HSC 2017-2018 Maths Marathon (1 Viewer)

leehuan · Dec 7, 2016

Re: HSC 2017 Maths (Advanced) Marathon

$\text{Let }f(x)\text{ and }g(x)\text{ be continuous, monotonic increasing functions.}\\ \text{1. Show that }f(g(x))\text{ is also monotonic increasing.}\\ \text{2. What happens if one and only one of }f(x),g(x)\text{ were monotonic decreasing?}$

Lugia101101 · Dec 7, 2016

Re: HSC 2017 Maths (Advanced) Marathon

$\begin{align*}\text{Let }h(x)&=f(g(x))\\ h^{\prime}(x)&=f^{\prime}(g(x))g^{\prime}(x)\\ \text{So }&f^{\prime}(x)>0,\forall x\in\mathbb{R}\\ &g^{\prime}(x)>0,\forall x\in\mathbb{R}\\ \therefore&h^{\prime}(x)>0,\forall x\in\mathbb{R}\\ \text{So }&f(g(x))\text{ is monotonic increasing.}\end{align*}$

If one function, but not both functions, is monotonic decreasing, then the product of their first derivatives will be negative, and so $f(g(x))$ will be monotonic decreasing.

leehuan · Dec 7, 2016

Re: HSC 2017 Maths (Advanced) Marathon

Correct, but small technicality error: Monotone means the inequality isn't necessarily strict (≥)

Lugia101101 · Dec 7, 2016

Re: HSC 2017 Maths (Advanced) Marathon

Thanks for clearing that up; none of my teachers were able to tell me definitively.

Lugia101101 · Dec 8, 2016

Re: HSC 2017 Maths (Advanced) Marathon

$\begin{align*}\text{Let }f(x)&=-x^3\\ f^{\prime}(x)&=-3x^2\end{align*}$
$\text{For all }x\in\mathbb{R},x\neq0\text{, }x^2>0\text{, so }-3x^2<0$
$\therefore\text{As }f^{\prime}(x)<0\text{ for all }x\text{ in the given domain, }x^3\text{ is monotonic decreasing.}$

athena13 · Dec 19, 2016

Re: HSC 2017 Maths (Advanced) Marathon

To add onto the marathon, here's a new question:

$The\quad diagram\quad shows\quad the\quad graphs\quad of\quad y\quad =\quad \sqrt { 3 } cos\quad x\quad and\quad y\quad =\sin { \quad x } .\quad The\quad first\quad two\quad points\quad of\quad intersection\quad to\quad the\quad right\quad of\quad the\quad y-axis\quad are\quad labelled\quad A\quad and\quad B.\quad \\ i)\quad Solve\quad the\quad equation\quad \sqrt { 3 } cos\quad x\quad =\quad sin\quad x\quad to\quad find\quad the\quad coordinates\quad of\quad A\quad and\quad B.\\ ii)\quad Find\quad the\quad shaded\quad region\quad in\quad the\quad diagram.$

pikachu975 · Dec 19, 2016

Re: HSC 2017 Maths (Advanced) Marathon

athena13 said:
To add onto the marathon, here's a new question:

View attachment 33649
$The\quad diagram\quad shows\quad the\quad graphs\quad of\quad y\quad =\quad \sqrt { 3 } cos\quad x\quad and\quad y\quad =\sin { \quad x } .\quad The\quad first\quad two\quad points\quad of\quad intersection\quad to\quad the\quad right\quad of\quad the\quad y-axis\quad are\quad labelled\quad A\quad and\quad B.\quad \\ i)\quad Solve\quad the\quad equation\quad \sqrt { 3 } cos\quad x\quad =\quad sin\quad x\quad to\quad find\quad the\quad coordinates\quad of\quad A\quad and\quad B.\\ ii)\quad Find\quad the\quad shaded\quad region\quad in\quad the\quad diagram.$

http://imgur.com/YqJJwLA

LC14199 · Dec 23, 2016

Re: HSC 2017 Maths (Advanced) Marathon

athena13 said:
To add onto the marathon, here's a new question:

