geometry-and-trigonometry

[hasty karma]

yep

i just seen similarity symbol more curved, almost like infinity

[static echo]

If I have a number in radians, how do I figure out what hyperbolic sine gave it?

Like hyperbolic sin of x gives 83 or 13 or something (or 0.<stuff>)

How do I find x?

[grim ruin]

the "axiom of Euclid" in Tarski's axiomatization of Euclidean geometry says that, if r,s are rays from the origin O, if ∠(r, s) is the angle between them, if C is a point in the interior of ∠(r, s), and if C is between O and another point C', then this point C' is also in the interior of ∠(r, s)

on the other hand, the "parallel axiom" in Hilbert's axiomatization of Euclidean geometry says that, if L is a line on a plane, and if P is a point of that plane not lying on this line, then there is at most one line on this plane that goes through P and is parallel to L

I know both of these axioms are equivalent to the fifth postulate, and thus to each other, over their respective axiomatizations of neutral geometry. my question is: what is the proof of the parallel axiom from the axiom of Euclid? or, in other words: how do I prove the parallel axiom from the axiom of Euclid?

[dark sparrow]

@static echo inverse hyperbolic sine

or fuck around with exponentials to solve a glorified quadratic

[static echo]

okay, ty

[spice thorn]

can someone help me with this exercise
its supposed to be something with Thales theorem or angle bisector

or triangle similarity

here it is

the value of EDA=90
AB is angle bisector for A (a1=a2)
BCA=2BAC
prove that BC X ED= BD X EA

i have come into some conclusions but not sure if they are handy

first of all i see that AD/AE=DB/BE bc of the angle bisector theorem
say that angle C=2a
that means that a1=a and a2=a and A=2a
and also angle EDA equals to angle EDC equals 90
that gives us triangle DCB is similar to triangle DAE because of the AA triangle similarity theorem
i dont know where to continue

[grim ruin]

you've done almost all of the work. since BCD and DAE are similar right triangles, then the ratios between their corresponding sides are equal. in particular, since BC and AE are the hypothenuses (as they are opposite to right angles), and since BD and DE are corresponding legs (as they are opposite to the angles C and A respectively, and you've shown them to be congruent), then the ratios BC/AE and BD/DE are constant and equal. multiplying both sides of the equation BC/AE = BD/DE by AE×DE, we have that BC×DE = BD×AE, which is what was required to show

[spice thorn]

so we didnt even need the angle bisector theorem in this section

[grim ruin]

you could do it by the angle bisector theorem as well, but then you'd have to do some algebra with the proportions of the sides, and simply proving the similarity of the triangles, as you've done, and inferring from it the desired proportions, works just as well

[spice thorn]

in the next section i have got to prove that BC X ED = AD X BE

could u help me with that please?

maybe this one has something to do with the angle bisector theorem

[grim ruin]

yeah. that one follows immediately from the theorem

[spice thorn]

how

i dont even understand

[grim ruin]

the theorem says exactly that, since AB bisects the angle, then the ratio BD/BE is equal to the ratio AD/AE*. note that BD and BE are just the two segments in which the line DE gets cut into by the bisector AB, and that AD and AE are just the sides of the triangle ADE whose angle A got bisected and which are both adjacent to that angle

since BD/BE = AD/AE*, then again, multiplying both sides by AE×BE, we get that BD×AE = AD×BE*

if this section uses the theorem, then it means you use this equation somehow

otherwise, you can go by the same route of triangle similarity as you've done before, which works too

[spice thorn]

isnt it supposed to be BD/BE=AD/AE?

with the angle bisector theorem

[grim ruin]

yeah

my mistake

[spice thorn]

okay

wait

how did we get from BD X AE = AD X BE to the answer

oh cuz we said earlier

that BD X AE = BC X DE

[grim ruin]

yep

[spice thorn]

i get it now

[grim ruin]

the rest is just a bit of arithmetic fiddling with the ratios

[spice thorn]

i dont even see when we used ratios as if we worked all the times with letters

when we compared between them

oh wait nvm

anyways thanks for your time

and explanation

🙂

[grim ruin]

[grim ruin]

.

[upper karma]

Is 1/sinxcosx = 1/sinx * 1/cosx?

[ionic vine]

yes

[cedar venture]

The large quadrilateral shown has 2 right angles, and two of the sides have lengths 7 cm and 10 cm. B and D are on the other sides so that AB= 2 cm and DC = 6 cm.
The area of the shaded quadrilateral ABCD in cm² is?

thanks

[wicked jay]

x=140

[celest gull]

Hey sorry to bother with a question. If I spin a sphere around its y-axis, is there a term for the very top and bottom y-axis of the sphere? Basically the poles. Something like zenith and nadir ?

[snow tide]

they’re just called poles

[fierce vigil]

what type of angle relates 24 to the other triangle

[smoky jetty]

seems correct for me

[woven fulcrum]

how do u find the value of k here?

[river vapor]

hey guys so I'm really only just a highschooler preparing for entrances and wanted to know if there's more to trigonometry

so far I've been taught how to calculate values of ratios of trigonnometric functions via double angle half anngle and other few methods

but I wanted to know what's really the best method to calculate these values? like I could only find methods with approximations on google

so what'ss like the best method to calculate any trigonometric function's value? without a calculator

[nocturne remnant]

can you give an example of such a question?

