geometry-and-trigonometry

[forest niche]

Anyone?

[forest niche]

Please someone

I got this as my answer

[shut crest]

I think that's quesion is flawed

Wait @forest niche Isn't the answer 0.00264105642257

[elfin heath]

That question is flawed. How many cards are you drawing?

[forest niche]

5 hearts in a row

[shut crest]

Then 0.00264105642257

[forest niche]

How

Wait I got this now

[shut crest]

Assuming you don't put the card back in the deck, there's a 13/52 chance of picking a heart the first time, 12/51, then 11/50, and 10/49

[elfin heath]

I believe your reasoning is correct.

[shut crest]

My advice is also right out what you assumed the problem to be, becouse it wasn't very specific

[forest niche]

This is what I did

[shut crest]

Oh that's right, I'm dumb

[forest niche]

Oh ok thank you

I think I got it

[north sandal]

A boat sails 20km north, then sails 40km on a bearing of 135 degrees.

a) How far is the boat from its original position?

I'm pretty sure the triangle isn't a right angled triangle and there isn't enough information to use the cosine rule, so I don't know what to do

[mystic shadow]

Split the 40km into a north component and a west component using the 135 degrees

Add the north component the the 20 km north

Then use pythagoras to calculate how for the boat is from its original position

@north sandal

This should be plenty of information

Going to sleep

[north sandal]

How do you split it into north and west components?

Quickly before you go to sleep, soz @mystic shadow

[elfin heath]

So, draw a diagram, starting from a point. Draw a line up and mark it as 20km. Bearing is 135°, or in the south-eastern direction. From the end of the first line, draw a line in the south-eastern direction marking 40km.

[north sandal]

done

[elfin heath]

You should have a 45° between the 20km and 40km lines. I believe that is enough for you to use the Law of Cosines.

[north sandal]

But I don't know the angle from where the boat finished to where it started

[elfin heath]

You don't need to. c² = a² + b² - 2ab cos C, where angle C is opposite c.

So a = 20, b = 40 and C = 45°.

[north sandal]

oh

ok thanks!

[pure pivot]

I got 1 over 100

Apparently that’s not right

[dark sparrow]

how did you get 1/100

[pure pivot]

Well since the radius is 1

[dark sparrow]

what radius

[pure pivot]

I did 1 square pie

[dark sparrow]

the radius of what

[pure pivot]

Thickness

Circle

[dark sparrow]

the radius of what circle

[pure pivot]

Oh the center

One

[dark sparrow]

you want the OUTERMOST section, not the INNERMOST one.

[pure pivot]

How would I get anything if they are both 10 pie

Every circle gets one more inch

[dark sparrow]

first off, it's pi, not pie. second, the other sections are not circles

[pure pivot]

Sorry 😂

[dark sparrow]

they are rings

circular rings to be more specific

[fossil lotus]

The whole circle has a radius of 10

[pure pivot]

So what am I doing wrong then

Yea ik

[fossil lotus]

Not 10 pi, just 10

[pure pivot]

I don’t get this at all

[fossil lotus]

The outermost section has an area equal to the area of the whole circle minus the combined area of the other sections

[pure pivot]

Minus the other?

[fossil lotus]

Yes

[pure pivot]

So everything but the outer?

[fossil lotus]

Yes

[pure pivot]

Ok I’ll try that rn

[fossil lotus]

(everything) - (everything except outer section) = (outer section)

[upper karma]

How do you find missing sides in a triangular prism?

Anyone?

,rotate 270

[spark stag]

You can't, based on that picture alone

Unless the triangle is assumed to be equilateral

In which case, all the sides of the triangle are just 8 cm

[upper karma]

So what you're saying is they're all 8?

Cause apart from that there's no more info

[spark stag]

If they're equilateral

[upper karma]

And I wanna be 100% sure I'm getting this right

[spark stag]

The question doesn't specify, so I'd assume they meant for them to be equilateral

[upper karma]

It's a triangular prism so I'd assume it's equilateral

[spark stag]

But they should've worded it better

Nah, triangle prisms don't have to be equilateral

Picture a cube, then cut it in half diagonally

[upper karma]

Wait, oh my God dude

[spark stag]

You just made 2 right triangle prisms

[upper karma]

My teacher put a note on the bottom right corner saying in worn out pen "problem 6 is equilateral"

[spark stag]

Well, that explains that

[upper karma]

i have a triangle HES
s = 18
h = 21
e = 24
how do i find H
i tried
21^2 = 18^2 + 24^2 - 2(18)(24)cos(H)
but it didnt work

If the radius of circle C is 8 and BD=100 degrees what is BD?

@spark stag do you know this too?

[spark stag]

...yes, what is BD

I need information about what BD represente

To tell what you're asking

@upper karma what do you mean, "it didn't work"?

God damn, having those characters in a message screws up discord

But like, did you get the wrong answer? Do you not know how to solve for H from there?

[upper karma]

i figured out the problem before you replied but thanks anyway

[spark stag]

Ah, kk

:)

[upper karma]

👍

Ok

A solid with a volume of 820cm3 is reduced by a linear scale factor of 1/2 what id it's volume? @spark stag

[spark stag]

All the sides are scaled by a factor of 1/2

If a volume is a product of 3 sides

That means it's scaled by (1/2)^3

So, 820 × (1/2)³

[and don't forget your units]

[upper karma]

I don't own a calculator

[spark stag]

,calc 820/8

Bleh

It's 102.5

[upper karma]

That's the answer?

[spark stag]

cm³, yes

[somber coyote]

Result:

102.5

[spark stag]

...

Thanks

[rustic cairn]

What do I do to solve this

[upper karma]

how would i find the value of h

i know how i could do it without cosine law but thats not what the teacher wants

the answer is 3.2m btw but i dont know how to achieve that

[olive solar]

wait what

h is 3.2?

