Circle Geo D: (1 Viewer)

[SpiralFlex]

Have a good crack at it. Join the lines. I will give you a clue, angles subtending from the same arc.

 

[kazemagic]

Have a good crack at it. Join the lines. I will give you a clue, angles subtending from the same arc.

kk thx ill think about it a bit more 2morro

 

[kazemagic]

Alright heres another question. I found x, but I had to use the sine rule. What is the other method to find x using circle geo? I tried looking for tangent properties but I can't identify any of them in this question

Thanks

 

[nightweaver066]

Is AT a tangent? If so..



Noticing that OTA is rt. angled triangle, apply pythagoras theorem so



Go from there

 

[kazemagic]

Is AT a tangent? If so..



Noticing that OTA is rt. angled triangle, apply pythagoras theorem so



Go from there

Alright thanks, so subbing that in:
(6+x)^2 - (6)^2 = (6+x)x
36 + 12x + x^2 - 36 = 6x + x^2
and i get x = 0

wats wrong with me...

 

[Rezen]

Alright thanks, so subbing that in:
(6+x)^2 - (6)^2 = (6+x)x
36 + 12x + x^2 - 36 = 6x + x^2
and i get x = 0

wats wrong with me...

Nothing, night weaver misquoted the theorem. If you extend the line OA to intersect on the other side of the circle which we label N. Then the theorem is: AT^2 = AM*AN.

EDIT: bonus points: It is actually proven using Pythagoras' theorem

.

 

[kazemagic]

Nothing, night weaver misquoted the theorem. If you extend the line OA to intersect on the other side of the circle which we label N. Then the theorem is: AT^2 = AM*AN.

EDIT: bonus points: It is actually proven using Pythagoras' theorem

.

thx, i tried solving the equation and i still cant solve it... O_O
AM x AN = x(x+12)
OA-OT = x
OA+OT = x+12

So subbing them in:
x(x+12)=x(x+12)

 

[Kurosaki]

Lol

 

[Kurosaki]

Try x(x + 12)=(x+6)^2 - 36
Edit: nvm this is wrong
Second edit: triangle there in the circle is equilateral due to equal radii. Then there's the tangent which is perpendicular to circle. That's all u need? Find angle amt as well

 

[kazemagic]

Try x(x + 12)=(x+6)^2 - 36

x^2 + 12x = x^2 + 12x + 36 - 36
Still unsolvable O_O

 

[Rezen]

thx, i tried solving the equation and i still cant solve it... O_O
AM x AN = x(x+12)
OA-OT = x
OA+OT = x+12

So subbing them in:
x(x+12)=x(x+12)

Thats because your not adding any new information using pythagoras theorem since the theorem already encodes it in a way. Notice that your not using the fact that OT=MT anywhere so it seems likely the answer depends on this condition.

 

[Kurosaki]

Someone check my working. It seems right to me

 

[kazemagic]

Try x(x + 12)=(x+6)^2 - 36
Edit: nvm this is wrong
Second edit: triangle there in the circle is equilateral due to equal radii. Then there's the tangent which is perpendicular to circle. That's all u need? Find angle amt as well

Yea, I got it using sine rule after doing that, but I thought there wud be a proper circle geo method to solve this

 

[Kurosaki]

That's all the circle geo u need and u don't even need the Sine rule its an isosceles triangle lol

 

[kazemagic]

That's all the circle geo u need and u don't even need the Sine rule its an isosceles triangle lol

lol crap i didnt notice it was a isos. triangle

 

[Kurosaki]

Haha no problem bro. Sorry for the slow reply speed, iPod is hard to type on

 

[kazemagic]

Haha no problem bro. Sorry for the slow reply speed, iPod is hard to type on

yea thx la

 

[Kurosaki]

yea thx la

No trouble, I enjoy doing maths and helping people

 

[kazemagic]

how do u do 3a

 

[Kurosaki]

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