geometry-and-trigonometry

[crude birch]

c seems to be linear

[wind heart]

how so?

[crude birch]

<11 and <12 formed 180 deg

[wind heart]

Is it because it forms a line?

[crude birch]

that's what i thought

i saw a picture of the same question lol

a is adjacent

it disregard the angle at the left so it doesnt make a 180deg

[wind heart]

For vertical angles, since they have to have completely opposites rays,

DGE and AGB would not be vertical pairs, correct?

[sly marlin]

yeah, not vertical

[upper karma]

how do i find x?

[sly marlin]

hmm

are those the arc angles?

[upper karma]

idk

what theyre called tbh

thats why i posted it herre

i wanted to look it up

but idk what its called

[sly marlin]

They are the central angles of that arc?

Arc measure

I think

you need to relate them with other angles

angles that are between lines already in the diagram

any ideas?

[upper karma]

idk I wasn't in class the other day so i don't know

i wanted to look iot up

but idk what its even called

lol

also need help with this if anyone knows

[spiral fog]

0<x<180

X is not 90

[upper karma]

how do i find this ? i thoguht it was 17x + 75 =360 but apparently its not

[silent plank]

what's 17x supposed to represent?

[upper karma]

the arc

the above arc

[silent plank]

construct lines to F and H from the centre

(let centre be O)

[upper karma]

i still don't really see it

[silent plank]

do you know anything special about OFGH?

constructing line OG may help you see it

[upper karma]

the bottom arc

is 75

right?

75 + 17x = 360

[silent plank]

no

[upper karma]

i see that 75 is the same on the opposite side

of the polygon

but thats about it

its 105?

[silent plank]

what's 105?

[upper karma]

the bottom arc

[silent plank]

and what's the reasoningvfor that?

[upper karma]

75 + 105 = 180

because the angles add to 180

on the polygon

[silent plank]

because...?

[upper karma]

because the two angles add to 180 on the polygon

the whole interior thing idk my math vocab sucks

[silent plank]

why do they add up to 180?
does that apply to any polygon?

[upper karma]

tbh i can't bel;ieve i didnt see this

no it doesn't

only rectangles / squares

[silent plank]

this is neither of those

[spiral fog]

That's not a rectangle but the reasoning seems right

[upper karma]

yes its not

LOL

fuck me man

idk how to explai nthis tbh

i learned this a while back

[silent plank]

you are looking for a very specific term to describe this type of quadrilateral

[upper karma]

i remember the concept

but idk how to exlain it

[spiral fog]

C-

[upper karma]

i can draw the angles that add to 180 but thats about it

LOL

[spiral fog]

It's cyclic if I remember

[silent plank]

well for this question, you can find it without knowing the name

[spiral fog]

True

[upper karma]

yeah the answer is x=15

i just saw it

thx for the help

how do i solve this one? I thought it was 7x + 3x = 80 because 40 is half of 80 then i can do 360-120 for CD but i think its wrong.

[spiral fog]

7x/2 - 3x/2 = 40 I believe

[upper karma]

what do i do to find the arc of CD though ?

[silent plank]

construct EA, ED and EC

[upper karma]

is it 160?

for DC

or CD*

[silent plank]

1sec,

[upper karma]

is just did 7x(1/2) - 3x(1/2) = 40

and found x = 20

then did 7x(20) + 3x(20) + DC = 360

[spiral fog]

It's good

[upper karma]

how would i write an expression for this ?

[spiral fog]

CD = 360 - 10x?

[upper karma]

last one, how do i think about this question?

[spiral fog]

P is half the difference between wz and xy

And wx equals zy

[upper karma]

is wz and zy equal?

[spiral fog]

Not necessarily

[upper karma]

how do i find what wx and zy is?

[spiral fog]

Wz + zy = 200

Wx + xy = 160

You don't need to find values for them, they don't have fixed values

[upper karma]

what do i do with the info of the 200 and 160 split between the two large arcs?

[spiral fog]

Take the difference, divide by 2

[upper karma]

wait

why?

i don't understand the concept

[spiral fog]

P = 1/2(wz-xy)

Zy = wx

When you take the difference, the zy and wx cancel, and you get wz-xy=40

[upper karma]

OK

i see

oh

i see it now

thanks bro

the two arcs are the same so we get the difference of xy and wz

[spiral fog]

Yeah

[fluid moat]

Yo

[spiral fog]

What

[fluid moat]

Can some one check my puthroym theorem

[spiral fog]

Ok I'll try

[fluid moat]

Sry I have bad handwriting

[spiral fog]

Seems good

[fluid moat]

K

Thanks mate

[spiral fog]

np

[terse sage]

You want to solve for x?

You can notice that these are similar quadrilaterals, which means that ratio of corresponding sides are equal

[sly marlin]

???

it's not similar if just the angles are the same

@terse sage

You can translate a side up and down keeping it parallel

and the angles will be preserved

[terse sage]

Hmm

Oh yes

You're right

[upper karma]

so are tarski axioms better than hilbert or?

and A is congruent to L in that pic? and M with B etc?

