crappy circular geometry! (1 Viewer)

[freaking_out]

i haven't done circular geometry and my teacher expects me to know from year 10. and i have my half yearly this friday!!!!!

if the above image doesn't load click on this link http://www.geocities.com/gamez_freak_4_eva/question.jpg

thanx in advanced

 

[InfiniteQ]

That link doesn't work.

 

[freaking_out]

yeah the link does work, i just clicked on it right now and it worked.

 

[Huy]

geocities doesn't allow hotlinking
so you'll have to copy the link into clipboard and paste it from there into your browser,

woohoo 1 mark for copying the diagram

i'm pretty crumby with circle geo.
i tend to fudge it alot

but we had a 3u Q similar to this one on tuesday,

i'm just guessing here,
but <PBC = <PCB (base angles of isos. triangle) for i)
can't really prove it though,

the tangent PT passing through the center (say O) bisects line BC. (using the chord theorem, the perpendicular from the centre of a circle to a chord bisects the chord)

you'd probably have to first prove that PT is parralel to BC (well that's given), then you'd say that

<TPC = < PBC (alternate segment theorem)
<TPC = <PCB (alternate angles, PT || BC)

therefore < PBC = < PCB (base angles of isos. triangle)

im not too sure though

but for ii) you use the tangent and secant theorem,
the square of the tangent is equal to the product of the whole secant multiplied by the part of it outside the circle.

so PB^2 = PA . PC?

i have no idea!
it has something to do with equal angles in the same segment or the tangent/secant theorem

but as you can see, maths is not my forte
i'll stick to what i know

 

[freaking_out]

Originally posted by Huy

but for ii) you use the tangent and secant theorem,
the square of the tangent is equal to the product of the whole secant multiplied by the part of it outside the circle.

so PB^2 = PA . PC?

i have no idea!
it has something to do with equal angles in the same segment or the tangent/secant theorem

but as you can see, maths is not my forte
i'll stick to what i know

yeah i tried using those theorems but it doesn't work coz PB is ot a tangent so i got confushed!!!! (shit i am dead for the test tommorow!!)

btw any tips for circular geometry solving anyone?????

 

[Huy]

for i)

it's one of those questions where you'd need a two-step approach to get the final answer.

this is what i've got so far:

i flipped the picture upside down so that it would make it easier for me to understand. (get a pen/paper out so it would be easier to understand my working )

PT is tangent to circle,
draw a dot in the middle of the circle to indicate center, O
draw a diameter from the point P, passing through O to another point called F.

join the line FB (from the end of F, the point on the 'other' side of P passing through the center to B)

let < TPB = x

. ' . < FPB = 90 - x
(as the radius meets the tangent at right angles)

Now < FBP = 90
(angle in a semi circle)

.'. < PFB = x
.'. < PAB = x
(angles in the same segment are equal)

Since < BPT = x and < PAB = x,
.'. < BPT = < PAB

you could prove < APT = < PBA in the same way,

once you've done that...

you can say that
since < APT = < PBA,
then < TPC = < PBC because it's an alternate angle, PT || BC

that's how i'd do it
but i'm kind of messed up with trying to get the angle at A to equal the angle < TPA

i've done the 'reverse' of it, which works in solving part I

as for II, i can only think of the tangent and secant theorem, which is what it looks like (same form, R.T.P 'PT^2 = PS.AP depending on what textbook you use)

 

[spice girl]

"PB^2 = PA . PC"

Whenever you see a ratio like this, think SIMILAR TRIANGLES. If you notice, the tangent - secant theorem is just a result of similar triangles.

Now a general rule for math extension questions: use the previous result if you can.

Given <PBA = <PCB (the proof of this is just alternate segment theorem to get <APT = <PBA, and then alternate angles equal for parallel lines to get <APT = <PCB)

We have similar triangles PBA and PCB since:
* angle P is common,
* <PBA = <PCB

Thus PA/PB = PB/PC (ratio of corresponding sides of similar triangles are equal)

cross-multiply to get PB^2 = PA . PC

 

[wogboy]

You can quite easily prove <PBA = <PCB, by using the following arguments:

PT || BC (data)

so angle PCB = angle CPT (alternate angles) ...(A)

since PT is a tangent to the circle,

angle PBA = angle CPT (angles in the alternate segment are equal) ....(B)

so from (A) and (B),

angle PBA = angle PCB

To prove the last part, (PB)^2 = PA * PC, you just need to use similar triangles as spice girl said, and that should work just fine with a little bit of algebraic manipulation.

 

[wogboy]

Oh and BTW, you might wanna use Fortunecity in the future for uploading math problem diagrams. They support remote linking.

www.fortunecity.com

 

[PG5]

u know what?
that's from the 2000 HSC Ext 1 exam...

i remember coz i did this question from the maths genie that i bought, to study for my half yearlies a few days ago

and yeah, there is no given centre O so you would have to use wogboy's answer for ii) and spice girl's answer for iii)

 

[m3.4..012]

Originally posted by freaking_out

i haven't done circular geometry and my teacher expects me to know from year 10. and i have my half yearly this friday!!!!!

if the above image doesn't load click on this link http://www.geocities.com/gamez_freak_4_eva/question.jpg

thanx in advanced

Yes I know what you mean.. My teacher does the same thing.
Expects me to know all these new information by the next lesson >.<*