geometry-and-trigonometry

[somber coyote]

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

[frozen ocean]

If I have $\cos\theta_{0} = \frac{v_1\sin\theta}{V + v_1\cos\theta}$ and need to plug it into another function that has an argument of $\sin\theta_{0}$, which identity should I use to make this easiest on myself?

[somber coyote]

Akira

[frozen ocean]

@tight lion here you go

[cunning acorn]

I don't know why the 4/3 part, but r^3 is because r^2 is for a circle, so making it 3d instead of 2d required r^3

[shadow forum]

how would I rotate a conic section around a certain point?

specifically an ellipse

[shadow forum]

im assuming one way of accomplishing this would be to graph the ellipse with an equation that uses the foci, and then i would only need to rotate the 2 foci around a point, but im not sure how to do that either

this is based on the fact that a sphere takes up 2/3 of the volume of a cylinder that has a height of 2r

[vivid dawn]

only have 3 of these questions left and need help

[sand ingot]

<@&286206848099549185>

<@&286206848099549185>

[sand cairn]

Hi

So here you have two octagons with areas 4 and 36

Do you know how to find such areas

what would be the scale factor to go from an octagon of area 4 → to area 36?

particularly in perimeter

A square's area is its side quantity squared
ex: 2² and 8²

the perimeter is its side multiplied by 4

ex: 2by4and 8by4

hope that helps

[primal field]

can someone help with this please

have a test in 30 minutes

<@&286206848099549185>

[foggy parcel]

since it's an equalieteral triangle, the three angles' value are?

@primal field

@primal field

nvm it's 60 degree I hope u know that

and another thing we know that the center of an inscribed circle & a circumsribed circle are same [ only in equaletral triangle]

So basically according to ur picture the big circle is circumsribed

and it's centre is O

so the inscribed circle's center should be O as well

And to add to that, we as well know that we have to divide the three angles equally, so getting the 3 eqators , the point they've crossed each other is the center of an inscribed circle

lemme show u how that looks ik u won't get it without seeing the drawing

the small guy is inscribed & the big guy is circumsribed]

now using trigonometry get the value of BC

Gradually get the value of the height

then multiply BC * height * 1/2

Mission passed

@primal field

[primal field]

thank you sm i’ll take a look

[foggy parcel]

surely u should

[primal field]

yeas

[kind cairn]

does the theorem about angles on the same arc being the same work in reverse? (for example if angle ABC lays on arc AC, and angle ADC lies on arc AC, can you in inscribe ABCD into a circle?)

[primal field]

what does bc equal?

@foggy parcel

[foggy parcel]

What do u mean?

Oh I get it

BC = Ab = AC

The sides of the equaletral triangle

[primal field]

u said get the value of b to c

which is?

[foggy parcel]

Yea

[foggy parcel]

.

[primal field]

like is it a value?

its

1?

[foggy parcel]

Yes

[primal field]

alr alr mb thanks

[foggy parcel]

no use trig

OC = 1 so BC = ?

Use trig

[primal field]

what trig like

im being slow rn

[foggy parcel]

Like draw a perpendicular from O to BC

[primal field]

huh

bro im so sorry im deadass so slow en

rn

[foggy parcel]

np

[primal field]

what would bc be then nd howd u get

sine law?

[foggy parcel]

use cos30

From here get the value of CD

[dark sparrow]

you know you can select a rectangle in MS Paint and ctrl-c ctrl-v it here right

[foggy parcel]

since CD = BD u can get BC

[foggy parcel]

sorry ma'am still a kid when it comes to MS paint

[dark sparrow]

now you hopefully know a bit more

[primal field]

cos30 = 1/x

xcos30 = 1

[foggy parcel]

CosTheta = adjacent/ hypotonus

[primal field]

mb

1cos30= x

[foggy parcel]

yea

so what's the value of cos30?

remember?

btw I have to leave now hopefully u 'll get this surely boss

Good luck

[primal field]

yea thank you sm

[upper karma]

i am lost

[agile siren]

Consider a cone with half angle of $\beta$ such that its base is placed on the ground. Assume that the angle that the direction of the Sun makes with the horizon is $a$. Show that a fraction $\eta$ of the cone will be lit, such that $\cos \eta \pi=-\tan \beta \tan a$.

[somber coyote]

hellfire

[agile siren]

i dont know if this is the right place to ask this but any hints?

[digital agate]

A (hypothetical)radio tower can only be functional if 3 other radio towers are in range. What configurations of 4 radio towers maximizes coverage? What about 5? Or 6? Is there an algorithm?

Is this even an appropriate geometry question?

[shadow forum]

is "in range" defined?

[digital agate]

Alright the defined version

Let there be 4 circles of radius 1. They are placed according to the rule that each circle must contain the center of at least 3 other circles. Define the configuration that maximizes the total area.

[grave pond]

That seems difficult.

There's the symmetric configuration where the centers are placed at the corners of a square with diagonal 1, but I don't think that gives the maximum swept area.

I would (handwavingly) expect one gets a larger area from placing three centers at the corners of an equilateral triangle with side length 1, and the fourth as far outside the middle of one side of triangle as it can be and still reach the opposite corner. But I haven't done the actual calculations.

[digital agate]

And I wonder what happens with more circles? Would a line-like shape be optimal or something that expands in all directions?

A line like shape benefits as more of each circle’s insides is unique and not shared and wasted

[primal field]

@foggy parcel test went well thanks

[foggy parcel]

My pleasure did u get that right?

[sudden quest]

Hello

can anyone provide solution for this

Two men are standing on the same side of 100m high tower. If the measure of the angles of elevation of the top of the tower are 20° and 30° respectively. Find the distance between them.

i have tried many resources they all are giving different answers

[foggy parcel]

first of all draw a picture of the situation

[slate herald]

Guys

how am I suppose to prove it's -1 if they haven't show me the numbers to work it out

[outer sinew]

technically this is false since for a line like y=0 and x=0 this is not true

But that aside, think about the general formula to find the slope of a linear equation perpendicular to another one which slope is given

[paper hull]

You need to use a geometric proof. Consider this figure

Notice how triangles $\triangle{XAB}$ and $\triangle{XDC}$ are congruent.

[somber coyote]

Cat's Schrodinger

[slate herald]

Then by default, it will be:

**X1 = 0
X2 = 0

Y1 = 0
Y2 = 0**

[outer sinew]

?