View attachment 33649
$The\quad diagram\quad shows\quad the\quad graphs\quad of\quad y\quad =\quad \sqrt { 3 } cos\quad x\quad and\quad y\quad =\sin { \quad x } .\quad The\quad first\quad two\quad points\quad of\quad intersection\quad to\quad the\quad right\quad of\quad the\quad y-axis\quad are\quad labelled\quad A\quad and\quad B.\quad \\ i)\quad Solve\quad the\quad equation\quad \sqrt { 3 } cos\quad x\quad =\quad sin\quad x\quad to\quad find\quad the\quad coordinates\quad of\quad A\quad and\quad B.\\ ii)\quad Find\quad the\quad shaded\quad region\quad in\quad the\quad diagram.$

Oh my god the number of times \quad is used in that TEX code... anyways, is that picture from the MIF textbook?

1729 · Jan 28, 2017

Re: HSC 2017 Maths (Advanced) Marathon

InteGrand said:
$\noindent Given $n$ real numbers $x_{1},x_{2},\ldots, x_{n}$ ($n$ a positive integer), a \textsl{weighted average} of these is a combination of them of the form $\sum_{i=1}^{n}\alpha_{i}x_{i}$, where the $\alpha_{i}$ are non-negative numbers that sum to $1$. E.g. $\frac{1}{3}x_{1} + \frac{2}{3}x_{2}$ is an example of a weighted average of $\left\{x_{1},x_{2}\right\}$.$

$\noindent Let $\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}$ each be weighted averages of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$ (where $K$ is a positive integer). Show that any weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$.$

$\noindent Now, the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is given by \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K}$\\\\But each $\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$. Thus, let us substitute $\sum_{i=1}^{n}\alpha_ix_i$ for $\overline{x}_1,\overline{x}_2, ... , \overline{x}_K$\\\\ Thus \\\\ $(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K} \\ = (\alpha')_1\sum_{i=1}^{n}\alpha_ix_i + (\alpha')_2\sum_{i=1}^{n}\alpha_ix_i + ... + (\alpha')_3\sum_{i=1}^{n}\alpha_ix_i $ \\\\ We factorise $\sum_{i=1}^{n}\alpha_ix_i$ and now we have \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=\sum_{i=1}^{n}\alpha_ix_i \left\{(\alpha')_1 + (\alpha')_2 + ... + (\alpha')_K \right\}$

$$\noindent But we know that the sum of these $(\alpha')_K$, where $K \in \mathbb{Z}^+$, is equal to 1 (by given data). And thus, \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i} = \sum_{i=1}^{n}\alpha_ix_i $\\\\ This means that the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$.$

InteGrand · Jan 28, 2017

Re: HSC 2017 Maths (Advanced) Marathon

1729 said:
Excuse my notation, but here we go (correct me if I am wrong InteGrand)

$\noindent Now, the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is given by \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K}$\\\\But each $\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$. Thus, let us substitute $\sum_{i=1}^{n}\alpha_ix_i$ for $\overline{x}_1,\overline{x}_2, ... , \overline{x}_K$\\\\ Thus \\\\ $(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K} \\ = (\alpha')_1\sum_{i=1}^{n}\alpha_ix_i + (\alpha')_2\sum_{i=1}^{n}\alpha_ix_i + ... + (\alpha')_3\sum_{i=1}^{n}\alpha_ix_i $ \\\\ We factorise $\sum_{i=1}^{n}\alpha_ix_i$ and now we have \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=\sum_{i=1}^{n}\alpha_ix_i \left\{(\alpha')_1 + (\alpha')_2 + ... + (\alpha')_K \right\}$

$$\noindent But we know that the sum of these $(\alpha')_K$ is equal to 1 (by given data). And thus, \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i} = \sum_{i=1}^{n}\alpha_ix_i $\\\\ In English, this means that the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$, as required.$