[river vapor]

like give me the ratio for sin 58(degree angle)

like if I could find literally any angle it would really ease up solving sums

[nocturne remnant]

It’s not constructible I think

[river vapor]

like how does the calculator give out the value ik there are methods with infinite series but still they take lots of time

@nocturne remnant

[river vapor]

so is there no way I can get the value without the calculator

[nocturne remnant]

Yeah maybe don’t tag someone just because they aren’t responding within 15 seconds; they’re not obligated to respond at all

But anyway

https://math.stackexchange.com/questions/395600/how-does-a-calculator-calculate-the-sine-cosine-tangent-using-just-a-number

[river vapor]

sorry for that

[nocturne remnant]

It’s pretty complicated, to put it briefly

[nocturne remnant]

Plus, why would you ever need to calculate sin58 for entrances?

[river vapor]

so like is this one the same as the one I found earlier

https://www.euclideanspace.com/maths/geometry/trig/series/index.htm

or is this one different

[nocturne remnant]

The stack exchange also mentions something called CORDIC

And loads of other things

But I don’t know anything about them either

[river vapor]

thanks anyways

[nocturne remnant]

just be familiar with compound angle formulas, you'll do fine

[river vapor]

yeah I can do those pretty much well

[trail dagger]

do i set my calculator to deg or radiants

im so confused

[jaunty quail]

can someone help me with a proof

[gleaming coyote]

can someone pls help me with this im so sorry

and this pls im sorry-

[timber cargo]

Area shaded = Area Large - Area Small

[timber cargo]

Calculate the total area of the rectangle and the total area of the triangle

and add the two

[gleaming coyote]

can u show me how to solve it? I'm very sorry I really need the answers and proof

[timber cargo]

Alright

Give me a second

[gleaming coyote]

thank u^^

[gleaming coyote]

tysmm

[gleaming coyote]

could u explain this one as well with answers ;-; im so sorry I really don't get it

[timber cargo]

@gleaming coyote Here you go

[gleaming coyote]

oh wait are we not supposed to multiply it with pi?

[timber cargo]

You could, you get the same answer

[gleaming coyote]

ohh

OH

TYSMM I DID

[willow sable]

Help me

Please

My grade relies on this test

[gleaming coyote]

just a guess i think its 70 as well

[willow sable]

I need a reason why

[gleaming coyote]

since it looks like it has come congruency

[willow sable]

What do i say

For y

[coral wraith]

Not sure how to do this mad stuck rn

[smoky jetty]

pheww, if there was a relatively easy way, I'd be revolutionizing my trigo comprehension again, lmao

[smoky jetty]

somehow the same thing I was thinking of earlier, nice stuff!

[upper karma]

hey, can anyone explain me what exactly a locus is?

[smoky jetty]

In geometry, a locus (plural: loci) (Latin word for "place", "location") is a set of all points (commonly, a line, a line segment, a curve or a surface), whose location satisfies or is determined by one or more specified conditions.[1][2]

In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property. - wikipedia

[timber cargo]

Great minds think alike

[rich linden]

im dumb with proofs can someone help?
<XOW =1/2 <YOX

there was other solutions but its either

Definition of angle bisector
or
**Angle bisector theorem **

[upper karma]

<YOX is 90 degree i believe, and line OW is bisecting it means cutting it into equal part ( or say equal angles i.e < XOW and <WOY )
it means <XOW is half of <YOX i.e 45 degree
and 45 degree = 1/2 * 90

[rich linden]

yea so angle bisector?

[upper karma]

ow is angle bisector

[rich linden]

oh no this isnt the question XOW =1/2 <YOX

this was one of the things i had to prove

[upper karma]

read it what i wrote

[rich linden]

yes you said 45 degree = 1/2 * 90

but which one is the proof i have to choose

Definition of angle bisector
or
Angle bisector theorem

[smoky jetty]

i'd say angle bisector theorem?

is that for the reasoning?

[upper karma]

theorem

i think we generally use theorems to prove things

[smoky jetty]

yeah

[rotund raft]

hi, does anyone understand herman-mauguin symbols for crystallographic point groups?

also is this the right channel for those

[ivory atlas]

Hey. Can someone help me with my work? Please

[tulip depot]

yo

need help

[rotund raft]

you multiply the distance from the centre of enlargement by the scale factor

so like, A is at (1,2)

and the scale factor is 3, so A' is at (1* 3,2 *3)

or (3, 6)

and the same for B

[tulip depot]

wow

howd u do it

[rotund raft]

this

[tulip depot]

had 1 more

[rotund raft]

its the same thing, but with a sf of four

so just use the same method

[tulip depot]

wait i mutipliy with 4?

dont really get it

can u givve me the answer for tht 1

[rotund raft]

so you scale the distance from the centre of enlargement by the sf

as the centre is (0,0), thats just the coordinates of the point that you are scaling

for A, the coords are (1,2)

so A' goes to that multiplied by 4

and the same for B

[tulip depot]

its showing i got wrong tho

[rotund raft]

oh

for which question?

