easy. (and remember, cos(90-x) = sin(x)).

Last edited by Dross (2006-09-06 06:05:03)

mathsyperson · 2006-09-06 07:40:37

I'll try the first one.

ab + bc = 44
ac + bc = 23

Take the second equation away from the first: ab - ac = 21
Factorise: a(b-c) = 21.

a and (b-c) must multiply to give 21, which means there are 4 possible values of a: 1, 3, 7 and 21.

By trying these values of a in the original equation, only two solutions can be found, and they are (1,22,1) and (21,2,1).

Dilbert · 2006-09-06 10:57:57

Correct. Can you not solve the first simultaneously too?

BIG_MAN · 2006-09-06 11:49:32

This is heldensheld here at school.

I tried Question 1 but by doing it simultaneously only gets you in circles and bigger circles.

I'll try question 3 but I still need help on it >.<!

Dross · 2006-09-06 22:28:50

With regards to question 3, there is a rule that will give you b[sup]2[/sup] = a[sup]2[/sup] + c[sup]2[/sup] - 2(ac)cos(2$), but at the moment I can't see how to get to the result given. Maybe some wacky formula for cos(2$) that I've forgotten?

(this is only a suggestion because it begins to make it look a little like the result...)

Last edited by Dross (2006-09-06 22:29:34)

mathsyperson · 2006-09-07 01:02:35

cos(2$) = cos² $ - sin² $

From that you can also get that it is identical to 1 - 2sin² $ or 2cos² $ - 1, if either of those are more helpful. I don't see how any of those makes it any simpler, but I might not be looking very hard.

Math Is Fun Forum

#1 2006-09-06 02:30:36

Need help...lots of help.....to tackle the Evil Maths Problems >.<!

#2 2006-09-06 06:03:58

Re: Need help...lots of help.....to tackle the Evil Maths Problems >.<!

#3 2006-09-06 07:40:37

Re: Need help...lots of help.....to tackle the Evil Maths Problems >.<!

#4 2006-09-06 10:57:57

Re: Need help...lots of help.....to tackle the Evil Maths Problems >.<!

#5 2006-09-06 11:49:32

Re: Need help...lots of help.....to tackle the Evil Maths Problems >.<!

#6 2006-09-06 22:28:50

Re: Need help...lots of help.....to tackle the Evil Maths Problems >.<!

#7 2006-09-07 01:02:35

Re: Need help...lots of help.....to tackle the Evil Maths Problems >.<!

Board footer