geometry-and-trigonometry

[olive cove]

Which you should be able to solve

[jaunty plank]

Did u mean 11.5 ? Or 11,5

[humble pulsar]

11,5 is 11.5

[olive cove]

11.5 sorry in my country you use a comma

I forgot about it

[jaunty plank]

Alright ty

[olive cove]

Np

[jaunty plank]

Alright I solved it I got 3.9196 then I needed to round to the nearest tenth foot. And got 3.9 but it’s wrong

@olive cove

[open spindle]

Can someone help me w these?

[olive cove]

,w solve cos(11.5)=x/40

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Jupieterrr#8346

[olive cove]

@jaunty plank I think you made a calculation mistake.

It should be 19.3

[olive cove]

Wait let me check everything

Oh sorry it's my fault

I got confused because you wrote the 40 at the hypotenuse but it's the vertical drop

You have to use tan

,w solve tan(11.5)=40/x

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Jupieterrr#8346

[olive cove]

Is 22.1 correct?

[jaunty plank]

Nope

I think there’s an error

[olive cove]

Try 196.6

That's what my calculator says

[jaunty plank]

Yes

It’s correct

@olive cove

Ty

[olive cove]

I just realized that Wolfram alpha uses radians as a standard so that's why it was wrong.

[jaunty plank]

Ohhh

So the equation stays the same

tan(11.5)=40/x ?

[olive cove]

Yeah

[jaunty plank]

Alright

[olive cove]

But 11.5 radians is something different than degrees

[open spindle]

Can someone help me w this one

[neat gale]

Who here is good with the Pythagorean theorem

[prime linden]

[lament cobalt]

Im confused on how to solve for this. I know that we have to use Pyhthagorean theorem, But Idk how to label which side is

[silent plank]

identify your longest side in the triangle

[naive tapir]

so we have a picture like this

We can take the base of our triangle to be the line segment connecting J and K

We know the area is base * height/2

I would try to determine the height in terms r, t for any point L

should involve finding the line orthogonal to JK

And then projecting L onto it

does that make sense?

Ping me if you have any questions shiv

[lusty osprey]

oh wow thanks boss I think I know how to the problem. It's easier to draw out the problem and think of it geometrically

ill finish the problem tmr and lyk if i have any questions thank you 😋

[green osprey]

how doe

i dont know where to even start

[green osprey]

<@&286206848099549185>

<@&286206848099549185> i need help please

[short arch]

The c should be the longest I believe

[green osprey]

damn

ffs

[green osprey]

can someone help me with this please i really dont understand how

[cyan shadow]

Does anyone know where they got the negative?

Just a quick question.

[hard geyser]

notice sin(a)=cos(90-a)

or some meme

90 degress ofc

[cyan shadow]

Yes

I understand how they got sin2wt

[hard geyser]

and sin is even

[cyan shadow]

Not the negative sine

[hard geyser]

odd*

so like

cos(90+x)=cos(90-(-x))=sin(-x)=-sin(x)

or something

[cyan shadow]

Right

But sine isn't negative

[hard geyser]

hm?

[cyan shadow]

cos (2wt +90 - 90) = sin(2wt)

[hard geyser]

call 2wt=x

and do that thing

[cyan shadow]

it'd still be sinx

There's no negative

[hard geyser]

here

dont ask me why there is an omega and a W tho

I assume w=omega W?

[cyan shadow]

Well omega, it's part of a formula

Was trying to just understand the trig portion on how they got neg

[hard geyser]

well yes

its what I said

[silent plank]

isolate x,
then its just a matter of entering stuff into your calculator

(in this context, you're working with degrees so make sure your calculator is set to degrees)

[green osprey]

I think i know it

is it 2(9x-7) = 17x-2

and i end up with 12?

[modern brook]

pls help

[wintry tundra]

@modern brook when u need help with something just share it outright instead of asking first

but sure ill help if i can

[modern brook]

this

[modern brook]

thank u :))

[green osprey]

can someone please help

please

[sand bay]

can someone help, imma litte confused

[green osprey]

bruh

i just posted here

srsly?

[sleek light]

Ummm

[wintry tundra]

@sleek light definition of trig functions

@modern brook as for u if ur still up what u should do is use triangle theorems to prove that the two triangles are similar

[sleek light]

❤️❤️❤️

[wintry tundra]

for the section under saying what the sinusoids on triangles are u have to use law of sines or law of cosines to prove each one

[modern brook]

oh yeah, that's what i know but where should i start

[wintry tundra]

well ED seems to be a section of AD

and DC seems to be a section of DB

and AD and DB connect

and so do ED and DC of course

which means they make the same angle at the same point

[meager carbon]

whats answer to question 1

[strange wigeon]

SOHCAHTOA

[severe latch]

Anyone help me

[olive cove]

What's the question?

[upper karma]

"At what time is the water level the deepest according to $y = 2+3.5sin(\frac{\pi(t-1)}{6})? The time t, is hours after midnight. Water levels may be negative".$

[somber coyote]

Fish of the sea

[upper karma]

I know that the lowest function value will be when the sin function = -1

i.e 1.5m (2-3.5)

after simplifying I get $-1 = sin(\frac{\pi (t-1)}{6})$

[somber coyote]

Fish of the sea

[upper karma]

however solving this gives me t = 12n+10

the correct answer should be t = 10

it only makes sense if n = 0

is it really that n = 0 or am I missing something?

[heady tartan]

How do you graph this?