[bright quarry]

Does somebody know a good YouTube video or thing that can explain more about Trig functions. Like I'm in calc rn, but I dont think I understand what sin or tan or etc. Is, as good as I should. I especially don't understand how Trig graphs (like graphing sin) work

[weary drift]

@bright quarry https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/intro-to-the-trig-ratios/a/opposite-adjacent-hypotenuse

once you finish reading the intro, head over to the video lectures and the practice question sections

[bright quarry]

I'm not a fan of Khan. They don't typically go into the depth I want.

Thanks tho

[patent idol]

I'm not sure how to do this. Using the cosine rule to find exact value of y

[dark sparrow]

can you write out the cosine rule

[patent idol]

a^2 =b^2 +c^2 - 2bcCos(A)

That was anoying on the phone

[dark sparrow]

ok great

well

$y^2 = (3\sqrt{2})^2 + 3^2 - 2 \cdot 3\sqrt{2} \cdot 3 \cdot \cos(45^{\circ})$

[somber coyote]

Ann:

[patent idol]

Yep I got that, I'd send a pic of my notes but it got messy with frustration

[dark sparrow]

i mean

it's all just a matter of arithmetic from there

[patent idol]

So 15-(18root2)cos 45

[dark sparrow]

cos(45°) = what

also

$(3\sqrt{2})^2 + 3^2 \neq 15$

[somber coyote]

Ann:

[patent idol]

Cos 45=1/root2

[dark sparrow]

so $18 \sqrt{2} \cdot \cos(45^\circ) = , ?$

[somber coyote]

Ann:

[patent idol]

18/root2 x 1/root2
Roots cancel out
=18

[dark sparrow]

you made an even number of mistakes that cancelled each other out and gave you the right answer in that last message

but ok

my other point, which i am not going to repeat, still stands.

[patent idol]

Hmm i think I did better here.

[dark sparrow]

,rotate

[dark sparrow]

oh ffs

,rotate 180

[dark sparrow]

ok yes good

[patent idol]

Fanks. Ann

Thanks*

[nimble cosmos]

not sure what im supposed to do here

[umbral snow]

sin(-3π/4), cos(-3π/4)...

Just fill the boxes with the exact answers

@nimble cosmos

[nimble cosmos]

how come sin((-3pi)/4) is negative even though it's in the second quadrant

im using the CAST rule and it says sine will be positive in the second quadrant

[dark sparrow]

-3π/4 is not in the second quadrant.

-3π/4 is in the third quadrant.

@nimble cosmos

[upper karma]

Help pls

[dark sparrow]

what have you tried and where are you stuck

[nimble cosmos]

ah i see

im having trouble with this

would it be 4/5 instead?

[dark sparrow]

how did you get 4/3 in the first place?

[civic lake]

yes

[dark sparrow]

and how did it being greater than 1 not set off a red flag in your mind?

@civic lake i wasn't asking you, nor is "yes" an appropriate response to my question.

[nimble cosmos]

oh so if my fraction is greater than up, it's a red flag?

cause it can only be from [0,1]?

[dark sparrow]

no

like

cos(x) and sin(x)

are always between -1 and 1

so if your value for the cos or the sin of something ISN'T between -1 and 1

[nimble cosmos]

ooh

[dark sparrow]

then that's a problem

[nimble cosmos]

i see

i got 4/3 cause it got my SOH CAH TOA mixed up

i was doing cos = adjacent/opposite i think

yeah

if sin and cos are from 0-1

then what is tan

[dark sparrow]

no

[nimble cosmos]

oh ya -1 and 1 lol

sorry

[dark sparrow]

"from 0-1" is not only incorrect, it's also horrible wording

[nimble cosmos]

ok

[dark sparrow]

but the range of tan is the whole number line

any number can be the tan of an angle

[nimble cosmos]

i see

[steep temple]

is there a way to get the magnitude, and the direction of the resultant force, in newtons and degrees respectively? I think I can use cosine rule somehow here, but I'm not entirely sure.
The problem is that I don't exactly know Vector 1's and Vector 2's direction

[coral sapphire]

Use trig on the triangles probably

[terse sage]

@steep temple You can use the cosine rule

[steep temple]

what

what do you think you’re doing

do you think you’re smart?

I literally mention that in my question

You’re not helping anybody here, buddy

[terse sage]

Chill, do you still need help, I can help further

[steep temple]

no it’s ok

I have my answer

[terse sage]

Cool

[steep temple]

yeah

[terse sage]

No need to be rude, just saying

I’m sorry if I said something wrong

You can also do it by vector components

Like, consider Vector 2 to be the x axis, break Vector 1 into components, add up the components, then use pythagoras for magnitude and tan for angle

[nimble cosmos]

would the period for this be :

$\frac{\pi }{10}$

[somber coyote]

The-Elite:

[nimble cosmos]

@steep temple damn, change your attitude bro. You asked a question and haddeji tried answering it.