[tulip depot]

the 2nd one i send

can u just give the answers

[rotund raft]

what did you put as the answers?

i think it should be
a) A: (1,2), A': (4,8)
b) B: (12,4)

[tulip depot]

omg

thnak u so much

i was stuck in this question

only this 1

[rotund raft]

np

[tulip depot]

omg tysm

[ruby jacinth]

isn't that just giving him homework answers though

[rotund raft]

it is a bit, sorry

I did explain the process to get there, though

[ruby jacinth]

it doesn't seem like he understood it though

[lament flume]

Idk how that was right it asked for B' not B

[rotund raft]

oh that was a typo

the coordinates given are for B'

[lament flume]

Oh I see

Yeah it is never mind.

[lyric walrus]

hello

[tribal elk]

can anyone teach me trigonometry

[hasty karma]

you had bad teacher?

it is actually easy operation

you scale something by 3

ill not dive into linear algebra

just keep in mind that point that had coordinates (x, y) before will have (3x, 3y) now

a' is transformed a
b' is transformed b

also scale (x, y) by 3 is same as scale (3x, 3y) by 1/3

also it is elementary transformation

scaling, without even zoom

oh, and last: you can get scale coefficient (i use k for it) by looking at same points, if you see that right angle of small has (x,y), and right angle of larger has (4x, 4y), then k is 4

[hasty karma]

so what you already know, do you have basic knowledge of right triangles and circles, pythagorean theorem and what you want?

you want to get basics?

or be able to solve some thick trigonometric equations?

[tribal elk]

circles yes triangles no pytha theory yes and no

[calm sandal]

help

[hasty karma]

so pythagorean says that in right triangle square of hypotenuse is sum of catheti squares

a^2 + b^2 = c^2

a,b,c - sides

cos A is how much b larger than c

imagine b is 1/2 of c

c = 1, b = 0.5
then cos A is 0.5

0.5/1 = 0.5

sin A is how much a larger than c

cos B is same, it is koefficient between c and b

cos B = sin A

soh cah toa
Sin = Opposite side/Hypotenuse
Cos = Adjacent side/Hypotenuse
Tan = Opposite side/Adjacent side

so imagine hypotenuse is now 2

and cos A = 0.5

it means b = 1/2c = 1

cos cannot be more than 1 or less than 1
sin too

triangle with cathet = 1 and hypotenuse = 1 is completely squished triangle, looks like line

right triangle with cathet larger than hypotenuse does not exist, if you see cos = -2 or sin = 5, something is definitelly wrong

but tan (or tg) can be any value

so, if b = cos A, a = sin A, then
(cos A)^2 + (sin A)^2 = 1

using this formula you can find sin if cos known or cos if sin known

also from definition of tan, it is true that
(sin A) / (cos A) = tan a

[hasty karma]

what means m in angle?

m<JKP

[nocturne remnant]

“measure of”

I believe

[hasty karma]

i see it has same font as letters in scheme or in angle

also isnt it obvious

like all you can find in angle is its meassure

[nocturne remnant]

Sometimes you want to refer to the angle itself, not its measure

[hasty karma]

ye, but how you can find angle?

[nocturne remnant]

?

[hasty karma]

what would change if take m out?
find <JKP

does it has two or more meanings?

[nocturne remnant]

Not really

The reason that some people use it is because they prefer to

[hasty karma]

@tribal elk also

ohshit embed fail

so рад means rad (radians), град - degrees

dont look at ctg, it is just 1/tg = cosA/sinA

it is quite rare and not really important at first

tg of 90deg is not defined, because zero division happens

you probably dont know what radian is.

1 radian is same angle as arc with length equal to circle radius would have, it is roughly 57 degrees

1 radian, text says

so perimeter of circle is 2pi*r

this means full circle is 2pi radians

half circle (0-180deg) is pi

90 degree is pi/2 radians

pi rad = 180 deg
1 rad = 180/pi ~ 57.29...

[lusty needle]

Damn this goes till 2pi pog table

[hasty karma]

you dont really need all up to 2pi

you can use formulas

or shift graph, since sin(x-pi) is just shifted horizontaly sine

2pi = 0pi, pi/6-pi/3 is same as mirrored 2pi/3-5pi/6

also cos is negative but it should be obvious

[lusty needle]

We need 2pi in defenite integrals tho lol

Gotta mug em up

[smoky jetty]

love this table!

I used to think that those were randomly there, barely memorizable. Not until I realized that they were just the ratios of special angles.

[upper karma]

How to get the rest of the sides

[timber cargo]

Sin of 30 is halve of the hypnethenuse. The side length opposite of 30 degrees is 4 cm, therefore the hypotenuse is 8.

To find the other side length, you could do another trick, but at your level, do it through the pythagorean theorem

[tulip depot]

i understood lol

i think

[short mason]

aldrin can you look at mine pls

[ruby jacinth]

You actually asked "can you just give the answers"

[tulip depot]

just a sec

[tulip depot]

ye coz i had to do it a little quickly

[short mason]

thanks m8

[tulip depot]

dint have time

[ruby jacinth]

Alright

[tulip depot]

gonna be honest idk wht tht means

[short mason]

can u look aswell pls

[short mason]

im lost aswell

[ruby jacinth]

Do you know what arithmetic progression means?

[short mason]

it adds k each time

k is a constant

[ruby jacinth]

So the result are separated by k?