[elfin heath]

That's not possible.

[olive solar]

doesn't that mean that the leg of a triangle is longer than the hypotenuse

[elfin heath]

Unless the h is in the wrong place.

[upper karma]

Two wires are supporting a tent pole as shown, how apart are the wires from the ground

the diagram dosent show h i just assumed that what h would be

[elfin heath]

I think h is the hypotenuse. There's no way that h can be 3.2m.

[olive solar]

well

no hypotenuse in this case

not a right triangle

but the third larger side yeah

because it cannot be the segment in the middle

[elfin heath]

,w sqrt(2.2² + 2.8² - 2(2.2)(2.8) cos 80°

[olive solar]

wolfram dead

wait

not wolfram

texit lol

all the bots dead rn

[elfin heath]

I'll manually type that into my calculator then.

[upper karma]

what he said gives 3.2 though

so no need

but how did you come up with that?

[somber coyote]

Display results online and refine query

Requested by Azuralytix#0267

[elfin heath]

Yeah. I think it's asking for the distance between the base of the two wires. Which would be the side opposite the angle.

[upper karma]

got it

thanks

you also need to find the angle each wire makes with theground

how would i do that

[elfin heath]

Law of sines?

You know 1 angle and all 3 sides. Should be able to use law of sines to find the other 2 angles.

[upper karma]

thanks!

[fleet lintel]

Anyone online able to help me out with Arcs? I’m terrible at geometry.

Please mention me if so.

[olive solar]

can't help with something you haven't asked

[elfin heath]

Feel free to ask! We'll try to help if we can!

[fleet lintel]

This, if possible. I can’t understand how to do it.

[olive solar]

ah

just look at the fact that DB is a diameter

this means it bisects the circle into two halves

DAB and DCB

[fleet lintel]

How do I find the answer with that information? I’ve read through the notes they give me but I can’t find any formulas.

[elfin heath]

Angles on a straight line add up to 180°.

[olive solar]

it bisects the circle, which is an angle measure of 360 degrees

and given that it's a diameter, you know it's a straight line

[sullen lantern]

Hey guys, here's a question I need help with

I don't understand what line I should be drawing, if anything

[cinder portal]

draw a circle whose diameter is the altitude of C, and AB

i think

[chrome vigil]

how to solve Q11

[dark sparrow]

are you familiar with the sum and difference formulae for sin and cos

[chrome vigil]

yes

[dark sparrow]

so then

what's giving you trouble there

[chrome vigil]

as in i get the formulas how do i apply it here?

[dark sparrow]

$8 \cos(x - \tfrac{\pi}{3})$

[somber coyote]

Ann:

[dark sparrow]

[chrome vigil]

?

[dark sparrow]

well if you say you "get" the formulas

can you state the one that could be applied here

[chrome vigil]

cosAcosB+sinAsinB i guess???

[dark sparrow]

that isn't a formula, that's an expression

[chrome vigil]

cos(A-b)

[dark sparrow]

so state the whole formula again

[chrome vigil]

cos(a-b)=cosacosb+sinasinb

[dark sparrow]

$\cos(a-b) = \cos(a) \cos(b) + \sin(a) \sin(b) \ \cos(x - \tfrac{\pi}{3}) = ; ?$

[somber coyote]

Ann:

[chrome vigil]

cos(x-60)=cosxcos60+sinxsin60 everything in deg dont know how to write pi

[dark sparrow]

also, no, that thing you just wrote is wrong

you're getting there but you shouldn't be writing things that are wrong

[chrome vigil]

cos(x-60)=cosxcos60+sinxsin60 everything in deg dont know how to write pi

[dark sparrow]

ok, great

so now

you're almost there

what are cos(60°) and sin(60°)

[chrome vigil]

root3over2 and 0.5

oops

yeh correct swap them

[dark sparrow]

1/2 and sqrt(3)/2, yes.

are you able to do the last step now?

[chrome vigil]

how do i find c and d the parts in front?

[dark sparrow]

well we've figured out how to rewrite cos(x - π/3) in the form they ask for

what about 8 cos(x - π/3)

[chrome vigil]

ohhhh 4 and 4root3

i get it

thanks

[rustic cairn]

How do I slove this

[twin prawn]

More info needed

[rustic cairn]

It’s a trapezoid

I need to find the missing length and angles @twin prawn

[twin prawn]

But without more info, there could be a lot of possible solutions

[elfin heath]

With only this, 13 < x < 17. There's literally insufficient information to solve for x.

Unless you're literally supposed to measure the length.

[neat grotto]

Guys, how do you solve an oblique triangle with one given side and angle

[idle bloom]

@rustic cairn does it say anything about the middle line besecting the sides of the trapazoid?

@neat grotto can you draw a picture of the triangle and send the picture? but it'll boil down to law of sines and cosines

[neat grotto]

One sec

[idle bloom]

kk

[idle bloom]

aight so

have you learned about the ambiguous case?

[neat grotto]

Nup

Also is there a way we could find theta (in second pic)

[idle bloom]

Uhhh

Maybe

but lets look at the first triangle first

[neat grotto]

Aight. First and second are the same triangles btw

[idle bloom]

ye

[neat grotto]

Ok first one now, how do we solve ee6

It

[idle bloom]

ee6?

oh

yea

[neat grotto]

I actually solved it but want a reliable method

So am asking for help

[idle bloom]

send yer work

or just tell me how you did it

cause I don't think you have enough info

one side and one angle isn't good enough

[neat grotto]

Yeah but if you could solve from that, calculating further objects would be much easier

[idle bloom]

well I don't think you can that's the issue

there are infinitely many triangles that have that configuration

[neat grotto]

Wym by infinite. ..