That’s the graph but I don’t understand how -11/2 graphs to that

[green osprey]

How do i do this?

[silent plank]

mostly apply properties of right triangles and radii of circles

[green osprey]

I know I learned this a while ago lol, I'll ask a specific question, how do i figure out where the line crosses

is it like 1/2 of 60 or something?

[silent plank]

apply perpendicular chord bisection theorem (which could be justified with triangle congruence)

[green osprey]

ok thank you so much!

wait how would i find YD though?

@silent plank ?

[silent plank]

construct YA and/or YB

apply properties of right triangles and radii of circles

[green osprey]

ok

Thanks!

[green osprey]

How would i do this?

i put the numbers there to try to guess

wait are AC and AB and BC equivalent?

<@&286206848099549185>

[warm sphinx]

College Trig question

On my end.

how do you go from:
Cos(x) = -1/2
to
Quadrant II --> 120deg x pi/180 = 2pi/3

and finding all the other quadrants

I dont understand

Quadrant III --> 240deg x pi/180 = 4pi/3

I just dont understand how you get there

(its the reference angle topic)

and on the other side... I am at the step of Cos(x) = 1
Quadrant I --> 0
Quadrant IV --> (would be 360 but you cant include it in the equation)

[warm sphinx]

actually better yet....

[humble pulsar]

it's a quadratic in disguise

[upper karma]

How do I do this, I have a few problems on it and Im lost.

[humble pulsar]

$1+\frac{3}{7}$

[somber coyote]

moshill1

[foggy talon]

Can someone help me with this one

[heavy timber]

woudnt it be SAS?

[waxen glen]

can someone prove it that AKC and DKC triangles are similiar triangles

[slender sonnet]

Will some one have time to help me really quick

[sullen jasper]

how do i do this

[olive cove]

You know soh cah toa?

[sullen jasper]

yup

but when I use it

I get the wrong answer

[olive cove]

What did you try?

[sullen jasper]

I got 48.8 for Z and 5.3 for X

I used TOA

[olive cove]

For which one?

[sullen jasper]

X

[olive cove]

Can you show your work?

[olive cove]

You have to make an equation

[sullen jasper]

I did Tan(30)x9.2 to get 5.3

[olive cove]

Oh

You have to multiply both sides of the equation by x

Because you are dividing by X and not 9.2

[sullen jasper]

oh ok so

Tan(30) x X

and 9.2 x X?

another answer i got for the hypotenuse was 4.6

[olive cove]

No the equation is
tan(30)=9.2/x
Multiply both sides by x
tan(30)*x = 9.2
And then divide both sides by tan(30),
x = 9.2/tan(30)

[sullen jasper]

Oh ok

[olive cove]

And for z you use the same technique but then with sinus.

[sullen jasper]

thank you

[olive cove]

No problem

[sullen jasper]

@olive cove so is it x=15.9 and z = 18.4

[slender sonnet]

Is someone one is willing to help I have a question

[lunar hull]

they'll never know if they're willing to help if they don't know the question

[toxic trench]

what is the problem?

[serene obsidian]

@toxic trench help?

[wintry tundra]

@serene obsidian use pythagorean theorem

And also don't mention people for help unless it's the helpers mention

[serene obsidian]

My bad

Ok

[wintry tundra]

ur alright

[serene obsidian]

ok so

i got sqrt 20

which is 4.47

and if i round to nearest 10th

4.5?

is that correct

[wintry tundra]

4+x = 49

x^2*

x = sqrt45

25 + x^2 = 45

x = sqrt 20

so

[serene obsidian]

Perimeter is 18 feet and area is 62.4

Correct?

[wintry tundra]

probably is

i know perimeter is

i got 2.598 for the area lol

[serene obsidian]

Wtf

How

[wintry tundra]

7.794*

forgot to add coefficient

[serene obsidian]

Still

[wintry tundra]

1/2 bh

base is 6

height is 3sqrt3 / 2

wait hold up

[serene obsidian]

...

<@&286206848099549185>

Oh shit

Whoops

[wintry tundra]

ok revised answer 15.588

[serene obsidian]

Ok so that would 15.6

For area

[wintry tundra]

height of a equilateral triangle is always a sqrt3 / 2, a being side length

[serene obsidian]

U 100% sure? Bc I got 36 root 3

For the area

[wintry tundra]

$\fracf{a \sqrt{3}}{2}$

[somber coyote]

hiidostuff
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wintry tundra]

wtf

oh i see the issue

$\frac{a \sqrt{3}}{2}$

[somber coyote]

hiidostuff

[wintry tundra]

ugh

i did it wrong my brain is bad

$a \sqrt{3}$

[somber coyote]

hiidostuff

[wintry tundra]

ok here it is

[serene obsidian]

K

[wintry tundra]

no no nvm first one was right

im so sorry

[serene obsidian]

LMAO

[wintry tundra]

i got home at like 6 pm

[serene obsidian]

Alr u 100% sure

It’s 15.6

Btw I rounded to nearest tenth

[wintry tundra]

thats fine

im uhh

kinda sure

[arctic mirage]

the area for that problem is 15.59 rounded

the first problem you sent

[serene obsidian]

It says nearest tenth tho

O

This one?