[short mason]

yh

[frozen stump]

Anyone know how I can find the next angle

idk where to go from here

[orchid oar]

Please help

[bleak blade]

congruent tringlels

FGI=FGH

FGI+FGH=180

thus FGI=FGH=90

[frozen stump]

can someone pls help me

[modest wigeon]

since BEFG forms a quadrilateral, its angles have a total sum of 360, and from there you can get the answer for angle FGB

and also BEF is equal to 110 degrees rather than 70 because instead its angle CED thats equal to 70 degrees

[upper karma]

Srry for the late reply but thanks

[white nova]

I am not sure if SSS is right

I just want to make sure

For the last reason

[green fern]

help please t.t

[stray schooner]

shouldn't there exist line AD and BC by axiom 1?

thus showing that AB, AC, and AD is concurrent?

[nocturne remnant]

I think the point is that AD and BC also exist but is not useful for the discussion

[stray schooner]

why though? wouldn't it disprove that the dual of axiom 3 exist?

[nocturne remnant]

Read the theorem again maybe

“there exists” is not the same as “for all”

[stray schooner]

OHHHH gotcha, i didn't catch that one. thank you so much!

i can finally move on to the next page after atleast 3 hours kek

[smoky jetty]

did u get the angle BEF by dividing the isosceles triangle CED into two right triangles? coz that's what I tried, dunno if it's right tho (but I got 110 as well)

[kind plinth]

hey guys, any help would be appreciated!

[hasty karma]

good question

imagine some random triangle

find second (if you sort them by height) vertex

draw line parallel to OX (y=const)

so, is it possible to find point or intersection wihout division

as i see, it is impossible, right?

[smoky jetty]

i wonder how

[visual wagon]

[tulip depot]

can some1 explain this question

<@&286206848099549185>

[timber cargo]

Again with this?

[tulip depot]

uhh ye just needed explanation

[upper karma]

@tulip depot you are given A and A', and B but not B'. You need to use the scale from A to A' and apply that to B to find B'. A is (2, 1) and A' is (6, 3) so the dilated triangle is just the original one but multiplied by 3. So if we take B, (3, 1), and multiple it by 3, we get B', (9, 3)

[crystal crest]

Is there some kind of formula for finding the Exterior Angle of a regular polygon? Is it just (n-2)180 or was it (n-2)180/2?

[crystal crest]

I got it now, I just needed to do (n-2)180/2 and then subtract it from 180. I forgot how to do it for a bit 😅

[balmy herald]

i got that <BPA is 60 (ofc), and that <PBC=30, but im not sure how to find <PBC

[nocturne remnant]

Notice that BP = BC, so <CPB = <PCB

[balmy herald]

how do you know that

and also that doesnt really help me if i dont know <PBC

oh i know how to solve this

why isnt it 75?

[glossy smelt]

can someone give me a step by step on how to solve for x

[zinc zealot]

trying to find x and y but im stumped and i forgot the formula thats needed for this

[zinc zealot]

did i get this right

[mighty cradle]

bdc is 45 bc bd is the angle bisector of adc so bdc must equal 1/2adc

since adb + bdc = adb

[mighty cradle]

for bda im not quite sure how you got 65

this is a 65 degree angle

[mighty cradle]

oh wait nvm i forgot rectangles dont have bisectors like that

lemme look at this again

[mighty cradle]

okay so since ebc = 65 abe = 25 i think because rectangles' interior angles add up to be 180 and 65(2) + 25(2) = 180
so yea you have it right

[ebon mantle]

i need to find PR, but what do i have to use? sin, cos or tan? i don't understand when use what

[timber cargo]

tangent

The length that is opposite of angle R is 15

and the side length that is adjacent to angle R is RP

So, tangent of 57 degrees = 15/x

I'm sure you know how to solve for x from there

[ebon mantle]

yeah but is there any possible way to solve this by PR=15tg(33?
we did this in class and i didn't understand anything. why 15tg(33
why not 15*sin/cos(33

[timber cargo]

tan of 33 degrees is the same as: sin of 33 degrees/cos of 33 degrees

[ebon mantle]

nono i meant why not 15*sin(33 or 15cos(33

[timber cargo]

Sin is opposite/hypothenuse

Cosine is adjacent/hypothenuse

But, since you're only given the angle and the side length that is opposite to that angle, it's convenient to use tangent

Sine also works, but then you'll have to do the pythagorean theorem to find the RP

[ebon mantle]

now im starting to understand something. when it's convenient to use sine or cosine?

[timber cargo]

So, lets say you have a 30, 60, 90 triangle. And lets say the side length that is opposite to 30 degrees is, 10

In that case, you can use sine to find the hypothenuse of the triangle

Taking the tangent of 30 degrees will only get you to finding the other side length and not the hypnothenuse

So, you'll know when to use each trig function whenever you're asked to find a missing side length

[ebon mantle]

so there would be hyp=10*sin(30?

[smoky jetty]

<CPB = <PBC? I thought it would be <CPB = 60?