[weary drift]

@nimble cosmos 👍🏽

[nimble cosmos]

is that a thumbs up to my answer being correct? @roek

@weary drift

[weary drift]

👍🏽

[nimble cosmos]

lol

okie dokie

[steep temple]

He tried answering it 2 hours later

That’s like trying to put a bandage on someone who died while attempting to put a bandage on them self

[nimble cosmos]

so i have this

and it told me to express it in terms of sinx and cosx

$\sin \left(2\pi \right)\cos \left(x\right)+\cos \left(2\pi \right)\sin \left(x\right)$

[somber coyote]

The-Elite:

[nimble cosmos]

so i did this

but it also told me to simplify my answer

so i was wondering if this was simplified

not really sure

[sly marlin]

not yet

[weary drift]

$\sin(2\pi) = ?$

[sly marlin]

@nimble cosmos do you know that sin 2pi is?

[somber coyote]

RokettoJanpu:

[sly marlin]

in factdo you know that sin is a _________ function?

[nimble cosmos]

wow roketto and element wrote the same thing lmao

nah i don't

[weary drift]

inb4 he says sin is a trig function

[nimble cosmos]

lol

[sly marlin]

what's sin of 100 pi?

[nimble cosmos]

0

oh i sin2pi is 0

i didn't read it properly lol

[weary drift]

$\cos(2\pi)=?$

[somber coyote]

RokettoJanpu:

[nimble cosmos]

1

[weary drift]

[nimble cosmos]

but then i would have no cos left

and the question says to express it as sinx and cosx

$\sin \left(2\pi \right)\cos \left(x\right)+\sin \left(x\right)$

[somber coyote]

The-Elite:

[nimble cosmos]

maybe that is simplified?

this question is very weird

[weary drift]

have you submitted your answer with just sin?

[nimble cosmos]

no

let me try

ok it is right

what a weird question

thx

[weary drift]

np

[nimble cosmos]

i have doubts on this Question

so what i did

was

found $sin\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

[somber coyote]

The-Elite:

[nimble cosmos]

which gave me that

then i found

$sin\left(\frac{\pi }{6}\right)=\frac{1}{2}$

[somber coyote]

The-Elite:

[nimble cosmos]

which gave me that

so now i just add $\frac{\sqrt{2}}{2}+\frac{1}{2}:?$

[somber coyote]

The-Elite:

[nimble cosmos]

oh wait

i got it nvm

i had to use the sum sinacosb+cosasinb thing

[nimble cosmos]

i'm having trouble finding cos(-2pi/3)

[umbral snow]

Which quadrant?

[nimble cosmos]

Q3

[umbral snow]

Very nice. It's a /3, so is it closer to the x axis, or y axis?

[nimble cosmos]

y

[umbral snow]

Okay! Something that is close to the y axis has a cos of ±1/2. But, q3 has a negative cos, so it's -1/2

[nimble cosmos]

yeah

wait you're talking about the x value that is -1/2 right

im trying to use the unit circle

but like i keep getting something else lol

not -1/2

[weary drift]

cos(theta) does give you the x coord if you walk an angle of theta around the unit circle

[nimble cosmos]

im trying to use that

[dark sparrow]

yeah

you even marked the right point

what are you getting if not -1/2

[nimble cosmos]

i thought we have to do opposite over Hypotenuse

right?

[silent plank]

what is the ratio for cos

[nimble cosmos]

adjacent over hyp sory

[weary drift]

you can still draw a triangle inside the unit circle and use trig side ratios if that helps

[nimble cosmos]

yeah im trying to

i keep getting -sqr(3)/2

adjacent would be -sqrt(3)/2 right?

[weary drift]

how?

[dark sparrow]

it's (cos, sin) not (sin, cos)

[weary drift]

cos(theta) does give you the x coord if you walk an angle of theta around the unit circle

[nimble cosmos]

what

so i made a triangle

i have to do adjacent over hyp now right?

[weary drift]

sure

[nimble cosmos]

so -1/2 is the x value right?

and y is -sqrt(3)/2 is y value?

[weary drift]

not the right y value

[nimble cosmos]

so like when you look at the circle, it's like that right?

sorry i meant -sqrt(3)/2 is y value?

[weary drift]

-sqrt(3)/2 is y value
yes

[silent plank]

1 sec, why did you mark the angle at the top?

[nimble cosmos]

cause that's theta

[silent plank]

no

[nimble cosmos]

oh

so where is theta?

[silent plank]

theta is the angle made with the x-axis

[nimble cosmos]

that is x axis tho

im looking at the 3rd quadrant

[weary drift]

are you looking at the point on the unit circle?

[nimble cosmos]

yeah

4pi/3

[weary drift]

put your pencil on the point on the circle and draw upward til you hit the x axis

[silent plank]

you're marking the angle made with the y-axis

[weary drift]

that will be one of your legs

for the hypotenuse, put your pencil on the point on the circle and draw a line connecting it to the origin

[weary drift]

now you have the right (pun intended) triangle

[nimble cosmos]

isn't that the x axis?

[weary drift]

that is

[nimble cosmos]

oh i get it

[silent plank]

and the yellow section you highlighted is clearly not the angle made with the x-axis

[nimble cosmos]

?

like that?