[arctic mirage]

o, wasnt paying that close of attention, then the answer is 15.6

[serene obsidian]

Wait which one

[arctic mirage]

the one where the area was 15.6 obviously

this one

for the problem you deleted for some reason i got x = 5.0, y=22.4, and z=11.2 might be wrong tho its been a while since ive had to solve a problem like this

[serene obsidian]

@arctic mirage

I got x = 5, y = 22.4, z = 11.2

Yep

GG

This is the last one I have to do

And it’s one I have no clue

I think it’s 300 smth tho

[wintry tundra]

uhh

im weird with geometry so idk

[serene obsidian]

Help pls

If you help me uhh

U get a cookie

[wintry tundra]

dang i want cookie

but tirangle

oh ohhhhh

find the hypotenuse of the bottom triangle

do some trig to get the angles

find the complement of the angle between the hypotenuse and the leg thats 91 ft long

then do some triangle things

[serene obsidian]

Still don’t get it :/

[wintry tundra]

ok

a^2 + b^2 = c^2

(91.25)^2 + (5)^2 = c^2

go

[serene obsidian]

Ok so

I got

c = 91.38 or -91.38

Dam

91.4 ?

Is that correct

Can someone double check with me and confirm 100%

[wintry tundra]

hmm

i think

sounds about right

oh so now do tan(x)= 5/91.25

do u understand why?

[serene obsidian]

That’s x = 0.05

K bc it’s my last question ima get confirmation

[serene obsidian]

Last question

[upper karma]

Identify the similar triangles here

[slender sonnet]

If someone could please help me it would be greatly appreciate

[west basin]

What's the question?

[wintry tundra]

i would assume its to find the volume

@slender sonnet can u be more specific

[merry fiber]

Could someone pls help me with this? I need to find those two angles

[copper marten]

idk how to find x but y is more easier to find:

since the whole triangle is a right triangle and there are already two given angle measures: 25 and 90. you can add them up which is 115, then subtract 180-115 because the sum of angle measures in a triangle is 180. 180-115 = 65. angle y = 65 degrees

@merry fiber

[merry fiber]

Ohh thank you so much

[copper marten]

no prob !

[dim cliff]

could someone help me with this problem?

[wintry tundra]

hmm

well theres a lot of identities to be used here

firstly, what is tan x equal to?

[dim cliff]

the directions only say prove the identity 😦

[wintry tundra]

yep

btw it is true

but what is tan x?

@dim cliff

[dim cliff]

tbh idk, thats all the sheet says

[wintry tundra]

well whats the definition of tan x?

is what i should say

[dim cliff]

opposite/adj.?

[wintry tundra]

yeah

but think about this

tan x is also sin x / cos x

[dim cliff]

oooh i thought you meant in general but yeah in this prob yes

[wintry tundra]

alright so now we have $\frac{\cos^2{x} - 1}{\cos{x}} = -\frac{\sin{x}}{\cos{x}}\sin{x}$

[somber coyote]

hiidostuff

[wintry tundra]

so what is cos^2 x - 1?

[dim cliff]

-sin^2(x)?

[wintry tundra]

yep

so now we have $\frac{-\sin^2{x}}{\cos{x}} = -\frac{\sin{x}}{\cos{x}}\sin{x}$

[somber coyote]

hiidostuff

[wintry tundra]

and i think u can see how it works from then on @dim cliff

[dim cliff]

yes i do! thank you so much!

[wintry tundra]

no problem

[storm elm]

Could someone help me with this x/tan 28 = 480-x/tan 34

This is my original problem if it will help

[vocal lodge]

help

[delicate hawk]

You sure this is all that’s given?

Or is an answer in terms of x acceptable?

[upper karma]

ABCD = rectangle, BEFG=square, then area PRC=?

[vocal lodge]

yep thats all thats given

you have to find x

so anyone know how to do it?

[delicate hawk]

I don’t think this works out, angle bisector theorem gets you one equation with two unknowns. I don’t see a way to find one of the unknowns, nor do I see a way to get another equation. Maybe someone else can spot sth

[vocal lodge]

should i ping helpers or somthing

[vocal lodge]

<@&286206848099549185> does anyone know how to do this

[deft dagger]

you have to find x
@vocal lodge let em know

[cold rune]

Yo can I get some help here

[sharp hull]

Hey! Do you guys know a site that can show what all of the possible angles on the unit circle of a certain function ?
So like if I input sin x = pi/3 or sin^-1(pi/3) it would show me both angles that get outputted

[orchid crystal]

hello guys

[twilit plover]

where do i go from 48 = r^2 * h/3 (volume of smaller cone)

am i supposed to set some sort of ratio?

[serene obsidian]

Can someone help me idrk how to do this

[twilit plover]

when the longest side is c and the other two sides are a and b:

acute triangle: a^2 + b^2 < c^2
obtuse triangle: a^2 + b^2 > c^2
right triangle: a^2 + b^2 = c^2

[serene obsidian]

thx

can u help for one more

x = 12, y = 20.8

round to nearest tenth btw

[humble pulsar]

,w 24sin(pi/3)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by moshill1#9771

[humble pulsar]

yes that's right

[twilit plover]

can someone help w this

[humble pulsar]

if they have equal volume, they must have half the original volume

oh it's asking for height

[twilit plover]

yeah

is the ratio of radius to altitude the same for each cone?

like would i use that somehow?

[humble pulsar]

yes

since the plane used to cut is parallel to the base, you will have similar triangles

[humble pulsar]

so find the volume of the cone half, then solve for the height

[vocal lodge]

#letemknow

[west basin]

What trig identity do you use for sides opposite and adjacent to an angle?