[slate herald]

For the slope

[paper hull]

You can't generalize it to $y=0$ because the slope is undefined. The rule only applies to slopes that are defined.

[somber coyote]

Cat's Schrodinger

[paper hull]

Sorry, x=0.

[outer sinew]

Yes, but it said "the slopes of 2 perpendicular lines"
Can't really multiply slopes when the slope is undefined can you

[paper hull]

Yes.

[sick hemlock]

Would you call trigonometric identities involving tan or any ratio other than sine or cosine false as well?

genuinely curious

[paper hull]

What do you mean?

[slate herald]

Probably means using Pythagorean theorm or trigonometric formula

In this case

[wide imp]

would it be appropriate to ask for help when it comes to translating a math problem into geogebra? i'm quite new to this so i'm not sure if i'm even using the right terminology. basically i have the problem done but i just can't work my way around geogebra. i can't grasp the program just by myself.

[outer sinew]

"False"?

[sick hemlock]

^

[outer sinew]

No. They are explicitly stated to be undefined at that point. But in this case they stated any perpendicular line

[sick hemlock]

right right

[lethal jackal]

how can we use sin cos tan to figure out this one?

[vagrant plinth]

I think you want to use the laws of cosine.

[lethal jackal]

uh ok

[dark sparrow]

make a diagram first tbh

thatll inform further actions

[lethal jackal]

i made it

the only problem is that I don't know if the triangle that formed is a right triangle or not

but will determine if sin cos tan can be use or not

[dark sparrow]

@lethal jackal still need help with this?

sorry, was away

[lethal jackal]

yeah

[lethal jackal]

its ok

I really suck at bearing

[dark sparrow]

show your diagram please

[lethal jackal]

ok sure

wait a sec

[dark sparrow]

ok this looks good

now you need to find the missing side

the law of cosines will help you

[lethal jackal]

ok I will try it

thanks

[tardy thicket]

Hi, I am looking for people that wants to read a geometry book in group
like to read, to get the information and to make a talk each week
What do you think about it?
you may decide the book

I am just starting, but the thing is to go forward, so you are able to chose whatever book

and it may be a pdf as well

[full mantle]

hi guys, I'm trying to figure out how to get the area out of this shape.
Since one of the sides is 1 and it's oddly placed on the coordinate system it's kinda throwing me off.
Anyone that can help?

[wide imp]

i guess you could look at it as two triangles and just calculate the area of each using determinants. draw an (imaginary) line from (-1, 0) to (2, 1) or the other two vertices, it doesn't matter, and you get two triangles.

and heres the formula

[wide imp]

hope this helps

[upper karma]

Have you been taught the distance formula

The above is technically applicable but it's much simpler imo if you just use the regular distance formula

[upper karma]

Also one of the sides is definitely not length 1 if that's what you meant

[wide imp]

right yeah since its a square you can just do a^2 with the distance formula

slipped my mind

[upper karma]

It's also like 50x easier if you just draw it yourself on a coordinate plane

You see immediately that it's a square and from there it's trivial

[slate herald]

Is this right?

[grave pond]

Impossible to tell without knowing what the two lines are.

[slate herald]

Oh

[wide imp]

yes its right

[barren ice]

how do i look for x and y on both figures?

[flat pawn]

Notice that each side can be obtained by the pythagorean theorem. Notice again that all sides are equal in length. You have the square. Voila!

[pastel vortex]

im a freshman and learning basic trig and this looks hard 💀💀💀

[sick hemlock]

For the first figure you could use alternate segment theorem or label the centre of the circle and work your way from there keeping in mind isosceles triangles and angle sum property

[idle orchid]

How i do this?

[storm portal]

What have you tried?

[solemn lagoon]

well you have to um first find the surface of area of the bottom of it using pi*r^2

and then you also have to find the circumference of the bottom and multiply it by 10 to get the surface area of the middle

and then you have to use 4pir^2 to find the surface area of the sphere

and divide it by 2

and then add all of it up together

@idle orchid does that answer your question?

[idle orchid]

Yes sir thank you

[idle orchid]

Yeh i did do it but i made mistake on the side they connect, i got it correct now

[ebon tinsel]

this is the way.

[idle orchid]

Does anyone know how to do this, im having trouble?

[ebon tinsel]

is there a figure?

[idle orchid]

Nope

[ebon tinsel]

i guess they are similar figures and for that reason they woudl have the same ratio between there volume and there surface area

SA/V = const for the similar shape

(SA_1 + SA_2) const= total volume . and total volume - (SA_2)const = volume_1

and SA_1/SA_2 =V_1/V_2

[idle orchid]

Ohhh ok thanks a lot!

[ebon tinsel]

||288||

[idle orchid]

Yess i got that thank you!

I was overthinking it 😭

[ebon tinsel]

idk if u were overthinking it. the solution mabye wasnt as intuitive

[idle orchid]

Yeh

[vagrant ore]

Just want to make sure my math is right but I think this is cos(37) 25/z correct?

[snow crystal]

Yes

[tardy thicket]

Is there any solution for this?

The radius of the small circles is 1

[grave pond]

What is the question?

[outer sinew]

[lethal jackal]

Help me find the angle

[outer sinew]

Probably to find the shaded area ig

[outer sinew]

Yes there is a solution, if you're asking for the shaded area

[grave pond]

The triangles are iscosceles. The bisector of the apex angle is perpendicular to the opposite sides. That gives you two (similar) right triangles to do trigonometry on.

(Beware that C looks like the angle is measured "the wrong way around", so they probably want a reflex angle there).

[true fog]

How do you find the height of a height of a right triangle with the bottom left angle and the bottom line length

The right angle being the bottom right

[upper karma]

Can we find the area of the shaded segment ?

[lethal jackal]

if given the radius of the circle

[upper karma]

yes they gave it

[upper karma]

^

[upper karma]

Cos(Angle)=Bottom line / Hypotenuse

this way we get the hypotenuse

from which we can to use Pythagorean rule to find the 3rd side which is the height

a^2+b^2 = c^2

[true fog]

We don’t have the hypotenuse

We have bottom line and all angles

[upper karma]

you get the hypotenuse from cos

[true fog]

Ohh

[upper karma]

i got it

[upper karma]

nvm

[gentle osprey]

I need help with my class work

I stupid and need help

[keen canyon]

cut n paste them into rectangles and triangles if you really can't remember any other formulae

[gentle osprey]

Wdym?

