Author Topic: Maths 3A/3B (Read 54774 times) Tweet Share

jamonwindeyer · « **Reply #90 on:** October 04, 2016, 11:52:07 pm »

Quote from: ml125 on October 04, 2016, 11:47:37 pm

Spoiler
$\text{From the diagram, we first need angles α and β to find the arc lengths}$
$\text{This can be done using the cos rule on the constructed triangle:}$
$\begin{align*}cosC&=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\\cos\frac{\alpha}{2}&=\frac{{10}^{2}+{15}^{2}-{7}^{2}}{2\cdot10\cdot15}\\cos\frac{\alpha}{2}&=\frac{276}{300}\\\frac{\alpha}{2}&=\cos^{-1}{\frac{276}{300}}\\\alpha&=2\cos^{-1}{\frac{276}{300}}\end{align*}$
$\text{Similarly, we can find a value for β}$
$\begin{align*}\quad\quad\quad\quad\beta&=2\cos^{-1}{\frac{174}{210}}\end{align*}$
$\text{Now that we have an angle, we can find each of the arc lengths:}$
$\begin{align*}l&=r\theta\\{l}_{1}&=10\cdot2\cos^{-1}{\frac{276}{300}}\\&=8.0543...cm\\{l}_{2}&=7\cdot2\cos^{-1}{\frac{174}{210}}\\&=8.3194...cm\end{align*}$
$\text{Adding these two values we get the perimeter of the region}$
$\begin{align*}\quad\quad\quad\quad{l}_{1}+{l}_{2}&=16.4cm(1d.p.)\end{align*}$

Legend! Awesome answer, questions like this are great revision for Circle Geometry

anotherworld2b · « **Reply #91 on:** October 05, 2016, 12:41:59 am »

Thank you for your help

I was wondering if i could also get help.with these two questions

Quote from: ml125 on October 04, 2016, 11:47:37 pm

Spoiler
$\text{From the diagram, we first need angles α and β to find the arc lengths}$
$\text{This can be done using the cos rule on the constructed triangle:}$
$\begin{align*}cosC&=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\\cos\frac{\alpha}{2}&=\frac{{10}^{2}+{15}^{2}-{7}^{2}}{2\cdot10\cdot15}\\cos\frac{\alpha}{2}&=\frac{276}{300}\\\frac{\alpha}{2}&=\cos^{-1}{\frac{276}{300}}\\\alpha&=2\cos^{-1}{\frac{276}{300}}\end{align*}$
$\text{Similarly, we can find a value for β}$
$\begin{align*}\beta&=2\cos^{-1}{\frac{174}{210}}\end{align*}$
$\text{Now that we have an angle, we can find each of the arc lengths:}$
$\begin{align*}l&=r\theta\\{l}_{1}&=10\cdot2\cos^{-1}{\frac{276}{300}}\\&=8.0543...cm\\{l}_{2}&=7\cdot2\cos^{-1}{\frac{174}{210}}\\&=8.3194...cm\end{align*}$
$\text{Adding these two values we get the perimeter of the region}$
$\begin{align*}\quad\quad\quad\quad{l}_{1}+{l}_{2}&=16.4cm(1d.p.)\end{align*}$

ml125 · « **Reply #92 on:** October 05, 2016, 09:38:47 pm »

Quote from: anotherworld2b on October 05, 2016, 12:41:59 am

Thank you for your help
I was wondering if i could also get help.with these two questions

For the first question:

Spoiler

$\text{By intersecting tangent and secant theorem, we know that:}$
$\begin{align*}{RP}^{2}&=RA\cdot RB\\{RP}^{2}&=12\cdot 32\\RP&=\sqrt{384}\\&=8\sqrt{6}\end{align*}$
$\text{Now, we can construct △POR, to find ∠POQ.}$
$\text{To do this, you will need to use the cos rule, as in the earlier question:}$
$\begin{align*}cosC&=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\end{align*}$
$\text{Once you have a value for ∠POQ, you will be able to find arc PQ:}$
$\begin{align*}l&=r\theta\end{align*}$
$\text{Then, all you will need to do is convert this value into a percentage of the circumference.}$
Not 100% sure about the next one, so I'll leave that for the others

anotherworld2b · « **Reply #93 on:** October 07, 2016, 11:02:01 am »

Can u get help with these questions please?

RuiAce · « **Reply #94 on:** October 08, 2016, 12:30:48 pm »

Quote from: anotherworld2b on October 07, 2016, 11:02:01 am

Can u get help with these questions please?

$\text{To force the equation to look like that of a circle we should complete the square.}\\ \text{This is fairly standard, as otherwise we don't even know it is a circle.}$
$\begin{align*}x^2+2x+y^2-10y+a&=0\\ x^2+2x+1+y^2-10y+25&=26-a\\ (x+1)^2+(y-5)^2&=26-a\end{align*}$
$\text{But to ensure it is a circle the radius must exist and it can't be 0.}\\ \text{Having a 0 or a negative radius is nonsense.}\\ \text{Hence }26-a > 0 \iff a < 26$

Syndicate · « **Reply #95 on:** October 08, 2016, 01:37:53 pm »

Quote from: anotherworld2b on October 07, 2016, 11:02:01 am

Can u get help with these questions please?