[arctic mirage]

tangent

SOH CAH TOA

[west basin]

Sorry I should of specified that was directed at Escort

I have indeed passed geometry

[upper karma]

Someone please help me

How do I figure this out

[twin prawn]

126 degrees?

[upper karma]

is it actually

[twin prawn]

I mean the angle of the circular arc is really just the angle between the radii that bound the arc

[upper karma]

so the answer is right in front of me...? 😐

[twin prawn]

As weird as it might seem, that's what it looks like to me

[upper karma]

so then

What would be the measure of ABC

Ohh it’s half of that?

[twin prawn]

No

[upper karma]

then how do i figure that out

[twin prawn]

There's a theorem that tells you the angle between a tangent and a radius that meet at the same point

[upper karma]

let me find it

Tangent Radius Theorem

a line is tangent to a circle if and only if the line is perpendicular to the radius drawn to the point of tangency.

[twin prawn]

Yes, perpendicular

[upper karma]

Ohhhh

my answer was radius = 9 . Diameter = 18 and circumference is 56.54 but I think that’s wrong

[twin prawn]

,calc 2 * pi * 9

[somber coyote]

Result:

56.548667764616

[upper karma]

oh

Thnxxxx😎

ONE MORE QUESTION

Do you know how I’d get the measure for C

I m really stupid

[twin prawn]

Calculate the three other angles of the quadrilateral ABCD and subtract them from the total angle sum of a quadrilateral

[upper karma]

uhhhh

i don’t get it I’ll just guess

[twin prawn]

[serene obsidian]

Help pls

[storm portal]

@serene obsidian wrong category, and also wait at least 15 minutes for someone to help you

read #❓how-to-get-help

[serene obsidian]

wait wut

oh probability

my bad

[storm portal]

all good

[upper karma]

https://cdn.discordapp.com/attachments/326138737606262786/817525717638053908/unknown.png

[west basin]

use a system of two equations

The corresponding sides will be scalar multiples of each other

[smoky crescent]

Ik this is probably a very easy problem but im stuck:

[upper karma]

what have you tried so far?

[smoky crescent]

Well ik a straight line is 180 so i tried doing something like: (11 * x + 19) + 44 + y = 180 but im pretty sure im on the wrong track

[upper karma]

no, that's totally correct.

now you may want to consider another information you have about the angle y

[smoky crescent]

yeah thats were i got stuck hehe

[upper karma]

yeah so, try staring at the diagram and look for pairs of angles that satisfy a well-known theorem

it may help to go over your notes over theorems you may have over pairs of angles.

[smoky crescent]

ok 👍 ill look

is the 2x and 44 degree angle a pair? @upper karma i cant tell if the line is centered or not

[upper karma]

a pair of what?

wdym by centered

there are different theorems/characteristics about certain types of pairs of angles

[smoky crescent]

Like are they the same angle in degrees? because i cant tell but i think they are

[upper karma]

2x and 44° aren't a pair of congruent angles, keep on searching and trying, you'll eventually come up with the pair that we are looking for.

we are almost there

[smoky crescent]

Ok thanks for all your help i looked over my notes and i think ive got the answer to what y and x =, thankyouuuuuuuuuuuuuuuuuuuuu

[cold rune]

Can someone please help

[upper karma]

https://cdn.discordapp.com/attachments/326138737606262786/817525717638053908/unknown.png I still need help lol

[wintry tundra]

well the triangles have to be similar to each other since they are dilations so just set some ratios

[hollow raven]

@cold rune Here's a start:

c^2 = (a^3 + b^3 + c^3)/(a + b + c)

c^2 * ( a + b + c ) = a^3 + b^3 + c^3

c^2 * (a + b) + c^3 = a^3 + b^3 + c^3 -> Simplify
c^2 * ( a + b ) = a^3 + b^3 -> Rewrite cubic binomial

[lament cobalt]

I know that I have to redraw the triangles, But I am bad at drawing them and I end up getting the wrong answers

[lethal mist]

LMN~LKM~MKN

[fast pulsar]

So this is just in case you don't have it yet. You are absolutely right, the angle of a line is 180°, but it seems you were looking at a line that isn't very useful yet.

Why not instead look at the line with angle
2x+11x+19+44=180?

Then by solving for x, you can then solve for y like you were wanting

[smoky crescent]

Yeah i ended up solving for x and then found y

[fast pulsar]

Okay good. Glad you got it!

[smoky crescent]

thanks 🙂

[lament cobalt]

thank you👍

[umbral viper]

Hey is anyone on?

[lethal mist]

Perhaps

[compact jasper]

i got close cubic spline https://i.stack.imgur.com/JvRpV.png . How best to get interpolation from center ? Just get center pointer?

[surreal bolt]

How many points do you have of data?

If you're assuming it's a circle, Take any one point of data and find its distance from every other point of data. Figure out the expected value of such a calculation for N points of data and work it backwards.

[delicate hawk]

What kind of interpolation are you looking for anyway? You can find the centroid and use the mean distance from it to approximate with a circle.

You can make it a bit closer using the max and the min distance to do an ellipse

Discarding extreme points can help make it more accurate for the most amount of points

[upper karma]

AB=AC, (...) then a=?

[fast pulsar]

Where is this a you speak of?

[main rivet]

that would be the angle thats shaded yellow pretty sure

[upper karma]

yellow

[upper karma]

yes

[main rivet]

hmm

well

a+b = 180- (16+28+30+14)

@upper karma

i feel like theres something missing

or im just dumb

do you know the length of any of the sides?