[keen canyon]

see that parallelogram? cut off its edge and paste it to the other edge

[keen canyon]

Oh wait I see for these things just use base* height

[keen canyon]

for the trapezoid use (top+bottom)*height/2

[gentle osprey]

Thank you

[quiet timber]

Y'all I need help with this, if y'all help me with this I'll have a d in math😭 I went through spinal surgery a month ago and they still forcing me to take the wpt even though I was out of school for awhile🤐

[blissful spire]

can someone pls help me with this problem? I thought the radius was 50 but now I can’t get an answer that isn’t a really weird fraction

[storm portal]

,rotate

[storm portal]

radius should be 50

[surreal hill]

He did take radius as 50

[torn wolf]

Can someone help with these

[surreal hill]

According to Pythagorean theorem x= under root 13^2-5^2

So x = 12

[torn wolf]

Tysm

[surreal hill]

For y I'm not sure but if u take hypotenuse = 37 assuming the triangle is an isosceles triangle then y = 35

[torn wolf]

God bless u🫡🙏

[surreal hill]

Area should be 480

And 9 will be 120

[torn wolf]

Tysm u r loved

[surreal hill]

Np

Happy to help

[grave mist]

i need help with this problem

[olive ferry]

HELP!!!!

[grave mist]

For problem number one you need to find the area of 3 rectangles which are 259, 918, and 189 units. Since there are two of each rectangle you multiply these values by 2 and add them to get the final area of 2732 units squared for the area

[warm sundial]

Number 2 just multiply 16*16 for each square. 🤙

And add them

[olive ferry]

OH

THANK YOU

[warm sundial]

Or 16x16x6 is quicker

[olive ferry]

so 1536 right

[warm sundial]

Yeah

✌️

[olive ferry]

THANK U

[grave mist]

let us know if we right 🙏

[olive ferry]

ok!!

idk what to do for these

😭

[grave mist]

the answer 828 units squared for problem 4

uhh

[grave mist]

you find number 4 by finding area of the outside triangle and mulitpling that by 3 then you find the area of the middle one and then add that to the previos product

i dont know what missing measurement the text is blabbering about

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[grave mist]

My apologies 🤚

[tardy thicket]

exactly, the blue segment

what is the best webpage that has Geometry problems?

I mean, for practicing and those stuffs

When I am on this server, it is like if you talk me in another language

[upper karma]

i know how to solve it

but its quite lengthy, if thats okay with you

[ruby fulcrum]

can someone help me with this problem

[tardy thicket]

do it anyway, if you prefer to do it at DM, go for it

[upper karma]

Step i :- Well First Connect the centres of any three adjacent circles which gives you are equilateral triangle . Find its area
Step ii :- to find the area between the 3 circles(red), we can subtract the segment area(yellow) from the equilateral triangle
Step iii :- Required Area =
(Area of big circle who's radius is 3) - 7(area of small circle)-6(Area between the circles) / 6

[alpine mantle]

An ornament is in the shape of a right hexagonal
pyramid. The height of the pyramid is 25 cm and
the sides of the base are 9 cm. The faces of the
pyramid are congruent isosceles triangles with
equal sides r cm. The length from the middle of
the hexagon to one of the vertices of the hexagon
is k cm. The trophy is made out of golden coloured
plastic.
Calculate how much plastic is needed to
make one ornament.

[upper karma]

looks like an elipse

just complets the squares

ecentricity is root1+b^2/a^2

the two square will be (3y-3)^2 and (4x-8)^2

[upper karma]

hi, i have an issue with 4.1, gr 12 mathematics analytical geometry

I equated the graphs like this: 2-3x=2-x^2, since i mean its an intersection

instead of that

the answer said i should sub y=3x-2 directly in the circles equation in y's place

what did i do wrong

[upper karma]

dude you equated it t wrong

the x^2 equation is for y^2 not y

just put the line"sequation into the circle

y^2=2-x^2 not y=2-x^2

[upper karma]

oh, i cant equate a parabola type equation with a straight line?

[upper karma]

dude ots for y^2

not y

yeah

im aware

they how can you equate

y is not same as y^2

i cant equate it if it has a square?

[upper karma]

yh lol

i need to revise gr 11 xD

i see, yeah then it makes sense

ty man

no probs

[lethal forge]

Hello can someone help me out with sliver question please

I tried to draw the bearing however I’m pretty sure I drew it wrong

[upper karma]

is it possible for a cylinder to spin both clockwise and counterclockwise at the same time?

[grave pond]

Depending on which end you look at it from, yes.

[thorny bough]

I tried using law of cosines to solve this but was unable to since I wasn’t able to simplify the equations

[rancid lake]

pls help me my sol is tmr and need help with this question idc how u want to help me just lmk

[grave mist]

wqould it be cde

[dense leaf]

if these two triangles are similar triangles, it means that the sides are proportional.
for instance, since 45/7.5=6, the scale factor from the smaller triangle to the larger one is 6.
therefore, x = 6*6 = 36, and the height of the geyser is 36 ft.

[rancid lake]

thank you but i got this one can you help with this one

im really blanking on what to do

[dense leaf]

this is geometry sol right? just want to make sure

[rancid lake]

yes

[dense leaf]

so since this is a kite, the diagonals bisect at right angles.
this means that we have a right triangle with sides 9 in, 12 in, and x.
hence, use the pythagorean theorem to solve for x: 9²+12²=225=15², so x=15 in and the answer is F

[rancid lake]

225=15² where did you get this from and thx

[dense leaf]

the square root of 225 is 15

[rancid lake]

oh ok im dumb

[dense leaf]

lol its fine

[timber cargo]

It’s a 3,4,5 triangle

[rancid lake]

ok thank you

[elder sundial]

Any Graph for p(x) = (x^2) -(3 * x) + 2 ?

[remote pebble]

Thats it?
No other information?

[short dagger]

Yes

[remote pebble]

Whoa

[remote pebble]

Where did u get this question?

[short dagger]

Test book

[remote pebble]

Damn

Can u send me a photo of the question?

[tardy thicket]

The six is because we only need one section similar to the blue required no?