[weary drift]

👍🏽

[nimble cosmos]

oh ok

so since i was drawing the wrong triangle

i was getting the wrong ratio

lol

how do i know where to draw the triangle?

so i just take the point

and connect it to the origin

and thats my hypotenuse?

[silent plank]

yes

[nimble cosmos]

i see

[silent plank]

(for unit circle it will always have length 1)

[nimble cosmos]

thanks guys this is making sense now.

okay

[silent plank]

and then draw a vertical line to the x-axis

[nimble cosmos]

ok

[upper karma]

guys need a bit of help

does altitude split the angle of the vertex into half

?

[dark sparrow]

...context?

[upper karma]

uhh

I am trying to do this geometery problem

[dark sparrow]

then post the problem

[upper karma]

if my phone had any charge I would

lemme check I kept it for charging

[dark sparrow]

i mean, no, the altitude doesn't bisect the angle at the vertex it's from. not generally.

the bisector does.

[upper karma]

sec

Ill send pic

I'm trying to find the angle which is adjacent to the 70 degrees

[sly marlin]

looks messy

let me try to figure out what it's saying

can you label some vertices?

[upper karma]

k I'll draw it again

I want to find $\phi$

[somber coyote]

epelta:

[unkempt jay]

wot

so

what is top right angle

and top left

700?

oh

70

[silent plank]

are you given any info about triangle ABD?

[unkempt jay]

what is top left

4?

[upper karma]

yes

I know measure of AC

and I know measure of AB

what do you mean by top left?

@unkempt jay ?

[unkempt jay]

nvm i sleep

@silent plank help him pls ty

[upper karma]

I hate you

[unkempt jay]

ok fine

[upper karma]

[unkempt jay]

what is the measure CBD

😄

[upper karma]

no idea on that

[silent plank]

you seem to be omitting some key piece of information

[upper karma]

it's not a problem that was given to me

it's a problem I made up

[unkempt jay]

what

what is the measure of CBD?

[upper karma]

I've been trying to find $\phi$ for last

[somber coyote]

epelta:

[upper karma]

30 minutes

what do you mean by measure of CBD

[silent plank]

as given, there isn't enough information to find that angle

[unkempt jay]

uh

[upper karma]

AC measure is given

[unkempt jay]

this angle

what is that

4?

[upper karma]

we dont know that...

it's the other way of writing $\phi$

[somber coyote]

epelta:

[unkempt jay]

ok

unsolvable

you dont have enough info

[upper karma]

well

this was the orginal question

[unkempt jay]

send

[upper karma]

https://cdn.discordapp.com/attachments/490971739061354506/629940368036855821/image0.jpg

it's a physics question

after reading I concluded either their method is wrong or it is implicitly meant that

wind exerts same speed in vertical and horizontal directions

if so then angle of $\phi$ is 45 degree

[somber coyote]

epelta:

[upper karma]

I was trying to prove that

@unkempt jay

[unkempt jay]

i dont

want to

read that

:c

[sly marlin]

it's just vector subtraction

[upper karma]

yuo told me

there is not enough information

[unkempt jay]

ok 45-38 = 7

[upper karma]

to deduce angle

....

[unkempt jay]

with the question you gave me

[upper karma]

if you can't deduce angle

[unkempt jay]

there isnt enough info

[sly marlin]

you didn't give the lengths

originally

[upper karma]

BRO

[unkempt jay]

what

OH

THERES LENGTHS

in the handwritten problem

[upper karma]

....

.-.

[unkempt jay]

what

[upper karma]

I told that there was measure before

[unkempt jay]

you gave us a triangle with one known angle

and thats it

[sly marlin]

yeah, should have given the explicit values too

[unkempt jay]

and half of another angle

@sly marlin take over

😄

[upper karma]

ok fine, let's just solve the problem

[unkempt jay]

ty

nighty night night

[upper karma]

@unkempt jay alright goodnight buddy

[unkempt jay]

thanks epe

[sly marlin]

so here, just decompose your 38 into a horizontal and vertical component

that makes it easier to do vector subtraction

[upper karma]

um

they say the vertical component is equal to the total horizontal component

rip

😦

[sly marlin]

what?

really?

argh

I don't see that, where is it?

[unborn bane]

No they don't that's a printing mistake

In 3.85 they meant $v_{wx} = v_{totx}$

[somber coyote]

Abrar:

[unborn bane]

and in 3.86 they meant $v_{wx} = -13$

[somber coyote]

Abrar:

[unborn bane]

the real value of $v_{wy}$ is given in 3.88

[somber coyote]

Abrar:

[upper karma]

@unborn bane How do u know

[unborn bane]

I saw the pic he gave me

I mean not me

the pic he uploaded

https://cdn.discordapp.com/attachments/490971739061354506/629940368036855821/image0.jpg

[upper karma]

Great

Thank y9u!

[steep temple]

The cosine rule is c^2 = a^ + b^2 - 2abcos(C)
so why the hell is this post
(https://www.quora.com/What-is-the-resultant-of-2-forces-each-equal-to-P-making-an-angle-x)
using c^2 = a^ + b^2 + 2abcos(C)

[mossy plinth]

Imagine a polygon with n-sides with each interior angles incrementing by 5 degree each time, from 120 degree until it becomes an enclosed polygon. How many sides does this polygon have? Answer: 9 sides.

