Vectors, on a circular arc

[lost gust]

I got the diagram, not the rest

[pulsar trout]

<@&1227988399579730072>

Note for OP

+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.

[undone lodge]

Yeah, the answer is right there. Take it as complex numbers or vectors.

Let OA be x component, OB be y component

[lost gust]

also, slapping a modulus on the given OC condition gives some weird 1 = alpha^2 + beta^2 + alpha*beta*sqrt(sqrt(3) + 2)

[undone lodge]

Don't do all that.

[lost gust]

hm?

of what

[undone lodge]

Oh wait nvm

It's AC 🤦‍♂️

[barren owl]

Well, you can do it by vectors or ...transformations?

[undone lodge]

Ok, take OA and OC as basis vectors, and express OB.
Then swap stuff around to get OB

[undone lodge]

2D vectors.

Just taking OA = (1,0) and OC = (0,1) works out

[lost gust]

ah

taking examples

genuinely good idea

[undone lodge]

No, just taking basis vectors lol

[lost gust]

lemme try this

[undone lodge]

OB = (cos75°, sin75°)
2√2OB = (√3+1,√3-1)

[lost gust]

ah brilliant nice

i got it

wtf have they done

[undone lodge]

Tf?

[lost gust]

yea 💀

[undone lodge]

Why dot products?

[lost gust]

not a clue

well i will safely be ignoring that monstrosity

thanks yall

[undone lodge]

Lol

Do that

[lost gust]

+solved Opt SirLancelotDuLac