[upper karma]

nope

[waxen glen]

I want to learn trigonometry and geometry, which books should i read first

i know basics

[surreal bolt]

You usually learn geometry first. Do you know what sine and cosine are?

[waxen glen]

no i dont know

[surreal bolt]

What grade are you in and what nation (if I may ask)?

Geometry is usually a high school freshman course. Khan Academy teaches math at a generally relaxed pace. If you think Khan Academy is boring, look into Art of Problem Solving books on geometry.

[waxen glen]

i knew Khan Academy but i want textbook too

an i am in 9th grade

it can be difficult textbook too i just want to learn geometry and trigonometry very well

[surreal bolt]

I don't have great rec's for textbooks. If I had to give you advice, I'd say go look at geometry books (if you're in the USA) at Amazon.com and check out the ones with at least 30 ratings with high number of stars. Then pick the top few and read the written reviews and see which book matches your wants.

But if you want a tough book ... The Art of Problem Solving series is usually nice.

[pure cape]

you should label some of the points to make it easier

[tawny cypress]

what do u mean

[pure cape]

and notice that all of the sides are rectangles, meaning they are normal to each other

[tawny cypress]

im quit confused on to where d cuts

[pure cape]

label as in writing A,B,C,...

[tawny cypress]

ohh

wait

but like it's cutting both the small and big rectangle right?

[pure cape]

what is?

[tawny cypress]

diagonal d

[pure cape]

no, its not going thru any of the rectangles, its INSIDE the rectangle, wait, lemme just write something down

,rotate

[tawny cypress]

yeah ik that but im basically saying like from what point to what point is it touching

[pure cape]

Not a very good drawing

But diagonal d is basically the line EB

[tawny cypress]

i still dont see a triangle

[pure cape]

Usually if you want to find a length of a line in spatial geometry, you would want to establish triangles

Yes

So you have to make one

For example, maybe something like triangle EBG?

Since BG is normal to the plane AFGH

EBG is a right triangle

That can be helpful

[tawny cypress]

wdym wait

like this?

[pure cape]

No

[pure cape]

This

Yes

That triangle

[tawny cypress]

is the 90 degree angle B?

[pure cape]

No

Remember, BG is normal to (AFGH)

Hence BG normal to EG

Tho you will need to prove that BG is normal to (AFGH)

[tawny cypress]

im so lost

[pure cape]

Do you know what normal means?

[tawny cypress]

nope

[pure cape]

Normal is kind of like perpendicular

[tawny cypress]

ooh

[pure cape]

Im not really familiar with english, but i think you would usually use normal for a line and a plane

And perpendicular for a line and a line

So basically the angle between BG and the plane (EFGH) is 90 degree

[tawny cypress]

i havent reallt got taught well with planes so i wouldnt know

[pure cape]

Hmmm ok

So basically when a line is normal to a plane

That line will be perpendicular to every lines in that plane

So if BG is normal to (EFGH), BG is perpendicular to EF or EG or any lines laying on the plane (EFGH)

[tawny cypress]

im using this as reference

[pure cape]

Yes sure

[tawny cypress]

how would EG and BG be perp

[pure cape]

like i said above

[tawny cypress]

i never learnt that wooahh

[pure cape]

Oh and i forgot

To prove that line A is normal to plane (d), you have to prove that line A is perpendicular to TWO INTERSECTING LINE on plane (d)

So in this example, BG is perp to HG, because ABHG is a rectangle, and BG is also perp to FG, because CBGF is a rectangle

and since HG and FG intersects at G

BG is normal to (EFGH)

[tawny cypress]

ohhhhh

i seee

that makes sense tysm

so to solve for d aka line EB, we first find line EG?

which is 3.6, so EB would equal 6.997?

[pure cape]

Yes

Wait, lemme check the numbers

But the method is right

For EG you should just keep it as sqrt(13)

And round your number when you found EB

(If needed)

[tawny cypress]

OHHH smart smart

okieee i get it now TYSM

you're a very good teacher btw

[unborn bison]

how does this middle one work?

number 2

[tiny snow]

@unborn bison $\triangle KLM \sim \triangle GML$ and why are the points renamed there?

[somber coyote]

Roenbaeck

[unborn bison]

oh

so how do I solve for x, @tiny snow

something to do with geometric mean?

[tiny snow]

@unborn bison I'd set up some equations, given that the triangles are similar, so their sides are proportional. If that still leaves you with some unknown, use Pythagoras as well, and you'd surely have an equation system that can be solved.

[unborn bison]

hmm

should I use Geometric Mean because I have 15 and 5?

[tiny snow]

@unborn bison If you are allowed to use the geometric mean theorem, you could use that, yes.

[sullen jasper]

I have to find AC how do I do that?

[severe latch]

Can anyone help me out ?

I nned help for the second one

[hearty iris]

thats AAS postulate

subtract 118 and 22 from 180 to figure out the angle c

and then use the law of sines to work it out

heres an example of how to do it

[quaint plinth]

Hi there. I am trying to find the area of my living room. However I don't know math well. I just measured the length and the width, but it is a trapezoid. so if I multiply the width with length, I get more space than what it is. Can someone help?

It says here if I take the length of the long side, as well as the length of the short side, then add those, then take the depth as well, and times that with depth, then divide by 2. I will get the correct answer.

[quaint plinth]

Yeah, that seemed like it worked.

[quaint plinth]

huh, its 350 square feet. nice. Now I know.

