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So, this gives:\[\cos \theta=\frac{-3+48}{\sqrt{7}\sqrt{73}}\]\[\theta \approx75.15 \]degrees.

You should be able to get the area of the triangle POQ now

|dw:1336587860891:dw|

let me know what you have so far

ok the page keeps refreshing thats why my replies are late

yeah sometimes this site chugs along bad

ah crap I made an error. the length of vector p should be sqrt(1+36)=sqrt(37)

ok im still working on the question

So the angle should be about 30 degrees

I got to get ready for work...good luck!

ok thanks so far bye