[idle bloom]

you were only given the one angle and one side

[neat grotto]

Yeah

[idle bloom]

like there is no "one triangle" that has that measure

I can draw, given enough time, an infinite number of triangles that have that configuration

basically there is no one solution

[neat grotto]

Can't we go on until we get an infinitely small number

Like 0.009272 which we could just ignore

I'm trying to make a project for my school function or smth

[idle bloom]

there is no "infinitely small number here"

[neat grotto]

The upper point is the light source

[idle bloom]

I literally mean we don't have enough info to get a triangle

wait so what are you trying to build?

[neat grotto]

Idk

Wait I'll draw another fig

[idle bloom]

but do you see why we need more info?

[neat grotto]

I guess

[idle bloom]

that doesn't sound confident

[neat grotto]

I'll pm

[idle bloom]

ok

[upper karma]

Can someone explain the cosine rule please

[olive solar]

so the cosine rule is a method of finding an unknown side in a non-right triangle given two other sides of the triangle as well as the angle opposite to the side you are trying to find

what do you mean by explain it, exactly?

[upper karma]

I have a test tomorrow

And I need to know this:

[olive solar]

what do you have that confuses you about it?

do you just not know what it is

?

[upper karma]

Joined the class later than most people and having a hard time understanding it

And when to use the different formulas

[olive solar]

ah

hm

this is going to be difficult to explain without drawing a diagram

https://www.mathsisfun.com/algebra/trig-cosine-law.html

so:

law of sines is for when you want to find the side

but all you're given is

two sides and the angle opposite to the side you want to find

http://jwilson.coe.uga.edu/EMAT6680Su11/McNeece/assignment 8/ABC triangle.png

so lets say you have this triangle

triangle ABC

tf embed is gone

w/e

given triangle ABC

you know the length of AC and AB

and the measure of angle BAC

and want to find lenght of BC

[upper karma]

Okay

[olive solar]

now

law of sines

is for when you know length of AB and measure of ACB

and know measure of any of the other angles

and want to find the sides

or

know the length of AB and measure of ACB

you can use the length to find the measure of the other angles

[upper karma]

Okay

[upper karma]

So I don't have a picture of this, but the problem has to do with a circle and it has arc lengths but it wants me to find x

Does anyone know what I'm talking about?

[thorn zinc]

Help

nvm

[upper karma]

If anyone knows what I'm talking about please message me

I can draw an example of it I think

[cinder portal]

just draw the picture and post it here, and we can answer your question!

[upper karma]

It wants me to find X

[cinder portal]

Oh

give me a second

let me pull something up

[upper karma]

I would pull up an example but I can't find it

It does have arc lengths too but I'm looking for what I need to do to get x

[cinder portal]

there is a theorem that states

the arc that is boudned by the two lines

is double the angle x

uhh give me a second

[upper karma]

Hang on, let me type the problem

[upper karma]

The arc lengths that are given are 92 and 101

[cinder portal]

Let me know if that makes sense or not

[upper karma]

So that would bean that the center is half of 92 or 101?

[cinder portal]

wait what

do you think you can take a picture of the whole problem given to yo?

you

[upper karma]

I can't, I drew it but here I'll put where the given lengths are

I'm given these two angles.

[cinder portal]

oh

what is the third arc angle then?

[upper karma]

There's no third arc angle but I'd assume I'd have to add them up and minus 360

[cinder portal]

yup

[upper karma]

So that whole bottom section is 167?

Now how do I find X?

[cinder portal]

Omce you get it, use the theorem i put above

[steady obsidian]

Someone help me with geometry with 3d figures
I just need to verify something
A triangular prism with a volume of (some number here), and a base perimeter of (some number here), the shape is then dilated by 2x. What is the volume and the perimeter.

[upper karma]

Wait, but the whole thing is to find X

[steady obsidian]

Oh your doing circles, with inscribed angles.

Inscribed angles are always 2x less than the arc length

I think^

Well not arc length, arc degrees.

[upper karma]

Wait.

Nyan

[steady obsidian]

Yea?

[upper karma]

So

What you're saying

I take it that it's just a triangle

And I thought from what Memes said that it was 92+101=193 360-193=167 divided by 2

I'm so fucking fucked

[steady obsidian]

Alright first the numbers you stated, were they in degrees?

Because I do not see any units there.

[upper karma]

They both are degrees, I think one of them has a degree mark the other doesn't but I just assume it's a mistake

[steady obsidian]

Okay so to find that arc that doesn't have a number. A circle has 360 degrees right?

[upper karma]

Ok

You need to find x

You're given the center

and the two arc lengths.

[steady obsidian]

Alright, so from what Memes said, an inscribed angle will always be one half of the length of the arc the angle is opening up to.

[upper karma]

Dude

Memes doesn't even know

I dmed him

[steady obsidian]

No it's true, inscribed angles opening to an arc will be one half it's length

So here, as you can see, the angle x is opening to the unknown arc.

[upper karma]

So.

It's 92+101=193 and then 360-193=167 divide by 2 and that = 83.5?

[steady obsidian]

Yes.

[cinder portal]

Im on mobile so i cant whip up any drawings

Like i said, read the theorem i posted above, that should help

Im sorry i got some of them numbers mixed up, it wouldve made it a bit easier if you couldve posted the whole question given to you

[upper karma]

Ok what about this

[upper karma]

I'm doing this right right?

A shape has a volune of 855 and is being decreased by a linear scale factor of 1/3 so is 855 × (1/3)³ right?

[shut crest]

Are conic sectsions grafhed on the plane, or from above?

[cinder portal]

Mind emphasizing? @shut crest

[shut crest]

So like you know how a conic secson is a plane running through a cone?

[cinder portal]

yes

[shut crest]

Well if you where to gragh a cone as x^2+y^2=z on a 3d grapgh, then grapgh mx+y=z on a 3d graph, then grapghed wear they intersect, could you acceritly transcribe it to a 2d grapgh by just knocking of the z part? Or would you need to figure out hoe to rotate it?

