Author Topic: Mathematics Question Thread (Read 1626121 times) Tweet Share

jamonwindeyer · « **Reply #3690 on:** August 13, 2018, 11:11:36 pm »

Quote from: amelia20181 on August 13, 2018, 03:41:50 pm

what would happen if you didn't write your number on the exam paper

They would catch it picking it up, usually! Otherwise they cross reference the student list with the scripts to see who is missing, and other similar logistical checks

either way, no reason to be concerned about it!

S200 · « **Reply #3691 on:** August 14, 2018, 10:24:49 am »

Quote from: billy.ohlmeyer on August 08, 2018, 04:42:33 pm

my textbook has this question:
Given that the wingspan of an aeroplane is 30m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' when it is directly overhead.
how do i do this?

thanks

Hey there!
Just stalked your posts.. (

) and realised that no-one had answered this...
Only 6 days late, but here goes.

Basically, a diagram of this would be an isosceles triangle, with a base of thirty and an angle of 14' opposite the base. By cutting it in two we can make a right-angled triangle, and with the base of this triangle (i.e: 15) and the angle given halved, we can find the height of the triangle...
So, by going
$tan (\frac {14'} {2}) = \frac {Opposite (15)} {adjacent} \\ \therefore \\ \frac {15} {tan (\frac {14'} {2})} = 7366.59m \therefore ~7367m $

Therefore, the plane is 7367 Meters above the ground...

darlo69 · « **Reply #3692 on:** August 15, 2018, 03:53:17 pm »

How do you do this?

Find the coordinates of the center of a circle that passes (7,2) and (2,3) and (-4,-1)

fun_jirachi · « **Reply #3693 on:** August 15, 2018, 04:42:59 pm »

Quote from: darlo69 on August 15, 2018, 03:53:17 pm

How do you do this?

Find the coordinates of the center of a circle that passes (7,2) and (2,3) and (-4,-1)

all explained on the sheet

amelia20181 · « **Reply #3694 on:** August 18, 2018, 04:52:39 pm »

how do you do

fun_jirachi · « **Reply #3695 on:** August 18, 2018, 07:06:27 pm »

Quote from: amelia20181 on August 18, 2018, 04:52:39 pm

how do you do

Cheers S200, saved me a bit of time on the last q there

Quote from: S200 on August 14, 2018, 10:24:49 am

Hey there!
Just stalked your posts.. () and realised that no-one had answered this...
Only 6 days late, but here goes.

Basically, a diagram of this would be an isosceles triangle, with a base of thirty and an angle of 14' opposite the base. By cutting it in two we can make a right-angled triangle, and with the base of this triangle (i.e: 15) and the angle given halved, we can find the height of the triangle...
So, by going
$tan (\frac {14'} {2}) = \frac {Opposite (15)} {adjacent} \\ \therefore \\ \frac {15} {tan (\frac {14'} {2})} = 7366.59m \therefore ~7367m $

Therefore, the plane is 7367 Meters above the ground...

Answers to the other two attached

fun_jirachi · « **Reply #3696 on:** August 18, 2018, 07:31:25 pm »

Also I just realised my photos are really unclear, because I take them on my phone, then send them over Messenger to myself on my laptop, then post them. I'm guessing my laptop does this weird laptop rendering thing?

Just wondering where you guys (Jamon, Rui, S200) get that really cool font that looks like it came off Symbolab so I can stop using photos

S200 · « **Reply #3697 on:** August 18, 2018, 07:35:17 pm »

Quote from: fun_jirachi on August 18, 2018, 07:31:25 pm

Also I just realised my photos are really unclear, because I take them on my phone, then send them over Messenger to myself on my laptop, then post them. I'm guessing my laptop does this weird laptop rendering thing?

Just wondering where you guys (Jamon, Rui, S200) get that really cool font that looks like it came off Symbolab so I can stop using photos

It's called LaTex.

The board has it inbuilt. Link in my sig is kinda exhaustive, and not all of it works. But if you just search the entire forum for Latex, you will get a few hits on tutorial threads

RuiAce · « **Reply #3698 on:** August 18, 2018, 07:37:51 pm »

Quote from: fun_jirachi on August 18, 2018, 07:31:25 pm

Also I just realised my photos are really unclear, because I take them on my phone, then send them over Messenger to myself on my laptop, then post them. I'm guessing my laptop does this weird laptop rendering thing?

Just wondering where you guys (Jamon, Rui, S200) get that really cool font that looks like it came off Symbolab so I can stop using photos

If you wanted to get started with $ \LaTeX $

Symbolab uses it, Wolfram uses it, a lot of things use it to display mathematics nowadays

I need to update it though. Two years or so later and I've become more proficient with it. Whilst not all of it, some of the stuff on there is a bit overkill

S200 · « **Reply #3699 on:** August 18, 2018, 07:43:36 pm »

Woah! I didn't even know about this!
I've been wanting to do inline for ages!

fun_jirachi · « **Reply #3700 on:** August 18, 2018, 08:04:09 pm »

Ok thanks guys!

amelia20181 · « **Reply #3701 on:** August 18, 2018, 08:22:30 pm »

how do you do

S200 · « **Reply #3702 on:** August 18, 2018, 08:24:10 pm »

Quote from: amelia20181 on August 18, 2018, 08:22:30 pm

how do you do

Literally ~~the same as~~ very similar to the last question, just with different numbers. Just sub the new amounts in!
This time you are solving for base, not height.

Edited working from last question

Basically, a diagram of this would be an isosceles triangle, with a ~~base of thirty~~ height of $1.49 \times 10^8$ and an angle of ~~14'~~ 31' opposite the base. By cutting it in two we can make a right-angled triangle, and with the ~~base~~ height of this triangle i.e: 15 ${1.49 \times 10^8}$ and the angle given halved, we can find the ~~height~~ base of the triangle...
So, by going
$tan (\frac {31'} {2}) = \frac {Opposite (\alpha)} {adjacent ({1.49 \times 10^8})} \\ \therefore \\ {1.49 \times 10^8} \times {tan \left( \frac {31'} {2}\right)} = \alpha \text { Where } \alpha \text { is the radius} $
Therefore, the sun is $ 2 \times \alpha$ km in diameter

amelia20181 · « **Reply #3703 on:** August 18, 2018, 09:14:02 pm »

how do you do this

Thanks!

fun_jirachi · « **Reply #3704 on:** August 18, 2018, 10:10:46 pm »

So just gonna answer your question and try out that cool text thing all in one

$\text{(a) The length of arc BD is found using one of the formulas on your reference sheet, specifically}$ $l = r \theta$ $\text{where}$ $\theta \text{is expressed in radians.}$ $\text{Therefore, the length of arc BD}$ $=11\times\frac{\pi}{9}$ $\text{or just}$ $\frac{11\pi}{9}$
$\text{(b) The area of ABD can be found by subtracting the area of the sector from the area of the triangle.}$ $\frac{4\times11}{2}-\frac{1}{2}r^2\theta$ $=22-\frac{1}{2}11^2\frac{\pi}{9}$ $=22-\frac{121\pi}{18}$
$\text{(c) This is just the two radii plus the length of the arc BD.}$ $11+11+\frac{11\pi}{9}$ $=22+\frac{11\pi}{9}$
Looks a little messy rn, but I'm getting there with the code!

Hope this helps!

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