I found the answer to this question, but I'm still not satisfied with the fact that I have to add the exterior angles with slowly until it reaches 360. I want to know if there is another quick way to solve it. I tried using the Arithmetic Sequence, but it doesn't work if I don't find the last value or the n-sides (which is pointless).

I know that the sum of the exterior angles equals 360, but I can't derive the number of sides given the 1st value and the sum.

[steep temple]

you have literally 10 question channels to ask this in

[mossy plinth]

they seem occupied, so I don't want to interrupt them.

[dusty coral]

hey guys, i have a question, i need to find the area of the shaded equilateral triangles, all 3 inside the trapezoid are equilateral

so i got 16, while others are arguing it is 8 sqrt 3, what is the correct answer?

<@&286206848099549185> insanely important question, sorry for the ping

[steep temple]

do you people realise help channels exist

[upper karma]

You ping 15 minutes after posting the question.

[steep temple]

"insanely important"

[upper karma]

Also each side of the triangle should be 4.

[steep temple]

🤦

[upper karma]

!15min

!15m

!min15

I don't think speyr working.

dambn it

The cmd is 15m lol.

!15m

there

I got it

[dusty coral]

yeah my bad thats why i broke the rule oof "insanely important", its for the greater good

[mossy plinth]

the area of one of the equilateral is 4 sqrt 3?

so it's 8 sqrt 3 because area of an equilateral is sqrt 3/4 (a^2)

a is the side of the equilateral

[dusty coral]

ok so you solved for height?

what is wrong with my method, combining both triangles into a square?

[silent plank]

the altitude isn't 4

[mossy plinth]

I think you are mistaking to base times height

[dusty coral]

yeah but the base is?

[mossy plinth]

An equilateral triangle area can be found by base x height or sqrt 3/4 x b^2

the base is 4

since all equilateral triangle are equilateral

[dusty coral]

i dont understand what i did wrong in my method

[silent plank]

the altitude isn't 4

[mossy plinth]

^

[dusty coral]

the base is though

[silent plank]

base is 4 thats fine

[dusty coral]

i combined the two triangles into a square

[mossy plinth]

ok can you distinguish a slant of a triangle is not the height of the triangle?

[silent plank]

it doesnt form a square but a rectangle

[mossy plinth]

you can find the height using the Pythagorean

[dusty coral]

why does it form a rectangle?

[silent plank]

the altitude isn't 4

[dusty coral]

shit i thought they formed squares

[mossy plinth]

you are basically using a base and diagonal to find the area.

[lone scaffold]

what is this property called?

[silent plank]

assuming those lines are supposed to be straight, vertical angles are equal

[steep temple]

of course theyre meant to be straight stop being sarcastic

[unborn bane]

lol

[steep temple]

@lone scaffold https://www.mathsisfun.com/geometry/vertical-angles.html

[lone scaffold]

@steep temple thank you

[novel flax]

they are called vertical angles

[mossy plinth]

Ur late.

[zenith ember]

@upper karma Here.
$sin(x) + cos(x) = \frac{O}{H} + \frac{A}{H} = \frac{1}{2}$

[somber coyote]

samanthaCS:

[zenith ember]

Also, $O^2 + A^2 = H^2$

[somber coyote]

samanthaCS:

[zenith ember]

And you want to find $\frac{OA}{H^2}$

[somber coyote]

samanthaCS:

[wind heart]

uh excuse me

I need a bit of help with 16

How do I find EOF if it’s not giving me any numbers?

oh wait I’m stupid

it’s 45 since it’s a right angle lmao

nvm

[mossy plinth]

lmao

[upper karma]

a diameter of a circle joins the points C (-7, -4) and D(-1, 10). What are the coordinates of the centre of the circle?

I need step by step for this question please

Please help me out

A parallelogram has vertices A(-2, -2), B(3, 3), C(7, 4), and D(2, -1). verify that the diagonals bisect each other.

I need help with bot of these step by step as well

[regal void]

I don't remember formulas, but use the midpoint formula to solve the first one

The center of the circle is the center of the diameter

[upper karma]

How?

Idk what to do

😦

[regal void]

Basically find the avg of the points

[upper karma]

OKok got that

[upper karma]

But how do I do for a circle

I know this formula already

but how do I find centre of circle?

with those coordinates

[regal void]

The center of a circle is the center of the diameter of the circle

[upper karma]

So...

what is the formula ?

[regal void]

You use the midpoint formula

[upper karma]

OHhh

[regal void]

Use those two points and plug them in

[upper karma]

its basically no need

for the circle

[regal void]

???

[upper karma]

theres no need right

[regal void]

No need for what

[upper karma]

Like forget the circle

Its the same as a line?

@regal void how bout the parallelogram?

A parallelogram has vertices A(-2, -2), B(3, 3), C(7, 4), and D(2, -1). verify that the diagonals bisect each other.
I need help with this one as well

[regal void]

No the circle is important, just understand that all points on a circle are equidistant from the center. If you were to draw a line between opposite points of the circle, it would surely go through the center of the circle.