[vagrant elk]

You'd have to rotate it in general, since a cone isn't necessarily neatly orthogonal to the usual xy plane

[upper karma]

@shut crest I suggest playing around with this applet

https://www.geogebra.org/m/GmTngth7#material/T8TV2JqG

[cinder portal]

YES THAT

I was going to tell him about it

pretty awesome imo

[idle bloom]

Yea

[upper karma]

855 × (1/3)³

Is that the correct equation for a shape with a volume 855 and decreased by 1/3 right?

855cm³

Did I right the equation

Is it right?

Anyone?

,calc 855 × (1/3)³

[somber coyote]

The following error occured while calculating:
Error: Syntax error in part "³" (char 12)

[quiet mason]

,w ,calculate 855 × (1/3)³

@upper karma

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

Requested by Lionel#5497

[quiet mason]

yes it is right i guess

but i think it needs calculus

,w calculate 855 × (1/3)³

[somber coyote]

Display results online and refine query

Requested by Lionel#5497

[quiet mason]

@upper karma

[fleet lake]

How does Euclids fifth postulate imply the existence if parallel lines

[fleet lake]

<@&286206848099549185>

If a lime l falls on line m and n such that it form interior angles with both sides . Now if the sum of interior angle of one side of Line m is less than 180 then the line when produced indefinelty then the lines will intersect but if the sum of is more than or equal to 180 the line will never meet .

This imply that there exist line which never intersect.

In that sense 5th postulate implies the existence of parallel lines

Am I correct @dark sparrow

I don't think so by the way I am correct

But with help of playfairs axiom I can prove this but is this allowed .

Playfairs is just equivalent version of Euclids fifth postulate

[oblique island]

Hey guys, just wanted to ask if there's any way to show that 4*sin(18)*cos(36) = 1?

[plucky marlin]

!15

[oblique island]

whoops

[urban oxide]

why do ppl use tan

am i weird

that i use tg

[dark sparrow]

yes you are

next question?

[dawn stone]

lol and don't delete the ping after pinging helpers, it just makes it worse.

[fleet lake]

My question remains unanswered

☹

[dark sparrow]

SORRY FOR NOT BEING ABLE TO DEVOTE EVERY FUCKING SECOND OF MY TIME TO YOU THEN

[fleet lake]

[dark sparrow]

WHAT

[fleet lake]

Tbh I didn't understand what did you mean?

English is not my first language.

Maybe you want to say that you didn't have time atm

[dark sparrow]

well given that it's me who you ping 80% of the time when asking questions, you're making me feel obligated to answer your questions, and then you make me feel guilty when nobody responds because, surprise, i sometimes have my own shit to deal with

[fleet lake]

I didn't pinged you. I pinged helpers

[dark sparrow]

you also pinged me personally.

[fleet lake]

Yeah sorry to check am I correct

Sorry i wont ping you must have told me before lol

I am so sorry that it disturbed you

,w obligated

[somber coyote]

Display results online and refine query

Requested by Mr.Why_(Krishna )#2893

[dark sparrow]

don't ask wolframalpha for english word definitions

those are usually bad

or inaccurate

you'd do better to google words you don't know

[near kayak]

i need help making a review for my geometry regent, can anyone help?

[fleet lake]

Ok ann

[wet hill]

Hey, how do I find ACB angle?

[wanton forge]

there is no more information ?

[wet hill]

nope :D

[wanton forge]

weird

[wet hill]

well O is the center but that should already be obvious

i can translate it, it's from 2015 lithuania's math exam :D

Point C belongs to a circle whose center is point O. From point M, that is located outside the circle, two lines are drawn, which touch the circle within points A and B, ∠AOB = 80° (look at the sample).

Calculate ∠ACB size.

[rotund swan]

I think it’s 40°

[wet hill]

the answer is 40, yes

how did you calculate it?

[rotund swan]

Do you know what’s the inscribed angle ?

[wet hill]

no :/

[rotund swan]

I use that theorem

[wet hill]

i see

[rotund swan]

Yep, that’s it

[wet hill]

thank you :)

[rotund swan]

You’re welcome 😉

[wet hill]

is AMB 80 as well?

[rotund swan]

No

[wet hill]

how come?

[rotund swan]

100°

[wet hill]

wth :o

[rotund swan]

Because OAM=OBM=90°

[wet hill]

i see....

80 + 180 + 100 = 360

forgot about 90 degree angles

thanks again :D

[rotund swan]

✋🏻😉

[near kayak]

is anyone good at sohcatoa?

[cinder portal]

[upper karma]

Thank you everyone for helping me

I passed my test.

Thank you Memes.

And thank you Nyan

[craggy spear]

Congrats

[cinder portal]

it was all you

😄

[wraith mango]

In-home assignment which is due tomorrow, kinda stuck here could someone help me out?

[cinder portal]

no.5?

[wraith mango]

Sure @cinder portal

[cinder portal]

draw ur triangle first

and then show me what you get

r is your hypotenuse

[wraith mango]

Well I’ve to figure out the exact value of “x, y and r” only knowing sin0 = 1/3

Which still doesn’t click into my brain

[cinder portal]

what does "sin" mean

[wraith mango]

it’s a function of an angle

[cinder portal]

mhm

[wraith mango]

Opposite of the hypotenuse correct

[cinder portal]

correct

now

construct a right triangle

in whic

the opposite is 1, and the hypotenuse is 3

[cinder portal]

label an angle as theta

[cinder portal]

you're doing just fine

now

the opposite of that angle

is 1

and the hypotenuse, is 3

label it please

[cinder portal]

that looks fine

now

how would you solve for the last piece?

