# Applying the Law of Cosines

• Dec 4th 2010, 10:21 AM
Ylani
Applying the Law of Cosines
This is supposed to be an easy problem that uses the law of cosines, but I must be overlooking something - can someone please give me a hint?

The problem: Point O is located at (0,0). Point A is located at (8,6). Point P (x,10) is moving along the line y = 10. Find the positive value of x when angle OPA is 60 degrees.

So far I have a = OP = [tex]\sqrt{x^2 + 100}[\math]
b = PA = [tex]\sqrt{4^2+(8 - x)^2}[\math]
100 = [tex] a^2 + b^2 - ab [\math]
Unfortunately this gives a 4th degree equation and no graphing calculators are allowed to be used for this problem. Using a graphing calculator I find that x = 5.634. If I use the tangent angle addition formula I also get x = 5.634, but this needs to be solved using the law of cosines. What am I missing here, and why can't I get LaTeX to work?
• Dec 4th 2010, 10:41 AM
pickslides
Somwtimes 4th order equations an be solved without technology,

I get

$10 = x^2+100+(8-x^2)^2+16-(x^2+100)((8-x)^2+16)$

Have you tried to exapnd this out completely? What did you get?
• Dec 4th 2010, 11:03 AM
Ylani
I get [LaTeX ERROR: Convert failed] , which gives x = 5.634 as a positive solution. The teacher said no graphing calculators, and the solution to the problem is really easy - you just use the law of cosines. I did not speak with the teacher; I am just trying to help the student.
• Dec 4th 2010, 11:35 AM
rtblue
If you want your latex to work Ylani, you should change the direction of the slash that ends MATH
• Dec 4th 2010, 12:59 PM
Soroban
Hello, Ylani!

Who said this was "an easy problem"?

Quote:

$\text{Point }O\text{ is located at }(0,0);\;\text{point }A\text{ is located at }(8,6).$

$\text{Point }P(x,10)\text{ is moving along the line }y = 10.$

$\text{Find the positive value of }x\text{ when }\angle OPA\,=\,60^o$

. . $OP \:=\:\sqrt{x^2 + 10^2}$

. . $AP \:=\:\sqrt{(x-8)^2+4^2} \:=\:\sqrt{x^2 - 16x + 80}$

. . $OA \:=\:\sqrt{8^2+6^2} \:=\:10$

$\cos 60^o \;=\;\dfrac{OP^2 + AP^2 - OA^2}{2(OP)(AP)} \;=\;\dfrac{(x^2+100) + (x^2 - 16x + 80) - 100}{2(\sqrt{x^2+100})(\sqrt{x^2-16x + 80})} \;=\;\dfrac{1}{2}$

Then: . $2x^2 - 16x + 80 \:=\:\sqrt{(x^2+100)(x^2-16x + 80)}$

. . which simplifies to: . $3x^4 - 48x^3 - 1396x^2 - 960x - 1600 \:=\:0$

Now what?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way, close your latex tags with [/math].

You were using the backslash, \

• Dec 4th 2010, 03:39 PM
Ylani
Thanks for the LateX help - I'll do better on my next post (Happy).

I told the girl to ask her teacher for help, and she e-mailed me today that the teacher just said it uses the law of cosines and it's really simple. That made me wonder what I was missing. This student misses a lot of school due to a life-threatening illness, so maybe the teacher lifted her ban on graphing calculators for this worksheet and she didn't know? She is in an IB program, so everything is supposed to be quite hard.

For the past two days I have been looking at this problem trying to find some simpler way of doing it. Maybe something to do with the geometry of it?