Since the two points are of equal distance from the center, then we can conclude that the center divides the segment (the diameter) into two equal diameters (radii). Therefore the center must be the midpoint of the diameter

[sly marlin]

you just need to find the midpoint of both diagonals @upper karma

[regal void]

^^

[upper karma]

@sly marlin

how do I tho?

[rapid nexus]

how is $1/tan(tan^-1(2)) = 1/2$

[somber coyote]

M3RCURY:

[pallid lily]

tan and tan-1 cancel and you get 2

1/2

[regal void]

^^

[pallid lily]

they're inverses of each other

like taking ln of e^x

[upper karma]

@regal void

[regal void]

Yeah

[upper karma]

how do i do that I am so confused on this midpoint

for parrallelogram

For the circle its just the formula

So nothing I do

just using the formula

[sly marlin]

@upper karma just take opposite vertices and find the midpoint

[upper karma]

But for the parrallelogram how do I find the midpoint

ohhh

@sly marlin ok what after

[sly marlin]

you are done

[upper karma]

then how do I "verify'

[sly marlin]

after you find both midpoints

find both midpoints

[upper karma]

?

affter your find midopoitns

[sly marlin]

then tell me if they are different

[upper karma]

find midpoints?

[regal void]

If midpoints are equal then I think they bisect each other

[upper karma]

😄

[regal void]

Wait no

Yes

NVM

[upper karma]

@sly marlin

if midpoints= its bisect?

or no?

[sly marlin]

what do you think?

[upper karma]

I think it does

[regal void]

Try visualizing it if it help

[upper karma]

@sly marlin

am i right

[dark sparrow]

$\frac{1}{\tan(\tan^{-1}(2))} = \frac{1}{2}$

[upper karma]

Just say I a,?

[somber coyote]

Ann:

[dark sparrow]

@rapid nexus typed that up properly

[upper karma]

@sly marlin its differnt

I checked

[sly marlin]

are you sure you took opposite points?

[upper karma]

Yes

2.5,1

[sly marlin]

what are your pairs of opposite points?

[upper karma]

A and C

B and D

[sly marlin]

and both of them gave you (2.5, 1) as the midpoint?

[upper karma]

yes

[regal void]

Wait, then why did you say it was different?

[upper karma]

OH,

I meant they were the same

Therefore, they bisect

So yes I was right 😄

Last question of the day: The endpoints of AB are A(10, 16) and B(-6, -12). Find the coordinates of the points that divide the segment into four equal parts.

[regal void]

Find midpoint of AB first

[upper karma]

ok

[regal void]

Call the midpoint C

[upper karma]

or m

[regal void]

It doesn't matter

[upper karma]

yes]

[regal void]

Just to simplify

Fine

Call the midpoint of AB m

Find M

[upper karma]

????

[regal void]

What's wrong?

[upper karma]

Mid points are (2,2)

I meant mid point is

[regal void]

That is correct

[upper karma]

ok

im done/

[regal void]

No you aren't

[upper karma]

No?!

[regal void]

We have to find the points that divide AB into 4 equal line segments

So far we only divided it into two equal line segments

1-->1/2 +1/2

[upper karma]

now I divide those midpoints

by 2 again

correct?

[regal void]

Find the midpoints of AM and MB

[upper karma]

OHhhh

I see

[regal void]

To make a total of four segments

[upper karma]

now I find midpoint of midpoint

[regal void]

Yes

[upper karma]

so howmany should I have

[regal void]

AM and AB

[upper karma]

3 right?

M

Am

AC

AB

[regal void]

I meant MB

[upper karma]

Not ac

So thre are 3 mid points

M

AM

BM

RIght?

[regal void]

So you should have AB midpoint, AM midpoint, and MB midpoint

[upper karma]

OH

And M

so thats 4

Oh kkkkk

[regal void]

No

three

[upper karma]

But I got one

already

[regal void]

AM midpoint is just M

[upper karma]

M

OH

[regal void]

So you need to find two more

[upper karma]

so I fount the mid point of ab

now I find the midpoint of A(and midpoint) and B(and midpoint)

@regal void RIght?

[regal void]

Yup

[upper karma]

thx

😄

[regal void]

Yw

[upper karma]

ok

[regal void]

I have to eat dinner

[upper karma]

so 1. (2, 2)

2. (6, 9)

3. (-2,-5)

@regal void

can you check if I got this right?

[regal void]

You got it right

[upper karma]

yessss

thanks

have a great dinner

[rapid nexus]

$3sin^{2}2θ + 8cos2θ = 0 for 0° ≤ θ ≤ 180°$

[somber coyote]

M3RCURY:

[rapid nexus]

i got this question in a past paper and have no idea what to do after the first step

and i have to solve it

my first step was

$3s^{2} + 8c = 0$

[somber coyote]

M3RCURY:

[dark sparrow]

your first step was to write an equation in 1 variable as an equation in 2?