[wintry tundra]

for future reference, try dividing an odd shape into shapes whose areas can be measure with simple geometry, like how a rectangles area can be measure with base * height and how a triangles area can be measured with half of that

@quaint plinth

[upper karma]

Can someone explain rq what law of cosines is

[wintry wave]

Hi! Um,,,,can someone help me out with this?

[autumn quiver]

For this, you just need to use trigonometry

SohCahToa

[tiny snow]

@severe latch You mean 7?

Oh, I was not looking at the most recent messages.

Introduce the green line as in this sketch, that should help you.

@WOWWA#3261 ^

[waxen glen]

https://cdn.discordapp.com/attachments/490557019623915520/818491072589594645/unknown.png

[gloomy mesa]

which line is 4 on

@waxen glen

[waxen glen]

EK is 4

DE is x

AD is 15

[gloomy mesa]

and you just have to solve this triangle?

prism i mean

[waxen glen]

i need to find x

[gloomy mesa]

start out with the numbers already given

and pythagora theroum for more numbers

[waxen glen]

but i cant

i need atleast 2 lines of triangles for this

but i know 1

[gloomy mesa]

are there any angles

[waxen glen]

only right angles

90 ones

[gloomy mesa]

hm

[waxen glen]

someone told me that i can solve this with similar triangles thing

[gloomy mesa]

you might have to create a few polynomials and once found the variable do multiplication ex 4x

[waxen glen]

but for that i need to prove ADC and BDC are similars

or AKC and BDC

[gloomy mesa]

yeah try the similar triangles first

[waxen glen]

I tried but i failed

[jaunty plank]

@olive cove could u help me ?

[olive cove]

Uhh I don't know how you know me but sure

Start by drawing the triangle

[jaunty plank]

cause of last time

And alright

Nvm I got it

[next jackal]

Why are inverse trig functions constricted to only two quadrants?

How are we supposed to know about the quadrants?

[lament cobalt]

I need help with geometry homework. I don’t know how I am supposed to label which side is opposite,Adjacent,and hypotenuse

[honest jolt]

@lament cobalt here is the same triangle from another perspective:

this help?

[lament cobalt]

I think I have it. When I tried that I got the answer as cos

[honest jolt]

nice

[lament cobalt]

thank you👍

[honest jolt]

np 😄

[unborn bison]

https://cdn.discordapp.com/attachments/802245352072675399/818632733001515028/Screen_Shot_2021-03-08_at_6.54.04_PM.png

how does #7 work?

[fast pulsar]

Shouldn't the triangles be similar, not congruent?

[unborn bison]

idk, @fast pulsar

thats what it says on page

yeah they should

[fast pulsar]

Okay so if I am not mistaken, the ratios of the sides should be the same.
So $\frac{6+2}{6}=\frac{5+x}{5}$
That should get you both your x and y values

[somber coyote]

dackid

[unborn bison]

ok

ty

[scarlet basin]

im so confused

how on earth do i do this

[upper karma]

for the below triangles the angles are 30 90 60

[wintry tundra]

wait how can we be sure kimo

are u responding to kold or asking ur own question

[upper karma]

responding but i also have question

[wintry tundra]

oh ok

we cant be sure the triangles are 30 60 90s tho

[upper karma]

361=490.69-489cos(A)

[wintry tundra]

what

[upper karma]

i just need help solving it

law of cosines

[wintry tundra]

so ur trying to find the cosine of angle a?

[upper karma]

law of cosines

not trig functions

[wintry tundra]

well i mean

a cosine is a trig function

and since cos a is unsolved im guessing u wanna solve for cosine

[upper karma]

nvm got it

thanks anyway

[wintry tundra]

cool

[fast pulsar]

Read my answer directly above this picture. It is the exact same type of problem.

[jaunty plank]

Someone help ?

[humble pulsar]

apply law of cosines like it says

[hybrid solstice]

can someone help please?

[humble pulsar]

pythagorean theorem

[hybrid solstice]

ohh

[jaunty plank]

I did I got 36.9396 and it still says wrong

[humble pulsar]

write out the calculation?

[jaunty plank]

1/2 times 10 times 12 sin 38

@humble pulsar

[humble pulsar]

um

no

that's area of a triangle

[jaunty plank]

ohhhh

[humble pulsar]

$c^2=a^2+b^2-2ab\cos{C}$

[somber coyote]

moshill1

[humble pulsar]

cosine law

[jaunty plank]

Ty

@humble pulsar

I got it

[swift osprey]

Can anyone help with #2

[mint sand]

Should be simple for some of you but I'm a little stuck on a

Worked the rest out no bother but I can't find a rule that suits a

[fast pulsar]

I do not encourage you to simply solve the problem for them. That does not help them understand it.

[hardy pasture]

I need help with these proofs if anyone can help. :)

I am unsure how to use the Reflexive Property or where to go from ST = TU

If anyone responds please ping me, I have tons of discord servers, thank you.

[fast pulsar]

@hardy pasture so in order to show two triangles are congruent, one way is to show all the sides are the same

We've looked at two sides for each triangle, but there is one more side we need to look at.

[hardy pasture]

Like this?

[slate pebble]

what is this website

[fast pulsar]

Ayyy, you got it!

[hardy pasture]

Would it be the same case here?

[fast pulsar]

You sent the exact same picture so yes. 😆

[hardy pasture]

sorry

[fast pulsar]

If it were different, why do you think it would be?

[hardy pasture]

I mean the angles and sides are different, but I was referring to the concepts.

