vectors help pls (1 Viewer)

sarbear.h

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theres sm questions idk how to do.. ill paste them below:
1744198401804.png1744198414749.png1744198429428.png
 

Scrambled

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Heya I think I’ve got the first two at least but not too sure bout the last :/IMG_4799.pngIMG_4800.jpeg
 

C2H6O

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I don’t have anywhere to work this out rn but for the third one maybe you’d have a circle with radius 4root5 centred at A, and C can be on the circle. Then use the side length of AB and AC and the area formula 1/2 a b sin C to find the angle BAC and where on the circle point C is (there should be 2) then use trig to find the 2 possible values of AC
 

C2H6O

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I don’t have anywhere to work this out rn but for the third one maybe you’d have a circle with radius 4root5 centred at A, and C can be on the circle. Then use the side length of AB and AC and the area formula 1/2 a b sin C to find the angle BAC and where on the circle point C is (there should be 2) then use trig to find the 2 possible values of AC
Actually I haven’t done any working there might be 4 total solutions, 1 acute angle, 1 obtuse angle, and then the same angles flipped iykwim
 

alphxreturns

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can't pm images :(

Screenshot 2025-04-10 at 9.19.52 am.png

1744240803150.png

ok we use these, as well as the point distance formula (for 4√5)

so you get x^2 + y^2 = 80 (from point distance formula) ----- (1)

and ±10 = 3y-x --------(2)

2 cases with each form of (2)

which, when subbed into (1), gets a quadratic that yields another 2 solutions each (from the ± in the quadratic formula)

and boom you got 4 solutions just like C2H60 said


** not sure if this is the right approach but there ya go... i didn't know those identities at the top existed before this LOL forgot everything vectors already
 

sarbear.h

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View attachment 47258

View attachment 47259

ok we use these, as well as the point distance formula (for 4√5)

so you get x^2 + y^2 = 80 (from point distance formula) ----- (1)

and ±10 = 3y-x --------(2)

2 cases with each form of (2)

which, when subbed into (1), gets a quadratic that yields another 2 solutions each (from the ± in the quadratic formula)

and boom you got 4 solutions just like C2H60 said


** not sure if this is the right approach but there ya go... i didn't know those identities at the top existed before this LOL forgot everything vectors already
is v1 being the x component and v2 the y component


(i meant like send a google drive link to ur pic lol)
 

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