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Mathematics 10 Online
OpenStudy (anonymous):

Trigonometry??? Find the angle between v and w. Round answer to one decimal place, if nec.? v=-5i+7j,  w=-6i-4j

OpenStudy (shiraz14):

Hint: Use Cosine Rule & Pythagoras Theorem

OpenStudy (anonymous):

@shiraz14 alright so i have the formula cosa= b^2 + c^2 - a^2/2bc, and the pytha theorem, what do i do with it?

OpenStudy (xapproachesinfinity):

use dot product w.v=||w||||v||cos(x) x is the angle we are looking for

OpenStudy (xapproachesinfinity):

cosx=w.v/||w||||v||= ?

OpenStudy (xapproachesinfinity):

what is w.v=? dot product you know how to do it don't you?

OpenStudy (anonymous):

no not really

OpenStudy (xapproachesinfinity):

hmm then forget this lol

OpenStudy (anonymous):

come on, i only have 2 questions left this, and this is one of them lol

OpenStudy (xapproachesinfinity):

let u=<s,t> w=<m,n> the dot product of w and v is define as follows: w.v=sm+tn

OpenStudy (xapproachesinfinity):

in your case s=-5 and t=7 m=-6 and n=-4

OpenStudy (anonymous):

alright i see, its pretty much just plugging them in, so which formula are we using? w.v=sm=tn?

OpenStudy (xapproachesinfinity):

just multiply the components and add them up

OpenStudy (xapproachesinfinity):

hmm formula, am i using any formula? i said definition of dot product was what i mentioned above

OpenStudy (anonymous):

ok so (-5)*(-6)+ 7(-4) which is 30+(-28) so 2

OpenStudy (anonymous):

so w.v = 2 right?

OpenStudy (xapproachesinfinity):

yes! that's one part next if finding ||w|| and ||v|| \[||w||=\sqrt{(-6)^2+(-4)^2}=?\]

OpenStudy (xapproachesinfinity):

do the same to find ||V||

OpenStudy (anonymous):

ok yeah i remember that, ||W||=sqrt(52) ||V||=sqrt ((-5)^2 + 7^2) = sqrt(74)

OpenStudy (xapproachesinfinity):

then next is multiply those two values you get after that you w.v/||w||||v||

OpenStudy (anonymous):

so it would be 2/(the value of ||w|| ||V||) ?

OpenStudy (xapproachesinfinity):

yes!

OpenStudy (anonymous):

ok so 2/(sqrt52) sqrt(74) = 31.01

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