[fast pulsar]

Does anything about this suggest the approach to solving it should be different?

[hardy pasture]

No, I do not believe so.

[fast pulsar]

And your intuition is correct.

[hardy pasture]

Thank you, one last question.

I tend to doubt myself even if I believe something is correct. Will you look over this, and let me know if you think its correct?

[fast pulsar]

It's pretty clear you don't need my help 😉

Have a little more confidence mate.

You're doing just fine

[hardy pasture]

Okay last last one.

What property would I use to state that two triangles share a set of angles or sides?

Wouldn't it be, SAS?

[fast pulsar]

Is this after you prove congruence?

[hardy pasture]

Also couldnt the final answer also be CPCTC?

[hardy pasture]

Yes.

[fast pulsar]

What is CPCTC?

[hardy pasture]

Corresponding Parts of Congruent Triangles are Congruent.

[fast pulsar]

Ah okay. That is my thought on your most recent question

[hardy pasture]

Wait what?

[fast pulsar]

I know SAS and SSS, and I also believe ASA is one.

However, both AAS and ASS are not ways you can show congruence.

[hardy pasture]

Perfect thank you!

[fast pulsar]

So first thing is first, what is the measure of a right angle?

[copper marten]

1. its a right angle so work from there
2. whats the sum of angles on a straight line? and work from there

[fast pulsar]

Okay, since the sum of the angles 75° and x are a right angle, how can we find x?

[fast pulsar]

See if this helps

Think about what you just said and ask yourself if that claim makes any sense.

p. s. I will absolutely help you, but I will not give you the answer

Why does what you said make sense to you?

You are in the wrong server for that

There are also rules that must be followed.

And this

This server is exactly what it's designed for. To band people that are doing math together and help out the people that are struggling.
Regardless of the language, I am still willing to help. But once again, I have no intention of giving away the answer

You can do what you will with that

[rough stone]

180-75= the missing degrees. If it is a 180 degree angle and they give you an angle then you need to subtract to achieve the answer.
This is an example not the answer. I hope that helps you find the value of x.

[fast pulsar]

Yes, x=15 in your problem

[rough stone]

So go back to your picture or the picture given and figure out what values you need to use

To find the value of x

[fast pulsar]

I think they just did shiro

[rough stone]

Oh okay

Hey dackid do you know trigonometry?

[fast pulsar]

Indeed I do

[rough stone]

Okay so I have a question if you don’t mind. I can break down the unit circle but when it comes down to like 5pi/12 I have issues with identifying what the angle is.

Do you know an easier way to identify the (x,y) or the angle.

[livid moss]

Isn't 5π/12 the angle?

[rough stone]

I mean identifying it on the unit circle itself

If that makes sense.

[fast pulsar]

Sure!

So 6π\12 is a quarter of the circle right?

[rough stone]

Do I need to multiply 2/2 by pi/6 to obtain it

So when I make the physical drawings of the unit circle I just get so confused

[fast pulsar]

Do we agree that is where 6π\12 should be?

[rough stone]

Yes

[fast pulsar]

Okay, now remove a 1/6th of that quarter of a circle and you have 5π\12

[rough stone]

Will I not get an x and y value

Or will it be really weird or calculator based

[fast pulsar]

Yes, but it can be particularly difficult to find

[rough stone]

Gotcha thank you

[fast pulsar]

There is a nice way to do it. But if you haven't learned yet, then don't worry about it

[rough stone]

I’m currently in college trig but I think I’m getting lost and behind. So I may have lots of questions. :/

[fast pulsar]

So you haven't taken calculus yet?

[rough stone]

Not yet

[fast pulsar]

Dang! Euler's formula makes some trig identities so easy to produce

[fast pulsar]

Don't be afraid to ask :)

[rough stone]

Thank you! 🙂

[onyx viper]

ye

[jovial axle]

@onyx viper

$\cos ^2 x + \sin ^2 x = 1 \to \cos x = \sqrt{ 1- \sin ^2 x }$

[somber coyote]

jan Niku

[wintry tundra]

Why are u guys mentioning specific people

[jovial axle]

its a secret

[wintry tundra]

Well Jan is fine cuz ur prob helping someone

Eh whatever

[jovial axle]

i am

i thought theyd come here

ill just use #bots tho

[wintry tundra]

@upper karma if 180 - 77 = 103 and 103 > 90 then ur all good

What are u trying to find

Well how many degrees is a circle

Not edge degrees

How many

Ok

So u have 250 of those degrees

And abc makes up the rest of the circle

Do u have the answer?

Huh

Well that just happened

[onyx viper]

did you also get a random ping

[wintry tundra]

Well I told the person to stop mentioning specific people

Then they told me to stfu and purged their messages

[onyx viper]

what

[wintry tundra]

Yeah

Idk

[onyx viper]

i can still see them though

[wintry tundra]

Huh

I was trying to help

[upper karma]

yea i just got randomly pinged

by that person

[onyx viper]

weird

[wintry tundra]

No its the apostrophe person

They canceled their messages

[onyx viper]

oh weird

[wintry tundra]

Yeah

[onyx viper]

your right

[wintry tundra]

I was trying to help them and gave em pretty good clues

Then they got mad

[onyx viper]

yea and it was not that bad of a question

[wintry tundra]

Cuz they kept mentioning other people

[upper karma]

yea

it was crazy

[wintry tundra]

Uh

[upper karma]

wonder where they went

[wintry tundra]

U know that you are the one who did it right @upper karma

[upper karma]

wth

[wintry tundra]

That's what I'm thinking rn lol

[onyx viper]

I think that they are trolling or hacked

[wintry tundra]

Prob

It's a pretty obvious question too so

That's prob the case

[rough stone]

@fast pulsar is this correct?