Because there are 7

And you need only 1

I don't know why you told me that would be long

other people answer me and they gave me hexagons and complicated stuffs

that is a good explanation

[upper karma]

yes

and there are only 6 not 7

and thank you

[fresh sedge]

How do angle of elevation

[wild zodiac]

I'd try to use the cosine theorem

X can be described in two ways, which will both sides of the equation

Nah, forget the cosine theorem

Tangens will do here

[short dagger]

sorry for making you wait

[remote pebble]

Angle D is 90
U never mentioned that

[short dagger]

I thought it was understandable that it was a quarter circle.

[remote pebble]

No its not

[short dagger]

ohh okey. My bad

[remote pebble]

Its ok
Never do that again

[short dagger]

Actually, your teacher wrote the question on the board in the lesson. I wrote the same on paper and took a picture.

I got the test book later

[remote pebble]

Bruh no one solved it yet??!💀💀

Im sry
Ill think of something

@short dagger
So draw CA and OD
And use Ptolemy's Theorem on OCDA(First prove that its a cyclic quadrilateral)
Then Pythagoras on ∆CAD

Oh ig he slept or something

Hope u solve it

[magic stag]

guys can u help me to solve this one please

just i have no idea how to solve it

i tried to use sin theorem , but it didnt work

[short dagger]

Thanks. I will try

[meager citrus]

For #8 the answer is 43(rounded) but I keep getting 37, what am I doing wrong

[wanton tangle]

guys how to calculate gradient

my ttest is in 2h pls help

(thanks in advance)

[magic stag]

please !!

[bitter veldt]

Is it gradient of a tangent to a curve or just a straight line

[magic stag]

did u solve it?

if u wanna my solution just let me know

[short dagger]

No

[short dagger]

I'm waiting your answer

@magic stag

[magic stag]

so the answer is 4

[short dagger]

The answer key says 6

[remote pebble]

The radius is 4√5 not CA
I think u missed that

[timber cedar]

Can someone help me

#1106310227084705794

[magic stag]

at first i also though that the radius is 4sqrt5
but after i have proven that OAD is an equilateral triangle , the radius will be 8

i dunno maybe i missed smth if the answer key says 6

[remote pebble]

Solution without trig

@short dagger

[short dagger]

Thank you for your solution, but can't the question be solved otherwise? Because I didn't learn Ptolemy's theorem before.

[remote pebble]

Nani

[short dagger]

Maybe by drawing the whole circle?

[remote pebble]

Ptolemy's theorem is not elementary?

[magic stag]

draw the whole circle with trig. and find the angles

so then u can use trig. rules to find the other sides

[magic stag]

so no one can help me to solve it?

[short dagger]

Can Ptolemy's theorem be applied in a quarter circle?

[remote pebble]

It can be applied in cyclic quadrilaterals

[remote pebble]

Oof
Wait ill think of something

I hate trig and am bad at it

[short dagger]

Can we use Euler's quadrilateral theorem on circular quadrilaterals?

[remote pebble]

Bro they taught you Euler's quad Theorem

But not Ptolemy's Theorem??

WTF

[short dagger]

Actually, I don't know either. So they didn't teach.

Because my country's education system sucks

[magic stag]

what country?

[short dagger]

Turkey

[magic stag]

ah

my country's edu also sucks xd

[short dagger]

What your country?

[magic stag]

Kazakhstan

[short dagger]

Maybe the bad education system is the fate of Turkic countries.

[remote pebble]

Its actually surprisingly simple

Tan (alpha) =(r/2)/r= 1/2

@magic stag @short dagger

Not better than Ptolemy's theorem but ok

Sry for the late😔

[short dagger]

Thanks

[short dagger]

No problem

[remote pebble]

:)

[magic stag]

gj bro)

[remote pebble]

I skipped a lot of steps
Ig u can understand xd
Ask me if u aren't clear with what i did

[magic stag]

how about this one?

can u solve it?

[remote pebble]

Maybe?

[magic stag]

please try to solve

just i tried sin theorem , but it didnt work as well

[remote pebble]

:/

Absolutely no idea

And its 2:45 am here
I have to sleep😔

Im sry

Gn

[magic stag]

np , gn

[timber cedar]

Some help please #1106310227084705794

[lone osprey]

I just finished geometry at my school, and since my school doesn't offer trig as a course (they moreover integrate it into pre calc), I wanted to study trig over the summer. I have seen that trig has some algebra 2 stuff, which I haven't taken yet

should I self-study trig?

and if so, what would be the best textbooks or resources to use

[surreal hill]

organic chemistry tutor has some amazing videos on maths, chemistry and physics

[timber cargo]

Organic Chem, Kahn Academy, NancyPi are some resources are highly regarded

[terse lynx]

i understand how to do the first part and proving its an isosceles triangle and i understand how to prove its a square but im lost on how to find point D. can please explain how to find the coordinates to D?

[tight solar]

segment BA will be parallel to CD.
parallel lines have the same slope. use the slope of BA to find CD. Or use the slope pf BC go find AD.

since the slope of BA is (-1/6) -> do the same thing from point C.
therefore D is (0, -5)

[upper karma]

Isn't d 0,-4

[tight solar]

theres a very good chance i miscounted or something

[upper karma]

Ye

[tight solar]

yes it is lol

ty

[twin acorn]

use cos law, I calculated final answer: square root 2

[magic stag]

did u use only cos law? just i can’t get it at all

can you send me ur full solution?

[dusky raft]

help

[snow crystal]

pythagorean theorem

[dusky raft]

thanks dawgh

[carmine ore]

Yeah. Is the objective to find the x on the lhs?

And then you have an x along the hypotenuse that I assume you want to know.

[weak orbit]

Also see "power of a point" which is especially helpful to solve more general configurations like this

[magic stag]

woow thank you bro!!!

[grand trout]

someone please explain this

I do not understand

bro hello

where r my mathematicians at

bro please my exam is in like a few days

[grave pond]

You should proably have started learning the material a bit earlier than that.

[jade kiln]

the answer is 8=x.

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[jade kiln]

Ok. Got it

[brave pebble]

lol

[thorny bough]

I know it’s a parabola

The graph is coming out something like this

It’s a rough draft although

But how do I find out the slope of l1 and l2?