[smoky jetty]

Foor x, remember that the median (or 2x) is the average of both bases of the isosceles trapezoid ( (B_1 + B_2)/ 2), thus, 2x = (8x - 43 + 10)/ 2. For the value of y, since the mid segment EF divides the legs into two congruent parts, then 2y + 3 = 6y - 1. For the value of AE, you must have gotten the value of x already from doing what I first said in the text. For both ED and BF, they are congruent as the midsegment divides the leg into two equal parts (ED = BF)

A key to not only remember what's said on the theorem is to basically analyze the diagram thoroughly

[smoky jetty]

seems right

[smoky jetty]

did u mean 15cotangent33? or im pretty much dumbo rn

[ebon mantle]

15tangent, we hadn't started learning about cotangents yet

[smoky jetty]

oh wait nvm

im used to using whatever func is necessary as long as x of the ratio is the numerator. Quite surprised that those works despite being in the denominator (yeah shame on me, lmao)

[smoky jetty]

u can also use 15cot57 deg right? cot57deg = x/15

actually, is it ok if u can elaborate why u used 15tan57deg?

coz 15tan57deg & 15cot 57deg doesnt give the same valur

though, using 15tan33 deg (as the cofunction of of 15cot57deg) gives the same value as 15cot57deg. So, Idrk how 15tan57deg can be used as well..

[errant lintel]

cot is just the inverse of tan

[nocturne remnant]

the *reciprocal of tan. the inverse of tan stands for something totally different

[jovial hatch]

can someone pls help with this

[heavy shale]

the figure is showing that 6 m side is equal to x m side

so x = 6 ig

is there something wrong in question or i am doing something wrong 😕

?

[jovial hatch]

on the answer sheet it shows this

i dont get how tho

[heavy shale]

i think they are referring to the whole side

so 6 + addition part at bottom

[jovial hatch]

its split in like 2 triangles so 6 is only a little part , the side with x would be x-6 i think

[smoky jetty]

afk, it's all about the congruent sides

[nocturne remnant]

Here’s how I did it but probably not the intended solution as I had to solve a cubic

[smoky jetty]

I could only grasp till the mid part, lol

actually nvm i dont

[coarse lance]

I don't understand this

[potent kelp]

any help pls

[spare palm]

Lowks jus need the answers please, Idk how to solve for x

[timber cargo]

The derivative of inverse sign is that expression: 1/square root 1-x^2

[coarse lance]

how do you get the exact form?

[orchid oar]

Yo

I need help

Angle ABC is bisected by like segment AD. If angle ABC measures 96 degrees and one of the smaller angles created by line AD measures (7x+6), what is the value of x?

[drowsy walrus]

what have you gotten so far?

[orchid oar]

I’m confused, I’m unsure on how to work this lesson question out

If u can help fast

[drowsy walrus]

what does it mean when an angle is bisected?

[orchid oar]

Angle bisector

[drowsy walrus]

well yes but

[orchid oar]

Triangle is divided

[drowsy walrus]

what does an angle bisector do?

right

a bisector splits an angle or a segment into two congruent parts

[orchid oar]

Ok

So does that mean that 96 gets split?

And 7x+6 is what is equal to half of 96

?

[drowsy walrus]

wait

are you sure angle abc is 96 degrees

or is it angle bac

[orchid oar]

WAIT I THINK KNOW THE ANSWER

it’s 96 degrees

I think what I do is

96 divided by 2 is 48

7x6 is 42

+6 is 48

Or am I confusing myself

[drowsy walrus]

no

96 is a different angle

the 7x+6 is one of the bisected angles

[orchid oar]

Hmmm

So guide me on the steps

Yo solve

If it’s not a problem

Nvm I’ll just

It’s fine

Thanks for the help

[deft galleon]

if the 96 degree angle is bisected then each of the smaller angles is 48 degrees. so you know 48 = 7x+6 and you solve for x

[digital totem]

Where did I go wrong?

[silent plank]

you rounded too early/ liberally

@digital totem

[digital totem]

Ah ok

Thank you

[coarse lance]

so 3/sin(50) = b/sin(110) = c/sin(20) ?

,w 3/sin(50) = b/sin(110) = c/sin(20) solution

[somber coyote]

Results provided by WolframAlpha

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

Requested by Price#6057

[coarse lance]

yeah I got the same thing when I worked it out in my head

[gusty lake]

Question is “Given the approximate value of one of the trigonometric function, determine the value of the remaining two trigonometric functions to four decimal places”
I’m really stuck on this

[smoky jetty]

Try making a rough representation of the problem

[upper karma]

16

[smoky jetty]

oh nice, welp I had a different equation to find x

[upper karma]

identify the two angles created by line AD. Since the angle ABC measures 96 degrees, the two smaller angles measure 48 degrees each. hen use the expression given (7x+6) and set it equal to 48. Solving for x, we get x = 16.

iirc

[smoky jetty]

oh wow

im wrong then

[smoky jetty]

but if x=16, would the sum of 7x+6 + 48 be >96?

[upper karma]

no

[smoky jetty]

can u elaborate it?

[upper karma]

no

https://tenor.com/view/gigachad-chad-gif-20773266

[coarse lance]

not sure how to solve this

[smoky jetty]

what is to find?

[coarse lance]

the remaining sides and angles

[upper karma]

c ~15.3?