Q1)
$sin(x+30)=cosx$
$=> sinxcos30 + cosxsin30 =cosx$
$=> \frac{\sqrt3}{2}sinx + \frac{1}{2}cosx = cosx$
$=> \sqrt3 sinx = cosx$
$=> 3sin^2x = cos^2x$
$=> 3sin^2x = 1 - sin^2x$
$=> 4sin^2x = 1$
$=> sin^2x= \frac{1}{4}$
$=> sinx = \pm \frac{1}{2}$
$\therefore x = 30, 150, 210,330$

However, only x = 30, 210 work as we squared the equation in order to get the solutions. Make sure you always check your solutions by substituting them back into the original equation (sin(x+30) = cosx).

RuiAce · « **Reply #96 on:** October 08, 2016, 01:42:02 pm »

Quote from: Syndicate on October 08, 2016, 01:37:53 pm

Q1)
$=> \sqrt3 sinx = cosx$
$=> 3sin^2x = cos^2x$
$=> 3sin^2x = 1 - sin^2x$
$=> 4sin^2x = 1$
$=> sin^2x= \frac{1}{4}$
$=> sinx = \pm \frac{1}{2}$
$\therefore x = 30, 150, 210,330$

However, only x = 30, 210 work as we squared the equation in order to get the solutions. Make sure you always check your solutions by substituting them back into the original equation (sin(x+30) = cosx).

Didn't have to square here. Just apply an auxiliary angle transformation, or divide by cos(x) and test x=90^o, 270^o separately.

I'd avoid squaring when solving equations, in general. Better ways out than just eliminating the mistakes.

More LaTeX advice: You can use \implies or \Rightarrow to create the arrow you need.

anotherworld2b · « **Reply #97 on:** October 09, 2016, 06:16:35 pm »

thank you all very much for the help
I was wondering is there a guide on graphing functions such as quadratics, cubics and trigonometric graphs on ATARNotes?
I am having trouble understanding how to graph these functions and finding the equation from a given graph

anotherworld2b · « **Reply #98 on:** October 11, 2016, 07:42:26 pm »

Could i get help with these questions please? I tried q5 but i couldnt get the answer.

Syndicate · « **Reply #99 on:** October 11, 2016, 10:16:40 pm »

Quote from: anotherworld2b on October 11, 2016, 07:42:26 pm

Could i get help with these questions please? I tried q5 but i couldnt get the answer.

Line 3: X(95-x) expands out to 95x-x^2 not 95 -x^2.

anotherworld2b · « **Reply #100 on:** October 11, 2016, 10:55:34 pm »

I tried q7 but couldnt get part b. I also wanted to check whether how I solved q8 was mathematically correct. I was also wondering if i could get help with q10 and 11 because im not sure where to start

anotherworld2b · « **Reply #101 on:** October 11, 2016, 10:58:40 pm »

Questions for above

Syndicate · « **Reply #102 on:** October 15, 2016, 04:16:35 pm »

Quote from: anotherworld2b on October 11, 2016, 10:55:34 pm

I tried q7 but couldnt get part b. I also wanted to check whether how I solved q8 was mathematically correct. I was also wondering if i could get help with q10 and 11 because im not sure where to start

Q8 seems fine.

Q7b)

$\text{Let the length of the rectangle = x}<br />$ $\therefore \space width \space = \frac{100-x}{2} \space = \space 50 - \frac{x}{2}$
$Area \space = \space width \times length$
$Area \space = \space 50 - \frac{x}{2}\times x$
$= 50x - \frac{x^2}{2}$
$= \frac{-1}{2} (x^2-100x + 2500 - 2500)$
$= \frac{-1}{2} (x-50)^2 +1250$

Therefore the maximum area = 1250 m^2, when x = 50. So the dimensions of the fenced area is 25 x 50 .

Q10) Hint: since the box is a cube, length would equal the width. Each side's length = 60 - 2x.

Q11) The second part of the question is missing.

anotherworld2b · « **Reply #103 on:** October 16, 2016, 04:45:26 pm »

Thank yoh for your help
I was wondering if i could get help with q30 and 32

Quote from: Syndicate on October 15, 2016, 04:16:35 pm

Q8 seems fine.

Q7b)

$\text{Let the length of the rectangle = x}<br />$ $\therefore \space width \space = \frac{100-x}{2} \space = \space 50 - \frac{x}{2}$
$Area \space = \space width \times length$
$Area \space = \space 50 - \frac{x}{2}\times x$
$= 50x - \frac{x^2}{2}$
$= \frac{-1}{2} (x^2-100x + 2500 - 2500)$
$= \frac{-1}{2} (x-50)^2 +1250$

Therefore the maximum area = 1250 m^2, when x = 50. So the dimensions of the fenced area is 25 x 50 .

Q10) Hint: since the box is a cube, length would equal the width. Each side's length = 60 - 2x.

Q11) The second part of the question is missing.

anotherworld2b · « **Reply #104 on:** August 23, 2017, 01:12:23 am »

Can I have help with q 21 and q22c please?

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Author Topic: Maths 3A/3B (Read 54774 times) Tweet Share

jamonwindeyer

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ml125

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Syndicate

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