I do know the shaded area will be found out by integrating it but how do I get there?

[fast moat]

On the picture you provided you already calculated L1 and L2. so i dont understand your question. Or do you want to know how/why thats correct?

Integrate f(x) from 0 to 2 and subtract the 2 triangles with each having an area of 1/2 (height and base are both one). (If you want, you could also integrate both tangent lines, so you integrate L1 from 0 to 1 and L2 from 1 to 2 and then subtract both from the integral of the parabola between 0 and 2)

[hardy obsidian]

does anyone know how to do 39? I can do it with calculus, but I have a gut feeling that I can do it with just basic geometry. I've worked at it and I'm stuck

[exotic yarrow]

There's only one intersection, so ||use discriminants||

[hardy obsidian]

@exotic yarrow what do you mean discriminant? I've only seen discriminant used for seeing if a conic was a certain type. Is there a different interpretation?

[hardy obsidian]

<@&286206848099549185>

[upper karma]

A circle is created when a sphere is intersected by a what plane?

I don't understand this simple question

<@&286206848099549185>

[grave pond]

Any plane that actually intersects it and is not just a tangent plane?

[vivid plinth]

There is a way to solve this by setting the discriminant to 0 and substituting to find the tangent line, but I haven't worked with that before so I can't explain that very well. This is my method of solving: Find the difference of the equations of the circles, in terms of either x or y. Substitute this into one of the circle equations (the 2nd one would be easier) to remove one of the variables and solve for the other. Now that you know the value of one of the variables, you can find the value of the other variable by using substitution into the circle equation again. At this point you have an x and y coordinate, and this ordered pair is the point of tangency. You also know the centre points of both circles, so choose one and use slope formula to find the slope of the line passing through the centre point and the ordered pair (x, y). This is perpendicular to the slope of the tangent line, so use -1/x to find the slope of the tangent line. Finally, you have the common point of tangency and the slope of the tangent line, so you can find the linear equation for the tangent line and then substitute for the x and y intercepts (a & b). Lastly, you add these together.

That was quite a mouthful, so ping me if you need clarification for anything.

[hardy obsidian]

@vivid plinth is there a way to do this with similar triangles?

someone told me theres a way but I don't see it

[vivid plinth]

I'm not sure, off the top of my head I don't think it can be done using similar triangles.

[hardy obsidian]

idk where he got these proportions

[tight solar]

https://math.stackexchange.com/questions/256100/how-can-i-find-the-points-at-which-two-circles-intersect

since they only intersect once, I don't know if you can use similar triangle trig for it (look at the second answer to this question on stack)

[vivid plinth]

Those values are correct, but I'm not sure how he got to that. This is my step-by-step process, if you can understand what I did.

[hardy obsidian]

ok so find the point of tangency, then its perpendicular to the two radii

[vivid plinth]

Updated picture ^

[vivid plinth]

yes

technically the difference of the circles has 2 solutions, but one passes through the point of tangency and one does not, so I removed the other one because we dont use it

theres a lot of extra steps I put in the table to make my process clearer, the actual process is fairly easy

all you really need is the slope of the line passing through both radii and the point of tangency

[midnight summit]

where tf do i find free material to learn about fractal geometry

[grand trout]

yeah we r still learning new things in my class right before the exam

[grave pond]

The distance between the two centers at (-6,-3) and (0,0) is sqrt(6²+3²) = 3sqrt(5). Their radii, which are sqrt(5) and 2sqrt(5), add up to this, so the circles are indeed tangent to each other, and the point where they meet is a third of the way from (-6,-3) to (0,0), that is, at (-4,-2).
The slope of the common tangent is perpendicular to the slope of ½ joining the centers, so -2.
Now it is easy to compute the intercepts (or just plot them out on graph paper).

[upper karma]

csc, sec, and cot are simply the reciprocals of sin, cos, and tan, respectively. So 1/sin(x) = csc(x) and so on

yes, csc is h/o

[grave pond]

They were convenient to have tables of back in the time before calculators.

Whether they are still taught at all differs a lot from country to country.

I never saw them during my high school or university days, but they seem to be standard curriculum in America.

Also: No, sin^-1 is the inverse function of sine (it takes a ratio and gives you an angle), whereas the cosecant still takes in an angle, and just gives you the opposite of the ratio you get from the sine. Multiplying by cosec is the same as dividing by sine.

Oh yes, if you find a calculator with an arccosecant button ;-)

[vivid plinth]

Oh that's much simpler, I didn't think of that.

[shadow swan]

is that uil general math?

[hardy obsidian]

yes

[cold wave]

anyone know gr 8 math 😭 lmfao, i like searched the internet for the formula but non seems to work anyone got acc formulas'?

the answer came oddly close with this one calculator but the the numbers after the decimal places were way off

[warm stag]

can anyone help me with deltamath?

[silent plank]

@cold wave the question wanted you to use an approximation for pi

[near harness]

The question asks you to use the value of pi as 3.14.

[mint garden]

Use pirl

Diameter is 26 so r is 13

So 13pi*19

Then add base of pi*r^2

[whole karma]

why is cos(x+180) = -cos x

[tight solar]

think about it in terms of numbers:
cos(45°)= sqrt(2)/2
cos(45°+180°)=cos(225°)=-sqrt(2)/2

adding pi (or 180°) to any angle value inside the cosine function will return it’s opposite

[sick hemlock]

the angle (x+180) lies in the third quadrant. if you draw a unit circle with the centre as the origin and take a point on the circle which lies in the third quadrant, it becomes more clear as to why it's negative

try to construct a right angled triangle using that point, by dropping a perpendicular on the x axis

[whole karma]

oh

and the adjacent side

becomes negative

and the hypotenuse stays positive

[sick hemlock]

yep

[whole karma]

so it ends up being negative

thanks guys

[sick hemlock]

I didn't think you'd catch on without a figure, impressive

[cold wave]

Oh dam ty

[devout glacier]

If you plug them in desmos you will see that the graph just gets “shifted” 180 degrees to the right

[thin locust]

to prove that AX⊥YZ, can I show that AYX = AZX?