[rapid nexus]

s is short for sin and c is short for cos

[dark sparrow]

ok well you haven't done anything then

beyond that abbreviation

[rapid nexus]

no

[dark sparrow]

consider rewriting sin^2(2θ) as 1 - cos^2(2θ)

also

when writing in latex, use \sin and \cos

[nimble cosmos]

https://cdn.discordapp.com/attachments/387449305885048852/630221963800870942/image0.jpg

for Question 12), would Opposite = -1 and Hypotenuse = -2?

[dark sparrow]

no because (-1)/(-2) = 1/2 != -1/2

[rapid nexus]

why is sin^2(2θ) the same as 1 - cos^2(2θ)

[dark sparrow]

... sin^2(x)+cos^2(x)=1

[rapid nexus]

oh yes

[rapid nexus]

trigonomic equations, do you always have to get rid of the sin ?

[steady forum]

Help R formula question part 1 2 3

Please dm me

How make 0<x+1<2pi Become 1<2x+1<4pi+1

[somber coyote]

JY1853:

[steady forum]

thank you

[maiden rain]

Guys, how am I supposed to solve this ?

[quiet mason]

sin(2x)=2sinxcosx

[dark sparrow]

let $x = \arcsin(5/7)$, then by definition $x \in [0, \pi/2]$ and $\sin(x) = \frac57$. you're asked for the value of $\sin(2x)$, which can be rewritten as $2\sin(x)\cos(x)$ as lionel just wrote

[somber coyote]

Ann:

[neat schooner]

hey can someone help me with simple trigonometry

I never learned it but I just need it for one thing im doing rn

[maiden rain]

Guys I know how to solve Sin(5pi/6) for example... But I don't understand what is Sin(1/3) ... is Sin(1/3) a thing ?

[dark sparrow]

sin(1/3) is absolutely a thing but you don't... care about that

it's sin(x) = 1/3, not x = sin(1/3)

[maiden rain]

So I'm just supposed to put everything in its place without actually solving the problem yeah?

oh

Aw Thanks A lot.. I didn't notice..

[dark sparrow]

you aren't supposed to find x and y themselves

[maiden rain]

Yeah Thanks a lot..

I just want to make sure I'm on the right path.. I figured out that Cos(x) = 1/3 ,right?

[dark sparrow]

no, cos(x) is not 1/3 in that problem

[nimble cosmos]

why does cotx go from left to right?

like its up then goes down

from left to right

[dark sparrow]

...are you asking why cot(x) is a decreasing function of x

[nimble cosmos]

yeah

[dark sparrow]

bc tan is increasing, and cot is its reciprocal? not sure what kind of explanation you're after

[nimble cosmos]

hmm true

okay

thanks

[upper karma]

The problem is: A and B are Alternate exterior angles formed by two parrarel lines cut by a transversal. Find mB if mA=38 degrees

How can I do that?

[dark sparrow]

make a picture.

[upper karma]

There is no picture

But how do I make it?

Like the shapes

[dark sparrow]

do you know what two parallel lines cut by a transversal look like

[upper karma]

Nope

That?

[dark sparrow]

less clutter but yes

[upper karma]

So I draw that?

[dark sparrow]

no

you draw

two parallel lines

and a transversal

[upper karma]

So this

[dark sparrow]

too cluttered

[upper karma]

That

So?

[dark sparrow]

...

[upper karma]

I have bad drawing

But I just need help

I just don’t understand the problem

[maiden rain]

Can someone explain to me how 6-3/15 = 1/5 ?

[random glacier]

(6-3)/15=3/15
From here, you can split 15 into 3x5, so
3/15=3/5x3. You can cancel the 3's, and you're left with 1/5

Very similar to how 2/4=1/2

you can divide both the numerator and denominator by 2

[maiden rain]

So I can't simplify this any further, right?

[gritty siren]

you can

6=3×2
15=3×5

[random glacier]

You can simplify that

[maiden rain]

This is wrong...

[long sapphire]

you didn't divide the 3

[maiden rain]

Oh got it now, Thank you guys 🖤 ❤

[maiden rain]

Tan(x)= -1/-3 = 1/3 right?

[nimble cosmos]

yes

https://cdn.discordapp.com/attachments/627656595358089218/630542621139533827/unknown.png

so im learning about the half angle formula

and i was wondering why the 2 beside the theta disappeared in the bottom one

[wary turtle]

Let's say x is theta, and you have "cos(x/2) = √[1+cos(2x)]/2". If you input x/2, what is your output? @nimble cosmos

[nimble cosmos]

x

x

[wary turtle]

Yes.

Because "2 × x/2 = x".

[nimble cosmos]

oh

i see

why did they do theta/2 tho?

why didn't they just square root both sides to get rid of the ^2

[wary turtle]

There should be theta/2 in the first equation, not only theta.

[nimble cosmos]

oh

i understand this better thanks to you

thank youuu

[wary turtle]

You're welcome.

[nimble cosmos]

why do we put the +- beside the square root for this?

[wary turtle]

Because it can result in both positive and negative results.

[nimble cosmos]

oh

true

[wary turtle]

For an instance: let's say you have "x^2 = 4".

"x = √2^2" or "x = √(-2)^2".