$\noindent Well what you've done here is use the same weighted average for each $\overline{x}_{j}$ ($j=1,2,\ldots , K$). This is because you let them all be equal to $\sum_{i =1}^{n}\alpha_{i}x_{i}$ (so you gave each of them necessarily the same set of weights $\alpha_{i}$). But the question was actually to prove it for the $\overline{x}_{j}$ being any arbitrary weighted averages, not necessarily all the same (so each of the $K$ weighted averages has its own set of $\alpha$ weights).$

1729 · Jan 28, 2017

Re: HSC 2017 Maths (Advanced) Marathon

InteGrand said:
$\noindent Well what you've done here is use the same weighted average for each $\overline{x}_{j}$ ($j=1,2,\ldots K$). This is because you let them all be equal to $\sum_{i =1}^{n}\alpha_{i}x_{i}$ (so you gave each of them necessarily the same set of weights $\alpha_{i}$). But the question was actually to prove it for the $\overline{x}_{j}$ being any arbitrary weighted averages, not necessarily all the same (so each of the $K$ weighted averages has its own set of $\alpha$ weights).$

$$ \noindent $\overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ $Now the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is given by \\ $\begin{align*} \sum_{i=1}^K \delta_i\overline{x}_i &= \delta_1\overline{x}_1 + \delta_2\overline{x}_2 + ... + \delta_K\overline{x}_K \\ &= \delta_1(\alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n) + \delta_2(\beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n) + ... + \delta_K(\gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n) \\ &= (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1)x_1 + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2)x_2 + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)x_n \end{align*}$

$$\noindent For this to be a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$ then \\ $(\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)$ must equal 1. Let us test. Now,\\ \begin{align*}(\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) \\ &= \delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \end{align*} \\ Now, we know that the the sum of the weights $\alpha, \beta$ and $\gamma$ is equal to 1.$

$\noindent Thus, we now have \\\\$\delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \\ = \delta_1+\delta_2 + ... + \delta_K.$ \\\\ The sum of these weights is also equal to 1, therefore we can conclude that \\ \begin{align*} (\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) = 1 \end{align*} \\ Since this is equal to one, we can conclude that any weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$, given the provided condition.$

InteGrand · Jan 28, 2017

Re: HSC 2017 Maths (Advanced) Marathon

1729 said:
Okay, I am going to use different greek letters for different weights, although not sure if this is reasonable notation. Now correct me if this is wrong.. although I am fairly confident that I got it.. this took ages to type my apologies

$$ \noindent $\overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ $Now the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is given by \\ $\begin{align*} \sum_{i=1}^K \delta_i\overline{x}_i &= \delta_1\overline{x}_1 + \delta_2\overline{x}_2 + ... + \delta_K\overline{x}_K \\ &= \delta_1(\alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n) + \delta_2(\beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n) + ... + \delta_K(\gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n) \\ &= (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1)x_1 + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2)x_2 + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)x_n \end{align*}$

$$\noindent For this to be a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$ then \\ $(\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)$ must equal 1. Let us test. Now,\\ \begin{align*}(\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) \\ &= \delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \end{align*} \\ Now, we know that the the sum of the weights $\alpha, \beta$ and $\gamma$ is equal to 1.$

$\noindent Thus, we now have \\\\$\delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \\ = \delta_1+\delta_2 + ... + \delta_K.$ \\\\ The sum of these weights is also equal to 1, therefore we can conclude that \\ \begin{align*} (\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) = 1 \end{align*} \\ Since this is equal to one, we can conclude that any weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$.$

Well done.

