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### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1174783 times) Tweet Share

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#### dc302

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##### Re: Specialist 3/4 Question Thread!
« Reply #30 on: December 05, 2011, 08:19:03 pm »
+2
yeahh thats right haha,
hence 'im stuck' :p

I know i shouldnt have even gone there..., so do you think it was something to do with not converting tanx to sin/cos ?
or just some arithmetic?

Cancelling is a big nono
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#### Bhootnike

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##### Re: Specialist 3/4 Question Thread!
« Reply #31 on: December 05, 2011, 08:21:29 pm »
0
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, $x=\pi$ or $x=2\pi$  or no solution

So x=0, $\pi$, $2\pi$

EDIT: forgot sols

legend! thanks bro!

just ah... wouldnt:
-8-4tan2(x)=0
-4tan2(x)=0 +8
tan2(x)= 8 /-4
tan2(x)=-2 ?

so yeah no solution because u cant take the sq.root of a negative number anyways!

And thanks alot guys, will remember this for next time!
+1

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#### b^3

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##### Re: Specialist 3/4 Question Thread!
« Reply #32 on: December 05, 2011, 08:24:32 pm »
+1
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, $x=\pi$ or $x=2\pi$  or no solution

So x=0, $\pi$, $2\pi$

EDIT: forgot sols

legend! thanks bro!

just ah... wouldnt:
-8-4tan2(x)=0
-4tan2(x)=0 +8
tan2(x)= 8 /-4
tan2(x)=-2 ?

so yeah no solution because u cant take the sq.root of a negative number anyways!

And thanks alot guys, will remember this for next time!
+1

YES! I did it in my head, and didn't type it, I've lost it already . Stupid holidays :|
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#### Bhootnike

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##### Re: Specialist 3/4 Question Thread!
« Reply #33 on: December 05, 2011, 08:26:27 pm »
0
ok sweet, atleast im getting something right ahaha... yaya.

thanks again guyss
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#### Bhootnike

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##### Re: Specialist 3/4 Question Thread!
« Reply #34 on: December 06, 2011, 07:43:41 pm »
0
Was wondering what other people's approach to this question is :

Use vector methods to prove that the midpoint of the hypotenuse of a right-angled triangle
is equidistant from all vertices.
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#### TrueTears

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##### Re: Specialist 3/4 Question Thread!
« Reply #35 on: December 06, 2011, 07:48:53 pm »
+1
let one side (adjacent to the right angle) be denoted by vector b, other side (the other side which is adjacent to the right angle) be denoted vector a, the hypotenuse is given by a-b, half of this vector is given by 1/2(a-b), from the right angle vertex to the midpoint; this vector is given by b+1/2(a-b) = 1/2(a+b)

compute the magnitude, the magnitude are also the same hence equidistant.
« Last Edit: December 06, 2011, 10:31:21 pm by TrueTears »
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#### kamil9876

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##### Re: Specialist 3/4 Question Thread!
« Reply #36 on: December 06, 2011, 10:10:07 pm »
+1
TT, $\frac{1}{2}(a-b)+b=\frac{1}{2}(a+b)$. Your attempt doesn't use the fact that it's a right angle triangle so that alerted me.

What you should now do is compute the length:

$||\frac{1}{2}(a+b)||^2=\frac{1}{2}(a+b).\frac{1}{2}(a+b)=\frac{1}{4}(a.a+2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Likewise we can show that $||\frac{1}{2}(a-b)||^2=\frac{1}{4}(a.a-2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Hence the magnitudes are indeed the same (notice this is false for certain triangles that are not right angled e.g: an iscosceles triangle with two very long equal sides and one very short side, this is a good way to check your proof, does it use all the assumptions?)
« Last Edit: December 06, 2011, 10:12:04 pm by kamil9876 »
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#### TrueTears

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##### Re: Specialist 3/4 Question Thread!
« Reply #37 on: December 06, 2011, 10:28:54 pm »
0
TT, $\frac{1}{2}(a-b)+b=\frac{1}{2}(a+b)$. Your attempt doesn't use the fact that it's a right angle triangle so that alerted me.

What you should now do is compute the length:

$||\frac{1}{2}(a+b)||^2=\frac{1}{2}(a+b).\frac{1}{2}(a+b)=\frac{1}{4}(a.a+2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Likewise we can show that $||\frac{1}{2}(a-b)||^2=\frac{1}{4}(a.a-2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Hence the magnitudes are indeed the same (notice this is false for certain triangles that are not right angled e.g: an iscosceles triangle with two very long equal sides and one very short side, this is a good way to check your proof, does it use all the assumptions?)
thats right, i got lazy and didnt end up computing the lengths lol but yeah in my head using a.b = 0 was quite obvious from the start
« Last Edit: December 06, 2011, 10:32:11 pm by TrueTears »
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#### Bhootnike

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##### Re: Specialist 3/4 Question Thread!
« Reply #38 on: December 06, 2011, 11:00:35 pm »
0
I don't get why the hypotenuse is a-b?
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#### dc302

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##### Re: Specialist 3/4 Question Thread!
« Reply #39 on: December 06, 2011, 11:41:24 pm »
+1
I don't get why the hypotenuse is a-b?

It's just a matter of 'how' you draw the triangle. I think the triangle is supposed to use the positive x, y axes, with the hypotenuse slanting down from the positive y axis to the positive x axis. Here I'm only saying x and y axes for illustrative purposes.

edit: actually a better visualisation: if vector b goes from origin to point B, and vector a goes from origin to point A, and AOB is the right angle, then the hypotenuse is from B to A, which is given by a-b. Draw it!
« Last Edit: December 06, 2011, 11:43:37 pm by dc302 »
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#### Planck's constant

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##### Re: Specialist 3/4 Question Thread!
« Reply #40 on: December 06, 2011, 11:57:47 pm »
+1
Easier to see why the hypotenuse = a - b, if you imagine the vertex corresponding to the right angle as the origin, O.
Therefore, the other two vertices will be represented by their position vectors,

OA = a
OB = b

And the hypotenuse, AB becomes,

BO + OA =
(-OB) + OA =
-b + a =
a - b

#### Bhootnike

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##### Re: Specialist 3/4 Question Thread!
« Reply #41 on: December 07, 2011, 12:16:33 am »
0
Thanks dc302,  making sense

Thanks to you too argonaut! Just checking, if you're going by the same diagram as dc302,  you've said ab = a-b,  but isn't that ba?  Ab would be - a + b wouldn't it?

Which leads me to asking,  the hypotenuse could be a-b or b-a yes?
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#### TrueTears

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##### Re: Specialist 3/4 Question Thread!
« Reply #42 on: December 07, 2011, 12:20:24 am »
0
Which leads me to asking,  the hypotenuse could be a-b or b-a yes?
yeh doesnt matter, but then the vector expression for the vector starting at the right angle vertex to the midpoint would also change, the final result is the same
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#### Planck's constant

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##### Re: Specialist 3/4 Question Thread!
« Reply #43 on: December 07, 2011, 12:25:31 am »
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Which leads me to asking,  the hypotenuse could be a-b or b-a yes?

Indeed it can be either.
But for the purposes of the (correct) solution posted by kamil, in order to find the relevant magnitudes, you end up forming the dot product,

(a - b) . (a - b)

which is equivalent to

(b - a) . (b - a)

#### Bhootnike

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