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March 29, 2024, 10:17:01 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164637 times)  Share 

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TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9555 on: December 22, 2019, 02:51:43 pm »
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If you can find which quadrant the angle, lies in then you can construct a right angled triangle in the unit circle and then use standard trigonometric ratios to prove the identities. (Hint: I recommend drawing a diagram to prove the identity by the way.

Hopefully this is helpful! :)

Thank you! I did draw one but I cant seem to prove these 2. Khanacademy broke the diagram down but I cant seem to verify these two identities

jnlfs2010

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9556 on: December 26, 2019, 11:13:11 am »
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Thank you! I did draw one but I cant seem to prove these 2. Khanacademy broke the diagram down but I cant seem to verify these two identities

Easiest you can do is to use the double angle for sine and cosine to prove the identities
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TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9557 on: December 30, 2019, 07:24:45 pm »
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Hey spesh heads

I am doing some spesh topics that I enjoy over the holidays and don't know how to break down part b, and have gotten x=4 for part a whereas the answers says x=3.
Part A) via similar triangles,  6/3 = 8/x   therefore x=4   (is there an error in the answer?)
 
Part B) ???

Part C) via similar triangles, 3/x= 2/8   therefore x=12 (correct)

RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9558 on: December 30, 2019, 08:30:34 pm »
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Hey spesh heads

I am doing some spesh topics that I enjoy over the holidays and don't know how to break down part b, and have gotten x=4 for part a whereas the answers says x=3.
Part A) via similar triangles,  6/3 = 8/x   therefore x=4   (is there an error in the answer?)
 
Part B) ???

Part C) via similar triangles, 3/x= 2/8   therefore x=12 (correct)
Is there supposed to be an attachment here?

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9559 on: December 30, 2019, 08:50:01 pm »
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Is there supposed to be an attachment here?

Sorry I forgot to add it

I've now attached it

RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9560 on: December 30, 2019, 09:19:55 pm »
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I had a glance at the study design and it seems that for Units 1&2 you're allowed to use the following results in circle geometry. You should be using these results here relating to intersecting chords and intersecting secants.
 

Although if you wish to do things bare-hand and use similar triangles:

a) seems fine. I don't know why \( x=4\) is not the solution to \( \frac63 = \frac8x\).

b) would require you to identify that \( \triangle ACD  \) is similar to \(\triangle PCB\). This then gives \( \frac{AC}{PC} = \frac{BC}{DC} \). Note that the similarity requires the result that opposite angles of a cyclic quadrilateral are supplementary.

cotangent

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9561 on: December 30, 2019, 09:21:29 pm »
+1
Sorry I forgot to add it

I've now attached it

Pretty sure ur answer for part a is correct. And for part b, you need to use one of the circle theorems. I have attached a photo of it. So you should end up with the expression "2x" times "x" = (12+3) times 3, so u should get x = plus or minus (3 root 10)/2. but since x>0, x = positive (3 root 10)/2
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TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9562 on: December 30, 2019, 10:05:46 pm »
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Pretty sure ur answer for part a is correct. And for part b, you need to use one of the circle theorems. I have attached a photo of it. So you should end up with the expression "2x" times "x" = (12+3) times 3, so u should get x = plus or minus (3 root 10)/2. but since x>0, x = positive (3 root 10)/2

Ohh! Thanks so much. You reckon I should know how to prove part b?


I had a glance at the study design and it seems that for Units 1&2 you're allowed to use the following results in circle geometry. You should be using these results here relating to intersecting chords and intersecting secants.
 

Although if you wish to do things bare-hand and use similar triangles:

a) seems fine. I don't know why \( x=4\) is not the solution to \( \frac63 = \frac8x\).

b) would require you to identify that \( \triangle ACD  \) is similar to \(\triangle PCB\). This then gives \( \frac{AC}{PC} = \frac{BC}{DC} \). Note that the similarity requires the result that opposite angles of a cyclic quadrilateral are supplementary.

Thanks for the breakdown! :) I don't quite understand how triangle ACD is similar to PCB?

RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9563 on: December 30, 2019, 10:17:08 pm »
+1
Thanks for the breakdown! :) I don't quite understand how triangle ACD is similar to PCB?
See this bit:
Note that the similarity requires the result that opposite angles of a cyclic quadrilateral are supplementary.
The cyclic quadrilateral you must consider is \(ADBP\). A pair of supplementary angles would be \(\angle ABP\) and \(\angle ADP\), but think about what other angle \(\angle ABP\) is supplementary to.

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9564 on: December 30, 2019, 10:35:18 pm »
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I don't understand :(

cotangent

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9565 on: December 30, 2019, 10:58:50 pm »
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I don't understand :(

I don't think we would get questions about proving circle theorems in the spesh course, the only thing we are gonna be asked to do is to apply them to other qs. So don't worry. I don't think the derivation of the theorems is that important.
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TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9566 on: December 30, 2019, 11:08:06 pm »
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I don't think we would get questions about proving circle theorems in the spesh course, the only thing we are gonna be asked to do is to apply them to other qs. So don't worry. I don't think the derivation of the theorems is that important.

Thanks so much!!! Both legends :)

RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9567 on: December 30, 2019, 11:20:38 pm »
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I don't understand :(
\( \angle ABP = 180^\circ - \angle ADP\) from opposite angles of a cyclic quadrilateral.
\(\angle ABP = 180^\circ -\angle PBC\) from adjacent angles on a straight angle.

Hence \(\angle ADP = \angle PBC\). But this also means that \(\angle ADC = \angle PBC\).

Now in \(\triangle ADC\) and \(\triangle BPC\),
- \(\angle ADC = \angle PBC\) as established above.
- \(\angle ACD = \angle PCB\) as this is just common \(\angle A\).

So indeed, \(\triangle ADC\) and \(\triangle BPC\) are similar, by the equiangular test.

Sine

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9568 on: December 30, 2019, 11:35:39 pm »
+2
I don't think we would get questions about proving circle theorems in the spesh course, the only thing we are gonna be asked to do is to apply them to other qs. So don't worry. I don't think the derivation of the theorems is that important.
Whilst I admit it is not likely you will get a straightup circle theorem to prove (you may have to do it parts where the question guides you) it is worthwhile learning the circle theorems.

e.g. the study design states
"vector proofs of simple geometric results, for example the diagonals of a rhombus are perpendicular, the
medians of a triangle are concurrent, the angle subtended by a diameter in a circle is a right angle."


sarah15

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9569 on: December 31, 2019, 08:36:16 am »
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Hello
Could someone please explain how to graph f(x) = x+1 / x^2-4 tech free?
I was trying to divide it using long division but I am unsure what to do.
Thank you :)