[wraith mango]

No idea to be honest with you I’d love to know tho lol

[cinder portal]

Pythagorean Theorem

[wraith mango]

do I have to substitute something

[cinder portal]

have you learned the Pythagorean Theorem?

[wraith mango]

Like it’s being taught in class but we have a sub for this week

But yes to answer your questions I know the basics

A2 + B2 = C2

so I would be using an angle and the hypotenuse to figure out the last piece correct?

[craggy spear]

Yeah

[cinder portal]

yup

[craggy spear]

Last piece would be sqrt( something )

[wraith mango]

2?

[cinder portal]

try again

[craggy spear]

Don't guess, solve

[wraith mango]

Am I not adding the angle and the side

One sec I think I understand

[craggy spear]

Your doing 3^2 - 1^2 = missing side^2 no?

[wraith mango]

Yea

so can we solve the powers to solve the last variable

Or do I leave them

[shadow anvil]

What do you mean @wraith mango?

You have to solve the powers, you can't just leave it like that

[wraith mango]

Okay

[shadow anvil]

Remember to take the square root of the sum too

Because you'll want a, not a^2

[shadow anvil]

Yep

So now what's a

[wraith mango]

a^2 = 3.16

[shadow anvil]

No, a=3.16

not a^2

[wraith mango]

Oh

[shadow anvil]

because we removed the power of 2 by square rooting it

And it'd be common to leave it as

,$ \sqrt{10}

[somber coyote]

Pseudo:

[shadow anvil]

Because exact form

[wraith mango]

Okay

[cinder portal]

wait no

its not root10

a^2 + b^2 = c^2

c = 3
a = 1

solve for b

c is always your hypotenuse

a and b are always your legs

in a right triangle

[wraith mango]

that’s wrong?

[shadow anvil]

Ah yeah

Thanks @cinder portal I didn't bother looking at the picture

Just the equation

[cinder portal]

ye ye np np

[shadow anvil]

So he's saying you're dealing with the pythagoras

And the hypotenuse (Longest side) is always c

[cinder portal]

i got distracted, was supposed to help him, but someone is going to break maths in prealg channel

[shadow anvil]

a and b are interchangable

[wraith mango]

like this 😄 and now we solve for b

sooooo

?

[shadow anvil]

Yep

Oh no

not quite

SO you've got c=3, a=1 and b=?

SO

[wraith mango]

:yea

[shadow anvil]

,$ a^2+b^2=c^2\1^2+b^2=3^2

[somber coyote]

Pseudo:

[shadow anvil]

Then you solve this for b

,$ \begin{aligned}b^2&=3^2-1^2\end{aligned}

[somber coyote]

Pseudo:

[shadow anvil]

Yep

[wraith mango]

so what’s b^2 then

[shadow anvil]

Solve for it

Make b the subject

[wraith mango]

1 + b^2 = 9

How does that work out?

We’re missing 8 in the middle

So something ^ 2 = 8

[cinder portal]

b^2 = 8

solve for b

basically

[wraith mango]

2.8

I’m gettin

[shadow anvil]

Leave it in exact form

,$ 1^2+b^2=3^2\1+b^2=9\b^2=9-1\b^2=8\b=\sqrt{8}

[somber coyote]

Pseudo:

[shadow anvil]

Leave it in exact form

So simply sqrt(8)

[wraith mango]

Okay

[shadow anvil]

Don't put decimals

[wraith mango]

Sec

[craggy spear]

Or as a surd if you've learnt those, 2* sqrt(2)

[wraith mango]

so I leave the sq root there on the triangle too?

[craggy spear]

Yup

[wraith mango]

Okay thanks 😃

Was the first question written correctly?

Answer *

[shadow anvil]

You can still simplify sqrt(8)

[shadow anvil]

Don't leave it as root 8

Should simplify it

[wraith mango]

What should I leave it as

[shadow anvil]

Do you know how to simplify surds?

[craggy spear]

Ofc it was :)

[wraith mango]

No i don’t know how to simply squares

[shadow anvil]

Ok

[craggy spear]

Then leave as is

[wraith mango]

Is it supposed to be taught in this level?

[shadow anvil]

I'd assume so seeming as you had some questions using cot and cosec below

[wraith mango]

I was put into this unwillingly it’s grade 11 functions

But I’m not good at math

[shadow anvil]

Then yes

Simplification is a requirement most likely

Basically break 8 up into two numbers that multiply to make 8

Ideally you want one of them to be a square

So what are some numbers that come to mind?

[wraith mango]

2(4)

[shadow anvil]

@craggy spear Please don't make any unconstructive comments

Ok

SO now

[craggy spear]

@shadow anvil i thought one of them HAD to be a square?

[shadow anvil]

,$ \sqrt{8}=\sqrt{2\cdot4}

[somber coyote]

Pseudo:

[craggy spear]

Sorry uWu

[shadow anvil]

Yes?

[wraith mango]

Yea makes sense

[shadow anvil]

Now because of the way exponents work

We can write this as

,$ \sqrt{2}\cdot\sqrt{4}

[somber coyote]

Pseudo:

[shadow anvil]

Now do you know what the square root of 4 is?

[wraith mango]

2

[shadow anvil]

Yep

So

,$ \sqrt{2}\cdot2

[somber coyote]

Pseudo:

[shadow anvil]

Which we commonly write as

,$ 2\sqrt{2}

[somber coyote]

Pseudo:

[shadow anvil]

which is now simplified

[wraith mango]

Seems good to me haha

[shadow anvil]

Yep

[wraith mango]

Is the last one worth trying? Looks too complicated for me tbh

[shadow anvil]

I'm assuming you have to prove that?