[coarse lance]

maybe, idk if I am supposed to use law of sines or cosines

[smoky jetty]

could be an ambiguous case? been a while since i've been there

[coarse lance]

oh wait

It asks to find if there is more than one solution

and solve for each one

[smoky jetty]

then it is

an ambiguous case

[coarse lance]

does that mean I use both law of sines and cosines

,w 7/sin(30) = 14/sin(B)

[deft galleon]

how tf you get x=16 from 48=7x+6

[upper karma]

weed

[deft galleon]

valid

[upper karma]

🤝

[smoky jetty]

not yet

sorry for the slow response, I had to get my trigo book for tht, lmao

[coarse lance]

lol, all good

[upper karma]

Hello can someone please help me with this trigonometry problem?

I tried this equation

but for some reason the function's x doesn't start with 0

[smoky jetty]

Basically, according to this book I have, there are 3 cases, where the first one states that there is no solution when a < b sin A (or the triangle's height). The second one states that there is only 1 solution if a = bsin A. And the third one states that there are 2 solutions if a > b sinA

for the first case, basically the opposite side to <A is too short to form a triangle, or to meet the other side

[coarse lance]

so this one I think a = b sin A

[smoky jetty]

yep!

u can find the value of b sin A with sinA = x/a

whats more interesting that I realized again, is basically you're finding for the opposite side, which is also the height of the triangle

[smoky jetty]

since the value of the opposite side to angle a is = to b sin A (also the triangle's height), what do you think would be the solution?

[coarse lance]

@remote plinth helped me solve it, this was the answer

I understand it now

[smoky jetty]

B = 90?

[static meadow]

can anyone help me solve this hard problem. Suppose point P, Q, R lies on a straight line and, there's a point O not on the line PR. and angle POR is 90 degrees . Draw a circle such that the point P,O,Q lies and another circle such that Q, O, R lies. if C is the center of the circle P, O, Q and D is the center of the circle Q, O, R. Prove that the measure of the angle C, O, D is 90 degrees

[coarse lance]

yeh

[static meadow]

or maybe some hint ☹️

[smoky jetty]

seems right

You may also refer to the Pythagorean theorem, as it's already enough to give u all the values (Case 1).

u can already find side c and angle C then

[upper karma]

my friend says x=15° but why? she couldn't explain why and it's very confusing

[smoky jetty]

yeah x is 15

[smoky jetty]

if u look at the diagram, line segment PQ is straight, and POR & ROQ are adjacent angles (next to) sharing a common segment (OR) where Point O is the midpoint of segment PQ. The common segment (or arm) OR divides the 180 degrees of POQ into two smaller angles, namely <POR = 2x, and <ROQ = 8x+30. From the diagram again, we can say that both angles are linear pairs of angles (Both of their sides form a straight line, and share a common line/segment/ray), thus being supplementary. Therefore, m<POR + m<ROQ = 180
2x + 8x+30 = 180 and so on...

[dry current]

Help

[grave herald]

Hey guys so i live in Denmark and im currently taking a course, on maths. On monday i have 24 questions i have to prepare for my end exam of the course, and i'm stuck with this one question, that makes absolutely no sense to me. The question is the following: Explain how x^2+4y^2-4b^2 makes an ellipse, and decide what the center is and define the ellipses semi-axes. I've never heard of b^2 in the actual equation. Maybe you guys can help me here

Any help would be a lot of help. My math class is in danish so i might be a little of with the right mathematical terms.

x^2 + 4 * y^2 - 4 * b^2 = 0

[soft flume]

Move 4b^2 to the other side then divide by it

And then look at what makes a equation an ellipse

[grave herald]

Can you further explain the details on the process from dividing 4b^2 to getting the ellipse. I see geogebra is giving me the right result, but i don't know how to quite explain it if im in a situation where i have to prove the equation

[soft flume]

Wdym by prove it's an ellipse?

You can see it's the equation of one assuming b is a constant

But actually proving that that equation is an ellipse is kinda hard

[leaden sparrow]

Hello guys !

Are there any tips for thought process while doing geometrical proofs

ok ?

i do that every question

[scarlet hatch]

label everything

helps me a ton

[leaden sparrow]

is there a way to be good at geometrical proofs ?

ok

How do i have good geometric intuition

Thank you !

[smoky jetty]

first, what do you about it?

specifically, the theorems regarding parallel lines cut by a transversal

[stuck sleet]

Are there still schools that offer courses about geometry (Just geometry, plane and solid)???

can you recommend some online courses about geometry....

thanks

[smoky jetty]

try khan academy, though it's not a school

[timber cargo]

The organic chemistry?

Maybe Professor Dave has videos on those subjects

[meager skiff]

No homework help but forgot how to fine lengths

[fallow tide]

Or if you want a useful answer, try to understand the intuition behind them so you’re comfortable with the concepts and not just the proofs

[ruby fulcrum]

can someone help me solve this awnser, help channels wont help

[deep vapor]

I’m really confused if this is either a parallel or neither

[ruby fulcrum]

its obviously not perpendicular and do both lines have the same slope?

[silent plank]

apply the slope formula and determine the slopes

is one way to determine if there's a special relation between those lines

[ruby fulcrum]

I tried

[deep vapor]

Ty it’s was neither

[smoky jetty]

yep

[midnight dock]

guys

i don’t understand how this works

[smoky jetty]

idrk anything, but, I think AB & CB are congruent?