boredofstudiesuser1 · Jan 28, 2017

Re: HSC 2017 Maths (Advanced) Marathon

1729 said:
Okay, I am going to use different greek letters for different weights, although not sure if this is reasonable notation. Now correct me if this is wrong.. although I am fairly confident that I got it.. this took ages to type my apologies

$$ \noindent $\overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ $Now the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is given by \\ $\begin{align*} \sum_{i=1}^K \delta_i\overline{x}_i &= \delta_1\overline{x}_1 + \delta_2\overline{x}_2 + ... + \delta_K\overline{x}_K \\ &= \delta_1(\alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n) + \delta_2(\beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n) + ... + \delta_K(\gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n) \\ &= (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1)x_1 + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2)x_2 + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)x_n \end{align*}$

$$\noindent For this to be a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$ then \\ $(\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)$ must equal 1. Let us test. Now,\\ \begin{align*}(\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) \\ &= \delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \end{align*} \\ Now, we know that the the sum of the weights $\alpha, \beta$ and $\gamma$ is equal to 1.$

$\noindent Thus, we now have \\\\$\delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \\ = \delta_1+\delta_2 + ... + \delta_K.$ \\\\ The sum of these weights is also equal to 1, therefore we can conclude that \\ \begin{align*} (\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) = 1 \end{align*} \\ Since this is equal to one, we can conclude that any weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is a weighted average of $\left\{x_{1},x_{2},\ldots, x_{n}\right\}$.$

This is HSC maths??

InteGrand · Jan 28, 2017

Re: HSC 2017 Maths (Advanced) Marathon

boredofstudiesuser1 said:
This is HSC maths??

It can be done using only high school algebra as 1729 did, but admittedly it probably wouldn't come up in a HSC paper (not without any guidance anway).

You can find out more about weighted averages at the Wikipedia article for it: https://en.wikipedia.org/wiki/Weighted_arithmetic_mean .

davidgoes4wce · Feb 3, 2017

Re: HSC 2017 Maths (Advanced) Marathon

I came across a question in the Cambridge textbook today and I haven't looked at the answer yet.

$\int^{5}_{4} \ dx=$

I'm assuming its a constant of 1 next to the dx when we integrate it? Somebody could confirm

InteGrand · Feb 3, 2017

Re: HSC 2017 Maths (Advanced) Marathon

davidgoes4wce said:
I came across a question in the Cambridge textbook today and I haven't looked at the answer yet.

$\int^{5}_{4} \ dx=$

I'm assuming its a constant of 1 next to the dx when we integrate it? Somebody could confirm

Yes, that's right (it's common practice to drop the '1' and write it like they've written it).

davidgoes4wce · Feb 8, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Congruency test

I came across this question tonight and wanted to clear a couple of things up. At times could we use either RHS test or SAS test to explain parts (b) and (f)? They are exactly the same shape as one can see and the Cambridge textbook gives 2 solutions:

The answer for part (b) was RHS in the textbook

The answer for part (f) was SAS in the text book

boredofstudiesuser1 · Feb 8, 2017

Re: HSC 2017 Maths (Advanced) Marathon

davidgoes4wce said:
Congruency test

I came across this question tonight and wanted to clear a couple of things up. At times could we use either RHS test or SAS test to explain parts (b) and (d)? They are exactly the same shape as one can see and the Cambridge textbook gives 2 solutions:

The answer for part (b) was RHS in the textbook

The answer for part (f) was SAS in the text book

I would say no (may be wrong).

In (b), the angle is not included within the two sides marked and therefore can only be RHS (SAS is side, included angle, side)

On the other hand, in (f) it is SAS as they are giving you two sides and an included angle but the hypotenuse is not one of the sides. Therefore it is directly SAS and I would say, indirectly RHS.

davidgoes4wce · Feb 8, 2017

Re: HSC 2017 Maths (Advanced) Marathon

^^ I can see it now . It's something that I never really looked into specifically.

SO I take it, if they give the hypotenuse length and a right angle, we can conclude its a RHS . (also one of the adjacent or opposite lengths)

If we are given a right angle, adjacent and an opposite length, then its a SAS.

davidgoes4wce · Feb 8, 2017

Re: HSC 2017 Maths (Advanced) Marathon

I was also looking to compare (b) v (f)....................earlier I mentioned (b) v (d)

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