[past nexus]

hello, I'll try my best to have good english here xD, I need a little help figuring out why a certain formula is used to get an angle. also why all of a sudden the value that gets me that angle which is the value of "a" changes , I will send the sketch and what I've calculated so far + solutions

btw the lession in which this problem is, is called trigonometry in planimetry (at least when translated to english)

this is the sketch. the information I have: a:b=6:5
, r=4 (the radius of the inscribed circle)
, α=106∘15′36′′

my calculations so far:

c1 = r/tan(Y/2)= 22, b1= r/tan(B/2)=8, a1= r/tan(A/2)=3 ,a=b1+c1=30, b=a1+c1="5, c=a1+b1=11, Y=180°-A-B=180°-106°15'36"-53°7'48"=20°36'36", so I do have the solution to B however I don't get it in the same way that my teacher showed me and told me how I should do it. basically what my teacher wrote was :

a/b=sin(A)/sin(B)=> 6/5= sin(A)/sin(B)=> cos B = 3/5 = 53°7'48"

so anyways, in my book it is written that in a right triangle, sin(A)=a/c=cos(B) , therefore I understand why the sin(A) was swapped with cos(B)

but what I dont understand is why did 6 turn into 3, in this way of solving it? or is it that we now used here a1? but how could it be? idk. if anyone could help I'd appreciate it.

[carmine siren]

En un triangulo isoceles ABC(AB ES IGUAL BC), LA MEDIATRIZ DE AB y la bizectriz del angulo interior A se intersectan en un punto de BC,Calcula medida del angulo b

[civic scroll]

One message removed from a suspended account.

[ruby fulcrum]

can someone help me with this problem

Put the conic section into Graphing Form: y2+14y−2x+59=0

[tight solar]

looks like a sideay parabola to me - shoot for the form:
x=a(y-k)^2 + h

[ruby fulcrum]

how did you know the x was what I needed to solve for

[tight solar]

so i’d say you could start by completing the square for y

[tight solar]

my precalc teacher made us memorize the formulas

yeah unfortunately all i can say i memory for me

[ruby fulcrum]

ok ty

[tight solar]

y is squared but x isnt, has to be a sideways parabola

gl 👍

[ruby fulcrum]

got it

[tight solar]

and you know, if youre on a test and run into a form you dont know - make a table and graph the points. hopefully a conic appears

[hot mortar]

the only geometry i like is geometry dash

[upper karma]

is tan(x/2)=secx tanx?

[wet blaze]

guys pls

help

de is angel bisector

it asks for ad

[austere mortar]

how did you get that |ED| = 2√6 and what does that word on the right mean thats written by |AD|?

[fresh wedge]

pythagoras theorem in traeng AZD ?

[south summit]

help

[fresh wedge]

have u got translate prolly?

[south summit]

it says that AD splits angle <BAC in half
and i need to prove that the triangle area of ACD is m^2xtanB

AC=2m, <ADB=<BAC=2B are given facts about the triangle

[fresh wedge]

uh...Have you tried to prove it through the similarity of triangles? sorry for late answ

[south summit]

no

how can you prove it with similarity of triangles

[fresh wedge]

uh..8th grade (Russian school exactly). U need to prove them with these signs: https://ru.wikipedia.org/wiki/Признаки_подобия_треугольников
idk why this article is on rus, but u can try to translate

[south summit]

i know what similiarity is but what triangles are similiar here?

[fresh wedge]

it's more complicated... because your ad is not perpendicular. Most likely bad and dac but I'm not sure I haven't done geometry in a week

[dusky raft]

#1 ?

I was absent

And my teacher is dumb

And doesn’t email back

[wet ledge]

Hey can someone help me with a few problems on a review for my geometry test I can’t fail I’m trying to graduate a year early and this will just make it harder

I just need help on a few problems I think it’s around 5 and that’ll be awesome

[neat python]

CB and CA make angles of 60° and 15° with parallel lines a and b, respectively. Find the angle ACB.
a) 45° b) 40° c) 35° d) 30
https://cdn.discordapp.com/attachments/961574968628510800/1108015815342706859/image.png

[opaque phoenix]

So i got it

It's 45°

Due to exterior angle property

45 + 15 = 60

And the exterior angle is 60°

Cause it's the corresponding angle to A and a || b

Oh wai

Yea

What happened to me this is like the easiest sum ugh

The answers 45 right?

@neat python

Wait is this a trick question

[neat python]

idk answer but ty for help

[neat python]

nah

[opaque phoenix]

Ok then it's 45

[neat python]

how did u get it

[opaque phoenix]

You understand right?

Ok so two properties come into play

Sum of all angles in a triangle = 180

[neat python]

yes ik that

[opaque phoenix]

The corresponding angle and the angle will be equal

[neat python]

yeah

[opaque phoenix]

Let's put D at the intersection of b and AC

And the other vertex of the line segment let's call it E

So

You understand?

Why did you delete it

[neat python]

Like i did like this

[opaque phoenix]

Ok continue

How do you proceed

[neat python]

Then u are telling me to add D where

[opaque phoenix]

Ok lemme just draw and show

[neat python]

Okay sure

[opaque phoenix]

It's linear pair btw

I couldn't properly take it

[neat python]

Okay let me see

[opaque phoenix]

@neat python

[neat python]

Ty btw

[opaque phoenix]

You understand?

These properties are pretty standard

[neat python]

yeah i understood

ty for help

[opaque phoenix]

No problem

[opaque phoenix]

Great!

Sorry for the terrible drawing and handwriting btw

[neat python]

Lol no problem

[weak orbit]

this figure 😭

[opaque phoenix]

Wtf is that

[wet blaze]

is 2x sqrt10

[austere mortar]

*"|ED| = 2√6", sorry

[austere mortar]

and yeah, thats better for |ED|, does |AD| split the angle of ADE in half?

[steel dust]

gonna ask a stupid question, you cant use cosine, tangent and sinus with a angle greater than 90 degrees, right?

nvm i found it

[analog quarry]

you cant use tan for an angle that IS 90 degrees because tan(90) is undefined

also tan(270)

repeat forever in 180 degree intervals

[steel dust]

ah yeah

[analog quarry]

or pi radian intervals

depending on how you feel

[sand cairn]

This one is a trigonometry problem that needs me to find the radius of the circles.

How do I know that the right triangles are 30-60-90?