[midnight dock]

i do not know

[timber cargo]

Since AB and CB are congruent, they are equal to eachother

So set them equal to eachother

15x-8=9x+10

Then solve for x and find the length of CB

[midnight dock]

i see

thank you @timber cargo

[midnight dock]

guys i don’t get this

also with this having some trouble

[smoky jetty]

they're congruent afaik

[pliant roost]

$\sin^2\theta+\cos^2\theta=1\$
If $-1\le x\le1$ and $\theta=\arccos x$, then:

$0\le\theta\le\pi,,\sin\theta\ge0$

Hence $\sin(\arccos x)=\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-x^2}$

[somber coyote]

DV Game

[limber summit]

guys how do you do circular inverse transformation(not the sin^-1 but geometric transform one that transforms lines to circles circles to lines) and how is it useful?

[pliant roost]

it's done using the definition of inversion and it's useful to map circles to lines and lines to circles

[zealous delta]

hello, sorry for the annotations, is there a way for me to find the distance between centre O and point C (aka the radius)? is there any formula i can apply

[meager frost]

Prove that sin(x+y)/ sin(x−y) = (tan x + tan y)/(tan x–tan y) I kinda need help with this

[boreal tiger]

help pls

[zealous delta]

nvm i found the solution lol

[eternal phoenix]

#❓how-to-get-help

[silver wolf]

jesus christ that looks like it was made by a parkinsons patient playing OSU!

[midnight dock]

hehehe

that is too crazy

does not even look like geommetry

guys does anyone know how to solve

[upper karma]

Can i use this place to get help with geometry questions?

[midnight dock]

i think we need to use the help channel

[upper karma]

I have

Nobody likes geometry enough to help me

[midnight dock]

same here

mabye i can help yo

[tame lintel]

Hey. Anyone able to help me on delta math geometry hw?

[timber cargo]

Can you expand the problem

I can't see the whole question

[upper karma]

Can you find an interior angle of a triangle using one exterior angle?

For ex

I'm finding y, so 180-110 = 70 right?

[midnight dock]

does anyone know how to do this

i have to find shades area

[zealous delta]

80/360 * pi*12^2

[zealous delta]

angle y and the bottom left angle add up to 110

if it's an isosceles triangle then it should be pretty easy
180-110 = 70
2*70+y=180

180-140 = y
40° = y

[midnight dock]

how did you figure out to do 80/360

[zealous delta]

to find a sector of the circle you use this formula

[midnight dock]

i got 90.47

but it isnt one of the answers

[zealous delta]

hmm

let me try to do it myself

[midnight dock]

alright i will try to solve it aswell

these are the correct answers

[zealous delta]

this is the right answer

[midnight dock]

oh i see

[zealous delta]

so it should be the dark blue one

the first

[midnight dock]

oh yes it was correct

[zealous delta]

great!

[midnight dock]

i will will need to learn thar theorem

thank you eda

[zealous delta]

hopefully its easy to remember

no worries

[midnight dock]

yes

thank you eda

[zealous delta]

sorry for double ping but yes, one example is the alternate segment theorem

if you come across this problem [not an alternate segment theorem example btw] if you do circle theorems you will see that the angles AOB and APB add up to 180 so you can take one of them away from 180 to find the other

the rule is that opposite angles in a cyclic quadrilateral always add up to 180°

not directly linked with triangles but i hope you get the idea

[upper karma]

No sorry! Thank you very much for your effort ☺️

[upper karma]

Interesting, I've never thought of that way

[zealous delta]

no worries

[rugged spear]

whats the correct answer

i think all of them are wrong

[zinc zealot]

how do i find BEC

[silver lotus]

It seems to be an isosceles triangle. So that must mean that angle EBC = angle CBE. Now having two angles of a triangle, you can just subtract them by 180° and you'll then get the values of angle BEC. Hope this answers your question

[zinc zealot]

that did help but now im stuck on BEA

[tawny vector]

EBC and ABE form a right angle

[silver lotus]

BD and CA are two perpendicular lines, which mean that BEA and BEC combined are 180°

Here's a good way to visualize it

[tawny vector]

O wait i cant read 🤣 sorry listen to bagel

[silver lotus]

113°+67°= 180°

[tawny vector]

Opposite angles

[silver lotus]

Thx for the assist either way 👍

[zinc zealot]

wait would bec be 50 degree

[tawny vector]

Yes

[zinc zealot]

ok so im finished with that problem

[tawny vector]

👍 nice

[zinc zealot]

i got a couple of these questions here done but im stuck on a few

wait a min

i think i see it

nope nvm

[tawny vector]

Wait

Idk if this is right

But wouldnt QRZ = 30?

[zinc zealot]

got me confused there

[tawny vector]

Because ZTS forms an alternate angle with ZRS

And then ZRS forms an alternate angle with QTZ

[zinc zealot]

QSR =60

[tawny vector]

Wait im stupid nvm

[zinc zealot]

nah you aint

just a mistake

[silver lotus]

So the fact that it's a rhombus means that all four of the triangles in the shape are identical

Which must mean our 30° also applies in these angles here

But that also means that these angles also equal each other

So what we really do here is solve for the rhombus's angles. We can start by adding angles QRZ and SRZ together to get one of the rhombus's angles. We know the values of ZTQ and STZ, so we can add those together too

Now for angles TQS and TSR, we know that they are equal to one another. In this case, we simply do some algebra to determine each value

Now we just split y into two, and we get our values for angles TQZ, RQZ, TSZ, and RSZ

Now finally, the middle angles can be found by subtracting the two known variables we previously found by 180°. And that's pretty much it

Hope it helped @zinc zealot

[zinc zealot]

it did thank you so much now i can get my grade up to 68

[silver lotus]

Of course my guy, just message is you need any more help

[upper karma]

Can I get some help with some geometry questions?

