Circle Geo D: (1 Viewer)

[Kurosaki]

Its Cambridge.

Cambridge is one of my favorites, but it's too big for me to carry lol( i have an old edition)

 

[RivalryofTroll]

Kurosaki, you'll definitely beat me in 3U and 4U. Mainly because I don't devote all my time to the subject and I'm just not gifted at the subject.

If you want a real challenge, try to surpass Realise or something

 

[Kurosaki]

Kurosaki, you'll definitely beat me in 3U and 4U. Mainly because I don't devote all my time to the subject and I'm just not gifted at the subject.

If you want a real challenge, try to surpass Realise or something

Mm ok. But first you are my major obstacle XD. I don't even do that much work on it tbh, I focus more on my English haha. I've heard that u're quite the English wizard ur self (strong HP fan u must be XD)? Thanks for the confidence though XD

 

[RivalryofTroll]

Mm ok. But first you are my major obstacle XD. I don't even do that much work on it tbh, I focus more on my English haha. I've heard that u're quite the English wizard ur self (strong HP fan u must be XD)? Thanks for the confidence though XD

I'm not really a HP fan nor am I a fan of books. My English is alright, for a person who doesn't indulge in books, nothing of brilliance or anything.

Trust me, find yourself a rival and it'll boost your competitive spirit.

Maybe a close friend in the same grade with similar subjects in the next 2 years.

I'm not that great or anything, I'm just someone who only knows how to try hard but doesn't have too much academic potential.

And there is only ONE obstacle, it's yourself

 

[kazemagic]

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

ahh kk thx thats 1 less question to ask my teacher lol

 

[kazemagic]

How do u do this question :/
EDIT: NVM

 

[kazemagic]

How do u find alpha and beta in this question lol

thanks

 

[twinklegal19]

Hint: construct AC and use the angle in alternate segment theorem.

 

[kazemagic]

Hint: construct AC and use the angle in alternate segment theorem.

ahh kk thx

 

[twinklegal19]

What exercise in Cambridge was that from? I still have to do hw from that chapter two terms ago haha (haven't started)

 

[kazemagic]

What exercise in Cambridge was that from? I still have to do hw from that chapter two terms ago haha (haven't started)

yr 12 camb 3u chapter 9e question 3d

 

[Kurosaki]

Kaze is a child prodigy. Is your chem tutor Peak Science btw?

 

[kazemagic]

Kaze is a child prodigy

dat sarcasm

 

[kazemagic]

can someone tell me what type of reasoning proof do I need for a)i) ? I can't seem to find any reasoning for this case :/

Thanks

 

[Aysce]

can someone tell me what type of reasoning proof do I need for a)i) ? I can't seem to find any reasoning for this case :/

Thanks

For a.i.

Notice how you know that angle PTA = angle BTQ (Vertically opposite angles)

Since there are two diameters in the individual circles, we can deduce that angle APT = angle TQB = 90 degrees (90 degrees in a semi-circle)

The remaining angles: angle PAT = angle TBQ since the others are equal.

Therefore, PAT is similar to QBT since they are equiangular.

====================

The angle sign isn't working for me.

EDIT 2: Also include the construction of AP and BQ, my fault.

 

[kazemagic]

For a.i.

Notice how you know that angle PTA = angle BTQ (Vertically opposite angles)

Since there are two diameters in the individual circles, we can deduce that angle APT = angle TQB = 90 degrees (90 degrees in a semi-circle)

The remaining angles: angle PAT = angle TBQ since the others are equal.

Therefore, PAT is similar to QBT since they are equiangular.

====================

The angle sign isn't working for me.

Ohhh thanks aysce

 

[twinklegal19]

can someone tell me what type of reasoning proof do I need for a)i) ? I can't seem to find any reasoning for this case :/

Thanks

angle PTA = angle ATB (vertically opposite angles)

construct PA and BQ and notice that AT and QB are diameters of the circles since they pass through the origin (o)

so angle APT= angle TQ=90 (angle in semicircle)

therefore the triangles are similar (equiangular)

EDIT: damn Aysce beat me to it haha

 

[kazemagic]

lol kaze

angle PTA = angle ATB (vertically opposite angles)

construct PA and BQ and notice that AT and QB are diameters of the circles since they pass through the origin (o)

so angle APT= angle TQ=09

therefore the triangles are similar (equiangular)

yea i kno im blind and stoopid

 

[twinklegal19]

Just keep practicing, you have plenty of time

this time in year 10 I used signpost instead of Cambridge to learn all of circle geo and eventually I could do quite a lot of the Ext 2 problems back then through angle chasing

You'll be fine

 

[kazemagic]

Just keep practicing, you have plenty of time

this time in year 10 I used signpost instead of Cambridge to learn all of circle geo and eventually I could do quite a lot of the Ext 2 problems back then through angle chasing

You'll be fine

if only i realised earlier how precious time is