[wraith mango]

Yea that’s what they ask of me

[shadow anvil]

We can go through it if you want

If you're familiar with a few of the identities

[wraith mango]

I’m not no, looks like a completely different concept I’ll wait and just ask my sub tomorrow I don’t think she’s expecting us to finish that question

I’ve got 2 more easy questions

[shadow anvil]

I mean I can explain it if you want

It's not very hard

It's just one step tbh

[wraith mango]

Yea go for it if you’d like sure

[shadow anvil]

Ok

So you're aware

[shadow anvil]

,$ \sin^2\left(x\right)+\cos^2\left(x\right)=1

[somber coyote]

Pseudo:

[shadow anvil]

Yes?

[wraith mango]

Yes

[shadow anvil]

So if we divide both sides by sin^2(x)

,$ \frac{\sin^2\left(x\right)}{\sin^2\left(x\right)}+\frac{\cos^2\left(x\right)}{\sin^2\left(x\right)}=\frac{1}{\sin^2\left(x\right)}

[somber coyote]

Pseudo:

[shadow anvil]

Yes?

[wraith mango]

Yea

[shadow anvil]

So

sin/sin cancels to 1

and 1/sin(x)=cosec(x)

,$ 1+\frac{\cos^2\left(x\right)}{\sin^2\left(x\right)}=\csc^2\left(x\right)

[somber coyote]

Pseudo:

[shadow anvil]

Now

,$ \cot\left(x\right)=\frac{1}{\tan\left(x\right)}=\frac{1}{\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)}=\frac{\cos\left(x\right)}{\sin\left(x\right)}\\cot^2\left(x\right)=\frac{\cos^2\left(x\right)}{\sin^2\left(x\right)}

[somber coyote]

Pseudo:

[shadow anvil]

Giving us

,$ 1+\cot^2\left(x\right)=\csc^2\left(x\right)

[somber coyote]

Pseudo:

[shadow anvil]

Ye?

[wraith mango]

Yes!

I don’t really understand the longer equation but I can connect how the sin/sin is cancelling out

How would I proceed with question 8?

[cinder portal]

What is the law of cosines?

[wraith mango]

The sides = the angles

[cinder portal]

You have to know the formula, and know what each variable correspond to

[wraith mango]

I have it saved

[cinder portal]

Ahh yes, use that

since you know your angle is G, or you are lookign for G

[wraith mango]

Yes

[cinder portal]

the hardest part is figuring which numbers are a, b, c

[wraith mango]

Alright I’ll mess with 8 tomorrow what about 7?

it’s given me 2 legs

So 2 sides

And one angle

@cinder portal does this work?

[cinder portal]

wtf calculator

dont use a calculator

do it on paper

and then do the computations with a calculator

[wraith mango]

I mean idk how to start calculating something I didn’t learn but okay I’ll watch a YouTube video on it

[cinder portal]

You won't learn if you use whatever calculators they have on the internet

[wraith mango]

I understand but I don’t wanna fail either

[shadow anvil]

But you won't have an internet calculator in any exams

[wraith mango]

I get what you mean though

[shadow anvil]

t!yt How to use law of cosines

rEE

https://www.youtube.com/watch?v=9CGY0s-uCUE

There we go

[wraith mango]

https://youtu.be/VKQfiJaEJvc

?

Yea I just saw that one too okay I’ll watch both then

Now how would I go about solving 7?

[shadow anvil]

Watch the video.

Do the other one first

[wraith mango]

Is it also using cosine law?

[shadow anvil]

no

[wraith mango]

okay I’ll start 8

batteries at 1% catch you guys in a bit

[craggy spear]

I mean, a graphical calculator does most of this...

Wait nvm didvnt see the trig calc

[shadow anvil]

[distant quest]

Can anyone help me to solve the first and second one? please

[craggy spear]

1 is AC

Purely from looking at it

2nd use cosine rule

So cos(theta) = stuff

Then cos-1(stuff) = theta

[distant quest]

How do I calculate AC?

[wanton forge]

law of cosines

[distant quest]

ok thank you

[wintry dagger]

does anyone know the vector equation of a cone and how to put it in 3d Geogebra?

[pastel bough]

Anyone know a good website to practice solving for angle and side on a right triangle?

[craggy spear]

as in app? or something that gives you questions (without a picture)?

mymaths.com has a few but yh

[upper karma]

Hi, I have a question about Buffon's Needle Problem, specifically the geometrical part of $ y \leq \frac{l}{2} sin \theta $

[somber coyote]

Sponge5:

[upper karma]

link https://en.wikipedia.org/wiki/Buffon's_needle_problem

I can't seem to figure out a way to get to that inequality

[fringe dirge]

Okay what's y and what's theta for you

[upper karma]

@fringe dirge sorry wasn't here

[somber coyote]

Sponge5:

[upper karma]

the needle crosses a line if $ x \leq \frac{l}{2} sin \theta $

[somber coyote]

Sponge5:

[upper karma]

as is specified at the end of the "Solution" part of the wiki page

[fringe dirge]

draw a line parallel to the parallel lines through the center of the needle

then you can form a right triangle by dropping perpendiculars from the tips of the needle to the line through the center of the needle

[ornate lodge]

Where do I begin in 1 - sin^2(5x) - sin^2(2x) = 0

[dark sparrow]

i'd begin by rewriting 1 - sin^2(5x) as cos^2(5x)

[upper karma]

actually id suggest using the half angle formulas maybe?

sin^2x = 1/2(1- cos2x)
1 - 1/2(1-cos10x) - 1/2(1-cos4x) = 0
1/2(cos4x + cos10x) = 0
cos4x + cos10x = 0
you can then try
the sum to product formula
cosx + cosy = 2cos(x+y/2)cos(x-y/2)
= 2cos(-3x)cos(7x) = 0

yep that gives you your solutions

@ornate lodge

[stoic wraith]

Need help solving this one. Can't recall or find the identity for this online for the life of me.