[upper karma]

thats a tricky one give me a bit

[sand cairn]

sure

[sand cairn]

Yes, average solve time for this problem is 2d12h30m

[safe tundra]

you can assume some things

start by assuming the values of the legs of the triangles: let the short be a and the long be b

then the side length of the square is b-a

write 2 equations, one using pythag, and 2nd being that the radii of the inscribed circles must of be equal

ping if u need more help

[trim plume]

I’m so lost

[smoky lantern]

@trim plume dm

[novel laurel]

what do we have to find here?

[upper karma]

actually i went sleep instead of breaking myx head

but i think i can solve it

[upper karma]

on god haha

[patent apex]

i hate this

[patent apex]

could it be (sprt(3) - 1 )/2

what no

i need to prove it's 90, 30 ,60 triangles

[austere mortar]

how does this work? Im genuinely curious

[nocturne remnant]

Me when the 1-0-i triangle is similar to the -i-0-1 triangle

🗿

[steady vault]

Can anyone explain why sin(a+b) = sinAcosB+cosAsinB?

[upper karma]

Simple

it's the law of sines

[steady vault]

ye but why is like that

[weak orbit]

you mean you want a derivation?

[austere mortar]

might be a bit heavy info wise becouse it might spoil the answer, but its a question: ||couldnt you try and prove it somehow by cheking the ratios of one of those trjangles sides since all triangles with the same angles are similar?||

[ashen jetty]

hello everyone, my little bro is asking for help for this assignment. and to be frank, i'm not good at math nor do i understand this. can anyone help in "finding the value of each variable used in the figures"?

[shell hazel]

1. t=8, r=8 √2

when angle is 45 degrees, the perpendicular and base are same.....and the hypotenuse has an extra √2

2. g= √2, l=2

4. n=o=15

5. e=3

m= 3√2

[ashen jetty]

thank you

how about this, sir?

[hallow flare]

WHAT IS THIS BULLSHIT

nah nah cnah

nah

[frozen ocean]

The question says "use the box to complete the equation"

That's what it asking for

[frozen ocean]

What grade are you by any chance?

[upper karma]

🤮 degrees? eww

[wind coral]

<@&268886789983436800> this seems a bit suspicious

[vivid plinth]

Yes, the exact same text was posted by a different account an hour ago.

The other account got banned.

[wind coral]

Probably a hijacked account lol

[vivid plinth]

Ya probably

Let's just wait until the mods deal with this account.

I don't want to bother them too much

[small ingot]

can someone help plz

[upper karma]

@opaque flicker

[opaque flicker]

Anyone wanna help me out pls?

[upper karma]

can u rotate it

[opaque flicker]

Sure

[upper karma]

for 1 a here's how you do it

you consider an angle and the side opposite to it

and the sin(angle)/side is the same for all 3 possibilities

so sin(x)/22 = sin(42)/17

now isolate x

[steady vault]

Yeah I want a derivation

[opaque flicker]

Can u show how bro I’m still lost

Then I can do rest

Anyone?

[upper karma]

I explained the process

[opaque flicker]

Do I multiply by 22?

Since it’s dividing

[sage lodge]

yo so like this is put weird like do I just plug in numbers for n

[upper karma]

Yes

[upper karma]

yes

for example let's look at 1

plug in n=2

a2 = 3a1

[sage lodge]

ohh ight so the 3 isnt a term

i was thinking the an-1 was separate

ty ty

[upper karma]

since a1 is 1 a2 is 3

a3 would be 9

make sure to pay attention to what happens to terms like n-1 when you plug in specific values of n

[opaque flicker]

Would this be correct?

[split mountain]

Hmm, i’ve got a velocity expressed in polar coordinates…

I’m not very familiar with polar coordinates. If i want the displacement, do i still integrate the velocity?

[azure pumice]

Hi, I need someone that can teach me how to get the area and perimeter of an pentagon inside a pentagon

[lament swan]

Hi I need the intersection of C

But I dont much information

[vivid plinth]

Are you given that angle ACB is 90 degrees? Otherwise I don't think you can solve this.

If AC is perpendicular to BC, then you can first find the slope of AC using the slope formula, and then you know that the slope of BC is the negative reciprocal of the slope of AC, and BC passes through (0, 5/2). Using this information you can make a system of equations and solve for the intersection points.

[weak orbit]

alright ill give you a figure and you try to derive it from there.

consider a rectangle ABCD in anti clockwise order. draw two lines from A to M and N on BC and CD respectively such that $MN \perp AM$. also let AN be equal to 1. let $\angle BAM = \alpha$ and $\angle MAN = \beta$. now try to name all the segments using combinations of sin, cos of alpha and beta.

[somber coyote]

prophetic potato

[torn gazelle]

Could someone help me with b and c?

[alpine pagoda]

for b use Pythagoras theorum

[torn gazelle]

that’s what I did for a

[alpine pagoda]

for c use the formula for volume of cone

[torn gazelle]

It was 3^2+9^2

So then a I got 90

[alpine pagoda]

but in a it asked for it's slant height

[torn gazelle]

Yes

AB is straight though

[alpine pagoda]

yeahh

[torn gazelle]

So I can use Pythagoreans theorem

Right?

[alpine pagoda]

yess

[torn gazelle]

Ok

[torn gazelle]

So then AB=90?

[alpine pagoda]

but for a I think you should use the formula for slant height

[torn gazelle]

Could you remind me that formula?

[alpine pagoda]

l= underoot height^square + radius^square

[torn gazelle]

Which would give us the same thing?

[alpine pagoda]

maybe?!

let me solve this briefly

[alpine pagoda]

yess 90 is correct

but I think it is more like underoot 90

[torn gazelle]

Ok

Then how would I solve for b?

[alpine pagoda]

for b we have to solve it using properties of triangle

as it is written there

its the sign of angle

[torn gazelle]

We could use tangent

[alpine pagoda]

try it '

I'm finding volume

c'

[torn gazelle]

Wait how would I do that?

I can’t use tangent cause I don’t have an angle measure rn?

[alpine pagoda]

yess

do you know about properties of angle of triangle

I have solved c

[torn gazelle]

Kinda

Oh yes I do

[alpine pagoda]

then solve using them

[torn gazelle]

Which ones?

[alpine pagoda]

I don't clearly remember it's been 2 years since I have solved these types of questions

In my syllabus I now only have trigonometry and calculas

[torn gazelle]

I’m lost rn😂

How did you find c?