[wintry tundra]

Wait ur trying to solve sin x / cos x - 1?

Cuz there's a lot to do with identities here

Trig is so fun

[rough stone]

Yeah I am

I’m not sure if I got the right answer here

[wintry tundra]

Hmm

Well using a conjugate is a good idea

Hmm

How did u get sin²x on the bottom

[rough stone]

From the identity

The pathagorean identity

[wintry tundra]

Hmm

[rough stone]

Sin^2theta + cos^2theta = 1

[wintry tundra]

Yeah

So sin^2 x * cos^2 x + 1 = cos^2 x - 1?

[rough stone]

Yes that’s what I assume

Through the identity

[wintry tundra]

I wouldnt agree

I dont see how

[rough stone]

I was told to do each section at a time so if cos^2 -1 if I were to set to zero I would get sin^2x+cos^2x=1

Hmm

[wintry tundra]

But it isn't set to 0

[rough stone]

Am I on the right path?

True

[wintry tundra]

We can't know

We are simplifying not solving

But hey I mean it was a good idea

[rough stone]

So then at cos^2x-1 can it be canceled by the numerator

Before distributive property is done

To sinx(cosx+1)

[wintry tundra]

well

If u cancel

U get $\frac{\sin{x}}{\cos{x} - 1}$

[somber coyote]

hiidostuff

[wintry tundra]

Which is unfortunately right where we started

Honestly I don't see why this would need to be simplified

Why are u simplifying it?

[rough stone]

It is apart of the hw question

[wintry tundra]

What's the whole thing? Because this seems already fairly simplified

[rough stone]

So this is the original equation

To multiply and then simplify

[wintry tundra]

Ahhh I see

So it asks u to multiply them together?

[rough stone]

Yes

[wintry tundra]

Huh

So simplify not expand?

[rough stone]

And I got this

[wintry tundra]

Yeah thats correct

[rough stone]

Yup it said to simplify

[wintry tundra]

Strange

That's already pretty simplified

Well I suppose cos² = sin² - 1

[rough stone]

Yeah which is why I had that at the bottom

Denominator

[wintry tundra]

So u then get sin² - 2 at the bottom

Yeah

[rough stone]

So circling back around to the picture am I still wrong

0.o

[wintry tundra]

Honestly

I have no idea

Because to simplify that u would get what the original thing is

Because $\frac{\cos x + 1} {\cos x + 1} = 1$ so uhh

[somber coyote]

hiidostuff

[wintry tundra]

Are u sure it doesn't say cos x - 1 / cos x + 1

[rough stone]

So I got a different answer and now it’s not simplified enough

So the answer was 1-cosx/sinx

X=theta

[wintry tundra]

Ohh cos x - 1

Not + 1

[silent plank]

which part?

[upper karma]

4 and 5

[sullen cipher]

use ur notes bruh

have u learned the altitudes of a right triangle

[rough stone]

I figured mine out ignore it.

[fast pulsar]

Glad you were able to figure it out

[vernal dawn]

does anyone know how to do proofs

i cant seem to get the rgith thing

[fast pulsar]

Care to share what you've tried/ the problem for what you have tried

[paper orbit]

Hey how would I solve an equation like this x√2=9

[wintry tundra]

what makes u think thats geometry

[spring forum]

ive been trying this for like 20 minutes and havent got anywhere

can soemone pls help

[wintry tundra]

hmm

no angle measurements?

[spring forum]

nope

[wintry tundra]

yikes

so ur right about the side length that u actually have

but

its only one side length

[spring forum]

yeaa so idk what to do from here

[wintry tundra]

i guess u could say $\tan{R} = \frac{f}{e}$ and $\tan{\gamma} = \frac{f}{d}

[somber coyote]

hiidostuff
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wintry tundra]

so $e\tan{R} = d\tan{\gamma}$

[somber coyote]

hiidostuff

[wintry tundra]

and u know e and d so $1.5\tan{R} = 9.5\tan{\gamma}$

[somber coyote]

hiidostuff

[wintry tundra]

and this is where it gets difficult...

i guess $\frac{1.5}{9.5} = \frac{\tan{\gamma}}{\tan{R}}$

[somber coyote]

hiidostuff

[spring forum]

🥲

[wintry tundra]

yeah this is pretty painful

but perhaps we can end up getting a single function

perhaps

[spring forum]

my brain is 🤯 right now

ugh my test is friday too

[wintry tundra]

f

alright so tan = sin/cos

so we have to look back at what tan gamma and tan R is

oooh

i think we can solve for f

hopefully

so we have $\frac{\tan{\gamma}}{\tan{R}} = \frac{f}{d} / \frac{f}{e}$

[somber coyote]

hiidostuff

[wintry tundra]

crap

the fs cancel out

f

hmm

its like midnight where i am i dont wanna be tired and screw up ur math so id mention helpers

sorry

[spring forum]

ookayy thanks for the help tho

i understood a little bit hehe

<@&286206848099549185>

[lament kraken]

what do the angles around a point add up to?

can you write down an equation linking the angles in the diagram to that?

[neon hamlet]

Would this be correct?

[blazing pond]

guys plz

were having a war

is this correct?