[vagrant elk]

No identity

just use the dang unit circle

[stoic wraith]

X is Sine and Y is Cosine on unit circle values right?

[vagrant elk]

no

[stoic wraith]

Or no I'm sorry, I've got it reversed, correct?

[vagrant elk]

yes

[stoic wraith]

Thank you

[spark stag]

(tip: if you want a rule-of-thumb for remembering)

(cosine and x both come before sine and y in the alphabet)

(that said, with enough practice, this should be ingrained into your skull anyway)

[stoic wraith]

thanks for the tip

[olive sinew]

I have a question

[devout shell]

[olive sinew]

Oh hi again

I know this is something simple but I need to know what I am doing wrong

71

[devout shell]

wondering where r=2 comes from

[olive sinew]

We know that csc(theta) = r/y but the given number is -1.45

I don’t even know

I just came up with it

[devout shell]

let's not do that then

[olive sinew]

Ok in this case what’s “r”

Any hint?

[devout shell]

I'm trying to work it out lol

I put -1.45 into a fraction, and I got -29/20

so knowing this, we can proceed now

[olive sinew]

Thank you I never thought about converting a decimal to a fraction

[devout shell]

took me a while to think of that too lol

[winter edge]

hello

[mighty narwhal]

Hewo

[ornate lodge]

What is the identity for sin(x)+cos(x)?

[upper karma]

@lusty trout What’s the definition of range? Does that give you any ideas on how to find the range of this function?

[lusty trout]

Range means all the possible values of f(x) where x belongs to the domain

[ornate lodge]

Is this correct? sin(x)+cos(x)=sqrt(2)sin(pi/4+x)

[dark sparrow]

yes

[mighty narwhal]

What's the difference between (a, b) and [a, b]?

[devout shell]

(a, b) says the end points are not included in the interval where as [a, b] says that they are included in your interval

[somber coyote]

DarK:

[dark sparrow]

that's the french way

[proper beacon]

That inward outward bracket seems convenient compared to ( vs [

[hard gale]

🥖

[gritty siren]

(i've seen spanish documents with that choice of notation as well)

[mighty narwhal]

(It's more like Europian stuff, not French or any countries)

[gritty siren]

anyway, reverse brackets are far sexier than parentheses

[mighty narwhal]

Yupp.

Also ]a;b] is better to look at than (a;b]

The second one looks like a typo

[gritty siren]

ikr!

[devout shell]

Tuong is in the wrong

[dark sparrow]

tuong is in the wruong?

[olive sinew]

I have to find all six trig values but before I move forward I just want to confirm with one of you guys

If I am going the right path

As you guys can see I got sec@=-2

[celest arch]

how would you get an upper limit of pi/6 out of that trig boy

my first thought is to just try values and work down from 1

[solar shale]

@me

Is this correct Im finding surface area btw

Oh shit ignore answer forgot to erase it, it would be 200pi

<@&286206848099549185>

[ornate lodge]

Why is sin(144)/sin(36) = 1?

[spark stag]

144 degrees and 36 degrees are equidistant from 90 degrees

that is, 144 - 90 = 90 - 36 = 54

therefore, they're the same value

[solar shale]

Wait so is my answer correct?

[spark stag]

i was replying to @ornate lodge but ill take a look

one sec

[solar shale]

Ok

[spark stag]

but yeah, you can draw a line of symmetry through x = 90 degrees, and sine is symmetrical about it

hence the sine values of the angles being equal

[solar shale]

Answer is meant to be 200pi not 153

[spark stag]

to clarify

this is a cone with radius 8

and slant height 17?

if so, then yes, 200pi is correct

[solar shale]

Ok 👍🏼

Yeah r=8 and height is 17

[upper karma]

If anyone is good at this type of stuff i have a bunch more that I really need help with

[olive sinew]

does anyone have an idea of how to work this out ?

im working on 89

[spark stag]

so, theta is between 90 and 180

which means 2theta must be between

2*90 and 2*180

or in other words

between 180 and 360

whats the sign of sin(theta) when theta is between 180 and 360?

[olive sinew]

negative since sin(theta)=y/r and quadrant 3 and 4 y < 0

but why 290 and 2180 ? how did you figure it out ?

2(90) and 2(180) i meant

[narrow sleet]

The question says sin 2(theta)

2(theta)

[olive sinew]

ohhhhh

damn i thought this whole time that it was a 20

[narrow sleet]

😂

[olive sinew]

and then i look closer and is 2 times theta

[native yacht]

can i post a problem that i made here

[olive sinew]

im sorry for the trouble guys

[native yacht]

well idk why not ig

ABCD is a quadrilateral with AB = 16, BC = 21 and CD = 11. If angle ABC is right and sin(BCD) = 24/25, find the area of ABCD.

[mighty narwhal]

Idunno. Figure out the third side and the angle than use
$$A=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2{\frac{\alpha+\gamma}{2}}}}

[somber coyote]

DarK:
Compile Error! Click the reaction for details. (You may edit your message)

[solar shale]

How do I know when to divide or multiply

@me

[mighty narwhal]

Why do you use $^\circ$ instead of $^{rad}$?

[somber coyote]

DarK:

[solar shale]

Tbh idk I just got the answers

[mighty narwhal]

But for the answer, you divide by 2 if you are looking for the angle, and multiply by two if you look for the arc

having the arc measured in degrees makes me shriek like a banshee :/

[solar shale]

Oh ok lmao

[somber coyote]

DarK:

[solar shale]

The answer is 195 I don’t see how tho

What is measure of XLZ

Nvm I got it

[mighty narwhal]

xl is not 90deg

[solar shale]

How come

Is it not diagonal

[mighty narwhal]

xl is opposite ky, not kz

[solar shale]

Oh my bad

[mighty narwhal]

because xm is the extension of ym, not zm

xl is 105 (90+15)

,calc 105+75+15

[solar shale]

I get it but then that makes zy=30?

