Circle Geo D: (1 Viewer)

[kazemagic]

OSP=beta equal radii isosceles triangle
Opt= 90 - half alpha
Spt 90 degrees angle in semicircle
Therefore beta= half alpha, complimentary angle, angle sum of triangle,mwhatever
Gamma is 90 minus alpha cuz tangent perpendicular ok. Now
Opt equals 90 minus beta= 90 minus half alpha
Ergo
Apt= 90+ half alpha
Angle PTA= half alpha
Angler sum of triangle PTA
U now no what alpha is now find the other angles by subbing alpha in
NB: going to eat now so cannot correct working for an hour at lest if wrong. Cya guys

O kk thx, but in the end I can't solve alpha cos it gets eliminated lol... i dunno i think i might have made a mistake

 

[Kurosaki]

I got it. Ok u see all try angles are cyclic, yea?
Ota has a 90 degrees angle.therefore OA is a diameter. Therefore
Alpha=90 minus beta
Beta= half alpha
Sub in. Solve.
Sub alpha In to beta and gamma
Done

 

[Kurosaki]

Basically inscribe OTA in a circle and OA is the diameter. With p as the center

 

[Kurosaki]

What's the verdict? It seems right to me...
Edit: suspense is killing me
Second edit: c'mon sy....

 

[kazemagic]

I got it. Ok u see all try angles are cyclic, yea?
Ota has a 90 degrees angle.therefore OA is a diameter. Therefore
Alpha=90 minus beta
Beta= half alpha
Sub in. Solve.
Sub alpha In to beta and gamma
Done

i understand how u get all those but where do you sub it in? in the triangle? but in the end it gets eliminated again doesnt it? O_O

 

[Kurosaki]

Angle pot= angle pto
And I already gave u beta and gamma in trams of alpha in the first post replying to this question
Was my spontaneity too much O__O?

 

[kazemagic]

Angle pot= angle pto
And I already gave u beta and gamma in trams of alpha in the first post replying to this question
Was my spontaneity too much O__O?

i dont understand it but yea thx lol, dw about it ill just ask my teacher , easier to explain irl

 

[Kurosaki]

Mm I gess. Rivalry and sy u guys have been lurking, just tell me, am I right?

Edit: fusty autocorrect screws up my words
Second edit: FUARK!!!
Third edit: my spontaneity better not be too much for u guys 2.....
Furth edit: age quitting! Nah jks continuing tour cya guys later

 

[RivalryofTroll]

I remember being stuck on the very same question a few months ago.

The relationships we form are:

alpha = 2 beta
alpha + gamma = 90
90 - gamma = 2 beta

iirc, alpha was somehow 60.

Thus beta and gamma would be 30 each respectively.

Sorry I can't figure it.

 

[Kurosaki]

Yes i am right

 

[RivalryofTroll]

I think it was intended that PT = PO?

Have no idea, even my teacher couldn't figure it out properly.

 

[kazemagic]

I remember being stuck on the very same question a few months ago.

The relationships we form are:

alpha = 2 beta
alpha + gamma = 90
90 - gamma = 2 beta

iirc, alpha was somehow 60.

Thus beta and gamma would be 30 each respectively.

Sorry I can't figure it.

i think we can all agree on that circle geo is gay

 

[RivalryofTroll]

i think we can all agree on that circle geo is gay

tbh its alright half of the time.

But some questions can be tedious.

but I never liked geometry so hmmmm...

 

[Kurosaki]

i think we can all agree on that circle geo is gay

Nh I ts cool bro. I'll pm u a fully worked solution when this'd ghey tour day is over

 

[Kurosaki]

Kaze post the picture again I celebs scrolling up XD.

 

[kazemagic]

Kaze post the picture again I celebs scrolling up XD.

here ya go

 

[Kurosaki]

how do u do 3a

Ok. Now.inscribef triangle ota in a circle, since ota = 90 degrees( tangent perpendicular)
We can safely say that's OA is a diameter yes?
Now PO=PT
Equal radii of circle
Therefore angle pot equals angled PTo
Therefore alpha= 90 mini s half alpha since we have established that beta is half alpha in my previous work.
Solve for alpha sub in identities for beta and gamma involving alpha to get beta and gamma, which I put in my first post concerning this question . Can't put it any simpler without more time

 

[RivalryofTroll]

Yes, Kurosaki is right.

Construct the circle passing through the vertices O, T and A of the triangle OTA.

Since OTA is right angled, OA will be the diameter and P is the centre.

Therefore, OP = PA = PT. (they're all radii of the constructed circle)

Good job Kurosaki.

 

[Kurosaki]

Yes, Kurosaki is right.

Construct the circle passing through the vertices O, T and A of the triangle OTA.

Since OTA is right angled, OA will be the diameter and P is the centre.

Therefore, OP = PA = PT. (they're all radii of the constructed circle)

Good job Kurosaki.

It only took like 10 mi utes of reading for u to get it down.....yo Kaze is this from Fitzpatrick? It looks similar to o e of the questions in the 3u book I was reading earlier. Or was it terry lee? I wanna no XD.
Thanksx rivalry it is appreciated. Shame my school only offers accel of 2U I would have liked to contest you haha.

 

[RivalryofTroll]

It only took like 10 mi utes of reading for u to get it down.....yo Kaze is this from Fitzpatrick? It looks similar to o e of the questions in the 3u book I was reading earlier. Or was it terry lee? I wanna no XD.
Thanksx rivalry

Its Cambridge.