HSC 2015 MX1 Marathon (archive) (2 Viewers)

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Three men Bill, Garry and Jason observe a vertical tower. Bill stands du North of the tower, and sees its top at an angle of elevation of A. Garry stands due East of the tower, and sees its top at an angle of elevation of B. Jason stands on a line from Bill to Garry, exactly half way between them.
If Jason observes the top of the tower at an angle of elevation of $\theta$
Show that $cot\theta =\frac{1}{2}\sqrt{cot^2A +cot^2B}$

Can someone do this -.-?

Crisium · Jul 5, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Can someone do this -.-?

Do you not know how to do it?

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

Crisium said:
Do you not know how to do it?

i get to a point were im stuck and it doesnt equal unless the towers height is 0

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Can someone do this -.-?

$Draw a nice neat diagram, label all right angles (note that $\triangle BOG$ is right angled (where $B$ is where Bill is, $G$ is where Garry is, and $O$ is the tower's base) because Bill is due North of $O$ and Garry is due East of $O$). Let the tower's height be $h$.$

$Using simple right-angled trigonometry in triangles $TOE$, $TOJ$, and $TOB$ (where $J$ is where Jason is and $T$ is the top of the tower), we see that $OE=h\cot \beta$, $OJ=h\cot \theta$, $OB=h\cot \alpha$, where we have relabelled the given angles so that $\angle OBT=\alpha$ (Bill's angle of elevation) and $\angle OBT= \beta$ (Garry's angle of elevation). Then by Pythagoras' theorem in $\triangle OBE$, $BE=\sqrt{h^2 \cot^2 \alpha + h^2\cot^2 \beta}$. So $JE=\frac{1}{2}\sqrt{h^2 \cot^2 \alpha + h^2\cot^2 \beta}$. Since the midpoint of the hypotenuse of a right-angled triangle is equidistant from its vertices, $OJ=JE\Rightarrow h\cot \theta = \frac{1}{2}\sqrt{h^2 \cot^2 \alpha + h^2\cot^2 \beta}$. Simplifying this yields the result. \blacksquare$

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$Draw a nice neat diagram, label all right angles (note that $\triangle BOG$ is right angled (where $B$ is where Bill is, $G$ is where Garry is, and $O$ is the tower's base) because Bill is due North of $O$ and Garry is due East of $O$). Let the tower's height be $h$.$

$Using simple right-angled trigonometry in triangles $TOE$, $TOJ$, and $TOB$ (where $J$ is where Jason is and $T$ is the top of the tower), we see that $OE=h\cot \beta$, $OJ=h\cot \theta$, $OB=h\cot \alpha$, where we have relabelled the given angles so that $\angle OBT=\alpha$ (Bill's angle of elevation) and $\angle OBT= \beta$ (Garry's angle of elevation). Then by Pythagoras' theorem in $\triangle OBE$, $BE=\sqrt{h^2 \cot^2 \alpha + h^2\cot^2 \beta}$. So $JE=\frac{1}{2}\sqrt{h^2 \cot^2 \alpha + h^2\cot^2 \beta}$. Since the midpoint of the hypotenuse of a right-angled triangle is equidistant from its vertices, $OJ=JE\Rightarrow h\cot \theta = \frac{1}{2}\sqrt{h^2 \cot^2 \alpha + h^2\cot^2 \beta}$. Simplifying this yields the result. \blacksquare$

i couldnt get the equidistant part D: Thanks !

Crisium · Jul 5, 2015

Re: HSC 2015 3U Marathon

InteGrand did I miss anything out in my working out for that question in pomsky's thread?

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

Crisium said:
InteGrand did I miss anything out in my working out for that question in pomsky's thread?

No, you did it correctly.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

Someone give me a head start on this question:

'When the polynomial P(x) is divided by x^2-a^2 the remainder is 2x+4 and when it is divided by x+a the remainder is a+3. What is the value of a?

I know the polynomial division process:

P(x)=A(x).Q(x)+R(x)

I tried to use simultaneous equations to solve it but have come unstuck with 2 unknown variables left in my equation.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

This is my honest attempt at it:

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Someone give me a head start on this question:

'When the polynomial P(x) is divided by x^2-a^2 the remainder is 2x+4 and when it is divided by x+a the remainder is a+3. What is the value of a?

I know the polynomial division process:

P(x)=A(x).Q(x)+R(x)

I tried to use simultaneous equations to solve it but have come unstuck with 2 unknown variables left in my equation.

$$P(x)=(x^2-a^2)Q(x) + 2x+4 = (x+a)(x-a)Q(x) + 2x+4$. From this expression, $P(-a) = -2a+4$. But from the remainder theorem, we also know that $P(-a)= a+3$. Hence $a+3 = -2a+4\Rightarrow a =\frac{1}{3}$.$

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
This is my honest attempt at it:

Also, we shouldn't be writing Q(x) in both of your last two lines, since the dividend polynomials are different in each case, as the divisors are different.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Also, we shouldn't be writing Q(x) in both of your last two lines, since the dividend polynomials are different in each case, as the divisors are different.

I noticed you wrote Q(x) in your line though

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I noticed you wrote Q(x) in your line though

That's fine, because that Q(x) is only referring to one polynomial. In your working out, you had Q(x) in two lines where the divisor was different, which is mathematically not correct because the dividend polynomials would actually be different (so maybe write something like $Q_1 (x)$ and $Q_2(x)$ to differentiate the two).

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

Im going to grab some Maccas, the study is burning me out a bit. Be back later

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

I understand how to use this doing it L'Hopital's Rule but not the '3 Unit way'

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I understand how to use this doing it L'Hopital's Rule but not the '3 Unit way'

$$Use the trigonometric identity $\cos \left(\frac{\pi}{2}+x\right)= -\sin x$.$

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

Is it cos of the whole thing ?

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Is it cos of the whole thing ?

Don't think so (I think this was meant: $\lim _{x\to 0}\frac{\cos \left( \frac{\pi}{2}+x\right)}{2x}$ ). If it's cosine of the whole thing, then the limit does not exist as $\frac{\frac{\pi}{2}+x}{2x}\to \pm \infty$ as $x\to 0^{\pm}$ and the limit $\lim _{\eta \to \infty}\left( \cos \eta\right)$ does not exist.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

I used the trig identity Integrand told me to use for cos(A+B)=cos(A)cos(B)-sin(A)sin(B)

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

the 2nd line on the right should be sin (x) / x

HSC 2015 MX1 Marathon (archive) (2 Viewers)

Well-Known Member

Pew Pew

Well-Known Member

Well-Known Member

Well-Known Member

Pew Pew

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)