[mighty narwhal]

zy is 180-(90+75)

the only error you made is that xl is 105, not 90

[solar shale]

Oh ok

[mighty narwhal]

105+75+15=195

[solar shale]

Tnx

[mighty narwhal]

,calc 105+75+15

[somber coyote]

Result:

195

[mighty narwhal]

Your welcome

[fallow notch]

i used tan(75) then multiplied by 5
what am i doing wrong

[dark sparrow]

are you sure $b = a \tan(75^\circ)$?

[somber coyote]

Ann:

[dark sparrow]

or in other words, that $\frac{b}{a} = \tan(75^\circ)$?

[somber coyote]

Ann:

[fallow notch]

i still dont understand

[rotund current]

What is the definition of tan?

[fallow notch]

Soh
Sin(x)=opposite/hypotenuse
Cah
Cos(x)=adjacent/hypotenuse
Toa
Tan(x)=opposite/adjacent

i use this

[rotund current]

Right, and in this triangle, which one is opposite and which one is adjacent?

[fallow notch]

tan(75) = 5/b ?

[rotund current]

Right, right

And now solve for b?

[fallow notch]

1.3398

[rotund current]

Right, and is that the correct answer?

For b

[fallow notch]

yep
i used online calculator tho, dont know how to do it on calc

[rotund current]

To solve for b? Or use a calculator to find 5/tan(75)?

[fallow notch]

to solve for b

[rotund current]

How comfortable are you with solving (linear) equations?

[fallow notch]

im okay i think

[rotund current]

Like, if I asked you to solve x=2/y for y?

[fallow notch]

oh then nope

[mighty narwhal]

Do y'all need any help? Or r u in control, TheKikko?

[olive sinew]

solving trig equations hasnt been covered in the book yet and this is one of the question i have to solve so i can finish my hw

what am i solving for here ?

theta ? this is confusing af

[rotund current]

Dark, I'm probably fine although I'm going grocery shopping in a couple of minutes, so if you'd like to take over its fine

Javakid, yeah, solving for theta

[olive sinew]

ok, drive safe man

[rotund current]

Croy: okay, so, generally the principle is that you should do the same thing to both sides, so for $x=2/y$, I'd multiply by y first, turning it into $x\cdot y = 2$ and then divide by x, giving you $y = 2/x$. If you treat $\tan(75^\circ)$ like x in this case, you should be able to do something similar

[somber coyote]

TheKikko:

[rotund current]

Thanks!

[olive sinew]

are you talking to me ?

oh yea i got 2 degrees

[rotund current]

No, the person who asked something before you ;p

[mighty narwhal]

1 minute and I will be here, ready to help

Okay, croy is afk I guess, JavaKid, one minute, I need to figure out how to do your problem

[fallow notch]

i able to get this writing it out helped
thanks kikko

[rotund current]

No problems, glad to help! :D

[mighty narwhal]

Okay, so JavaKid. if you have $\tan{x}={\tan{y}$, if $0<x,y<180^\circ$, then $x=y$

[somber coyote]

DarK:

Okay, so JavaKid. if you have $\tan{x}={\tan{y}$, if $0<x,y<180^\circ$, then $x=y$
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.11 ...so JavaKid. if you have $\tan{x}={\tan{y}$
, if $0<x,y<180^\circ$, th...
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.

[mighty narwhal]

$\tan(3\theta-4^\circ)=\tan(5\theta-8^\circ)$

$3\theta-4^\circ=5\theta-8^\circ$

[somber coyote]

DarK:

[mighty narwhal]

Oh, just realized he already got the answer
nvm then

[north crescent]

Sup guys

[mighty narwhal]

hewo

[north crescent]

Wutcha doin?

[mighty narwhal]

nothing really. waiting for someone to have a problem

[north crescent]

K lol

I’m not gr8 at maths but I do like it

In high school atm

[mighty narwhal]

I'm in 8th grade right now, not sure if it counts as highschool or elementary school at your country

[north crescent]

Ye I’m in twelfth

Technically post twelfth

[mighty narwhal]

Oh, Cool

[north crescent]

Taking gap year for stats ioi college application

I’ll go to uni next year

I think imma give ap phys stats calf and eco if it’s not too hard

From the course I have seen, it all seems similar to stuff I already know so that shud be cool. Wbu?

U like maths?

[mighty narwhal]

Well, I don't like it specifically, I'm just kinda good at it, I guess. At least I'm not bored when I'm helping others

I am going to learn electronics mainly, But it needs some math skills so.. I will also learn math,

[north crescent]

Cool cool 😃

Well best of luck on electrical eng

[olive sinew]

ok i have a question

it says determine if csc(theta) = -100 is possible or not

i say that it is possible since the range for csc is from (-inf,-1]u[1, +inf)

am i correct ?

[mighty narwhal]

,w csc^-1(-100)

[somber coyote]

Display results online and refine query

Requested by DarK#2404

[mighty narwhal]

It is possible

[upper karma]

By applying Pythagoras's theorem (the usual two-dimensional version) twice over, prove that
the length r of a three-dimensional vector r = (x, y, z) satisfies r² = x² + y² + z².
im struggling a lot, i think that the idea is to start workin on a two dimensional plane but...
well if anyone has an idea ^^

ah just found the solution sorry ^^

[mighty narwhal]

uuuuuuuuuughhhh

I just took a lot of time drawing it

$a=\sqrt{x^2+z^2}$

$r^2=y^2+a^2=y^2+x^2+z^2$

[somber coyote]

DarK:

[olive sinew]

well it has a solution, so im guessing nothings wrong

right ?

[native yacht]

@mighty narwhal nope, extend AB and CD to meet at point P such that PBC is a right triangle