[upper karma]

$x^2=4 \ x=\pm\sqrt4=\pm2$

[somber coyote]

leviosa:

[nimble cosmos]

true

[fringe breach]

i don’t think so

[crude hornet]

Find the set of values of k for which the line y=2x - k meets the curve y = x^2 + kx - 2 at two distinct points.

I can't wrap my head around this one

[keen scaffold]

So uh

Anyone up and awake enough to help me?

[dark sparrow]

there are at nearly all times people awake enough to help you. this is a pretty big server.

what do you need help with?

[keen scaffold]

Hih i suppose so since you're responding to me lol

I just need a lil help with Geometry stuff

Like definitions of stuff i mean

Imma send a photo

[dark sparrow]

yes, please do that

[keen scaffold]

As you can see i got all of them wrong :)

[dark sparrow]

ohhh fucking yikes those things

[keen scaffold]

?

Postulate flashbacks?

[dark sparrow]

no, this entire exercise is just... really bad by design

but then idk how to make it better

and like

this just reeks of two column proofs

which are their own can of worms

just... yuck. sorry, i cannot help with this.

[keen scaffold]

Ah its cool

How is it bad though?

If i may ask

Also anyone else up for it? :(

[weary drift]

https://mathwithbaddrawings.com/2013/10/16/two-column-proofs-that-two-column-proofs-are-terrible/

[keen scaffold]

Ah

Not really helping but its alright

[upper karma]

Idg what the worksheet is asking.

Wait nvm.

[keen scaffold]

?

[upper karma]

I'm seeing like random things.

[keen scaffold]

Oh my

Oh well

Looks fine here

[reef flower]

Need help with this I'm struggling to get a right answer... I think I know the way to get the answer it's just not correct.

Find the line k through T(0,0,3) that is perpendicular to the plane P: 5x-2y+13z=-14. Find coordinates for the point U where the line k intersects the plane P.

[reef flower]

Still stuck on this, this is my working so far:

The direction vector for the normal to P can be found by looking at the coefficients of P; (5,-2,13). We let U = (a,b,c), a point on P. UT = T-U = (0,0,3)-(a,b,c) = (-a,-b,3-c). Also, UT = t(5,-2,13) where t can be any real numbered parameter. Therefore (-a,-b,3-c)=t(5,-2,13). Hence, a=-5t, b=2t, c=3-13t. Using b=2t, t=b/2. Soooo... a=-5(b/2), b=b, c=3-13(b/2). Since U is a point on P it must satisfy P therefore 5(-5b/2)-2b-13(13b/2)+3=-14. Solving for b, b=17/99. Plugging b into a,b,c in terms of b the answer I get is incorrect. I have no idea why im pulling my hair out rn eek.

[weary drift]

@reef flower remember, all you need to construct line k is a single point and a direction vector

you found P's normal vector and you know k must pass through T. how can you construct k using just that?

[tawny tendon]

@crude hornet pretty late but if you still need the help:
Since the two curves intersect, 2x - k = x^2 + kx - 2.
Rearrange to get a quadratic in x
When does a quadratic equation have two distinct roots?

[maiden rain]

How come the function Cot(x) is an odd function ??

[weary drift]

cot(x) = cos(x)/sin(x). check whether those functions are odd/even

[maiden rain]

cos(x) is even

sin(x) is odd

[dark sparrow]

yeah

even/odd is odd

[maiden rain]

Ok, But isn't right that just by looking at cot(x)s graph you'd say that it's neither odd nor even ?

[dark sparrow]

no, it's odd just fine

rotate it half a full turn around (0,0) and it lands right back on itself

[maiden rain]

So even/odd is odd...

tan(x) is odd as well and it's odd/even...

[weary drift]

odd/even is odd as well

[dark sparrow]

1/odd is odd.

[zenith ember]

I usually find it helpful to work with the definition

For this sort of thing.

[wind heart]

Uhhhh I kind of need help

Usually I don’t have a problem with these

But since there’s more than one variable it’s kind of throwing me off

[olive solar]

this is a combination of using vertical angles as well as supplementary angles

[wind heart]

So would I do it like

9x + 20 + 7x = 180 to find x?

And then divide x by 2 to find y?

or no?

[olive solar]

you can find x and y like that yea, as well as z

[wind heart]

let’s see

[olive solar]

just try it out, i can check your work if need be

[wind heart]

Is that correct?

Also, is there an easier way of finding z then just guessing what would fit in with the variable?

Like, for the next question it’s not as easy since it’s harder to guess what would fit in the variable

[olive solar]

sorry, i didn't see the messages

look at the value of z again

the vertical angle shows that 5z = 110

so z would not be 110

@wind heart

[wind heart]

oh

[olive solar]

also for the second question, use vertical angles to find x

7x-26 = 3x+34

[wind heart]

I got 79

For x

But I’m kind of confused on how to find y and z for the second question

[olive solar]

you got x = 79?

or you got 7x-26 = 79

[wind heart]

oh wait 15*

[olive solar]

lololol

anyways, to find y and z

[wind heart]

((I think it’s 15))