[alpine pagoda]

using volume of cone

formula

[torn gazelle]

V=1/3Bh=1/3pir^2h

[alpine pagoda]

V = 1/3 x πr2 x h

put the pi's value 22/7

[neat python]

@opaque phoenix

i have question

[opaque phoenix]

Yea go ahead

[torn gazelle]

So then it’s 1/3•pi 3^2•9??

[alpine pagoda]

no

1/3 x 22/7 x 3 x 3 x 9

now after solving this we get

[torn gazelle]

Why 22/7?

[alpine pagoda]

it will be easier

you won't need a calculator

[torn gazelle]

Then pi?

That’s the same thing?

[alpine pagoda]

it is the value of pi

[torn gazelle]

Ok

[alpine pagoda]

pi has two values

3.14 and 22/7

[torn gazelle]

Ok

So then we get

1/3• 22/7•81

[alpine pagoda]

yess

[torn gazelle]

Then what’s after that?

[alpine pagoda]

solve them completely

we get

594/7

after dividing

final ans is

84.85 cm^3

[torn gazelle]

Ok thank you

[alpine pagoda]

most welcome

you understood what I meant?

[torn gazelle]

Yess

I just need to find the measure of angle BAO now 😭

[alpine pagoda]

🥲

[torn gazelle]

I’m confused with a-f

[alpine pagoda]

f) trapezium

last year frustum part was cut out from our syllabus

and I didn't even help learning it myself

so I have no Idea

[neat python]

i will dm u

[barren lantern]

b² represents the area of the bottom square?

[edgy apex]

help or imma fail😫😫

[alpine pagoda]

2nd option 308 pi

using total surface area of cyclinder

[alpine pagoda]

126 pi

[edgy apex]

for #2

?

God bless you

[alpine pagoda]

yess

[edgy apex]

alr and what abt the 126 pi

where that at

i’m p sure the others are wrong

like the ones i did

[alpine pagoda]

question 3

[edgy apex]

🙏

are the others correct?

8,9,10 not done

[analog quarry]

it has the same value of pi to 3 significant figures

if you need an exact answer just leave it in terms of pi

22/7 will give you an estimate

[alpine pagoda]

question 1 is 226 m

[edgy apex]

❤️

[alpine pagoda]

question 4 , 200

[lusty needle]

Ya sure we should give em direct answers?

[alpine pagoda]

the questions are direct

and he is solving them wrong

[edgy apex]

🙏❤️

[lusty needle]

Well we can atleast provide em with formulas and ask em to try to solve atleast

[edgy apex]

nah years over

i don’t have anymore test

i can just do these for free points

so i can PASS

😭😭😭

[lusty needle]

Yeah and that's what we r supposed to do ryt? "Teach" Em how to do instead of just telling em direct answers

[alpine pagoda]

you're right

[edgy apex]

nah this is my last semester of geo

i jus need

BRO PLS

I DONT TAKE GEO ANTMORE

I DONT NEED THAT

pls no teach

[lusty needle]

Well

[edgy apex]

i just need the point bro

bless up for me

🙏🙏

[lusty needle]

That's not allowed here

Acc to server rules

[analog quarry]

that would be academically dishonest

[edgy apex]

oh my goodness

[analog quarry]

and this is an academic server

[edgy apex]

UNREALLL

[analog quarry]

people come here to learn math

not pass math tests

[edgy apex]

this is a practice quiz

for free points

not a literal test

it’s just free points

it’s in test format tho

[analog quarry]

then learn the concepts and get the points legitimately

[vivid plinth]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[edgy apex]

praying on my downfall fr

[vivid plinth]

We can't give you direct answers

[analog quarry]

we're praying on you learning the concepts and how to apply them to practise questions

[analog quarry]

and we're all offering our time as volunteers to help teach you the concepts you need to learn

[edgy apex]

again it’s not test or exam

i’m willing to learn the concepts

but i have sooo many of these to do

😭😭

[lusty needle]

And we are willing to teach.

[analog quarry]

if you learn the concepts and methods, each one will only take a few minutes

at most

[edgy apex]

so its if y’all are willing

[alpine pagoda]

yeahh

[lusty needle]

I mean this makes it sound its a test.

[alpine pagoda]

we all are here for you

go ahead

[analog quarry]

what specific concept are you struggling with

what part of the questions are getting you stuck

we'll see if we can explain it in a way that's easier for you to understand

[lusty needle]

Bro disappeared

[edgy apex]

all of it bro

i literally don’t comprehend

Geometry

[analog quarry]

okay so start at question 1

or at least the first question you sent in

do you understand what surface area means

[edgy apex]

Yea

i had a formula paper

of vol sa and other stuff

[analog quarry]

that's a good place to start

explain how you'd find the surface area of, say, a cuboid

[edgy apex]

SA = B•H

p sure

[analog quarry]

that's the formula for the area of a rectangle

surface area is the total area of all faces of a 3d shape

you understand what faces are when talking about 3d shapes?

[edgy apex]

yea

congruent or facing each other

?

[analog quarry]

face literally just means a surface

3d shapes, or polyhedrons, are made up of 2d shapes, or polygons, joined together

a cuboid, the polyhedron, is made up of 6 rectangles, which are polygons

are you happy with that idea

[edgy apex]

yea

[analog quarry]

ok cool

so the question's asking you about a cylinder

how many faces do you reckon a cylinder's got

it's not a super intuitive answer so don't worry if you don't immediately get it

[edgy apex]

2

[analog quarry]

you're not far off

a cylinder's actually got 3 faces

[edgy apex]

☠️

[analog quarry]

it has the 2 circles at the ends, and a rectangle that wraps around the shape to join them together

a lot of people dont realise that the curved part is a face but it is and you can find its area

do you know the formula for finding the area of a circle

[edgy apex]

yea

C = pi diameter

[analog quarry]

that's circumference which is the length of the line around the outside of the circle

[edgy apex]

yea nah bro what the hell

[analog quarry]

but you will need that in a minute so keep that fresh in your mind

[edgy apex]

i’m so slow

[analog quarry]

the area of the circle is πr^2

pi times radius squared

[edgy apex]

ah ok x

[analog quarry]

you're comfortable with what pi and and the radius mean, yeah?

[edgy apex]

yea

[analog quarry]

i assume you know what pi means because you know how to use it to find the circumference