Number 7

Now I think it's B. that conjecture 2 is correct as is and

C maybe?

But since they all are equal

I think they would both be right tbh so A and B

R can equal T

Q can equal U

Really trying to get my last 6 questions done before school so i can study

[upper karma]

Damn can't even get my questions answered in the geometry section

https://tenor.com/view/cry-sad-gif-25488313

[spiral bough]

It’s B and C

[worn sun]

does somebody know how to find the orthocenter 😭

[sonic barn]

How would I prove the identity?

[smoky jetty]

use pythagorean theorem to find the value of the hypotenuse i.e c^2= a^2 + b^2. Then once you have the value, you can proceed to find the angle where sin theta = o/h (when o is the opposite side and h is the hypotenuse

[smoky jetty]

what do translation and reflection mean there?

[brittle mesa]

hey, i might be stupid, but i need a bit of help. i need to prove that BM=2 √3

[zealous delta]

where does the idea of sin, cos, tan come from? how do we know what sin/cos/tan(x) equals to

[timber cargo]

I believe it dates back to ancient babylonians

[desert delta]

Did i do this correctly

Did i find the right solution?

[polar sentinel]

how do i do this

[prime jasper]

Do you know how to find x

For questions 2

Think of a straight line

[polar sentinel]

yeah but i got a decimal

i’m kinda contemplating

[prime jasper]

What was your method

[polar sentinel]

57<4x+25<180

and then solve it like an algebra problem

[prime jasper]

Oh eh

[polar sentinel]

?

what

[prime jasper]

Now I'm confused lol

[polar sentinel]

what did u do

what answer did u get

[prime jasper]

I was thinking 180 - (4x+25)

Oh nvm the question is asking

The restrictions

[polar sentinel]

yeah no

[prime jasper]

For x

[polar sentinel]

yeah

it is

now do yk or not? @prime jasper

[prime jasper]

Idk lol

I was interested

[polar sentinel]

ph ok

[prime jasper]

What topic is this?

[polar sentinel]

geometry

[prime jasper]

Ye but like specifically

4x+25 should be less than 180

Therefore

x<38.75

[prime jasper]

I think I got it

[polar sentinel]

alr

ty

[prime jasper]

57 was there just to trick you I guess

[ruby fulcrum]

I don't get this explination, where did the x come from

[coarse lance]

tldr

[tame lintel]

Hey can anyone help me out with this.

[coarse lance]

I haven't done this stuff in a while, but wonder if this would work

[tame lintel]

use L2+L2=H2 to get your answer

[bleak blade]

pythagorean on coordinate plane

for all of them

[tame lintel]

5(5)+6(6)=H2/2=J

I meant =H

yes dont do divided by 2, my bad. Square root it

[calm sandal]

#help-0

#1021175428326633542

[pliant roost]

your question

you had it in x

[upper karma]

How would you describe the geometric place given by this set with f a lineal function from R to R2

[rapid tiger]

Can someone help me understand radians? I know that 1 degree is equal to pi/180, but when we’re talking about pi, is pi in terms of degrees(180) or in terms of its actual value(3.14159)?

[tame raft]

Pi is a rotational value

Given that pi is half a full rotation

So angle pi in radians converted to degrees would be 180°

But pi degrees would be some comically small radian value

[smoky jetty]

Moreover, a radian is approximately equal to 57.296°, thus equivalent to the arc formed by the rotation of the terminal side of the 2 rays that makes up the angle. pi being equivalent to 180deg is basically the simple way of doing pi * 57.2966°, try doing it on a calculator and you'll see. Pi's 3.14... value is its constant value (being the ratio of diameter/circumference)

afaik, the arc value of approx. 57.296° is = 1 radian, as it's the equal length to the circle's radius... correct me if this is all vague

[smoky jetty]

For Problem no.3, I'm not sure if this is the correct way, but the way I found x is by doing this: extending the lengths of the lines inside the parallelogram, thus creating 2 transversals that also form vertical angles & also two triangles. Then, find the supplementary pair of 130deg, giving u the value of one of the triangle's angles. Then, with the Alternate interior angles, the value of the second angle would be = to 30deg. And the last would be the angle to make the sum of all 3 angles = 180. With the last angle of the triangle, find x by subtracting it(the top angle) from 180, as it's supplementary with x (linear par of angles theorem).

[smoky jetty]

for quadrilateral ABCD, 3x is = 90 (as one of the properties of a rhombus is its diagonals are perpendicular), thus you can find x from there now. Then, you'd have the value for x to multiply to angle EDC (assuming the point of intersection is labelled point E). Then, find the rest of the angles of the triangles either with alternate interior angles or the angle missing for the triangle to have 180 degrees.

[kindred stream]

this is for geometry

hu is anyone available to help?

[still pelican]

yo bro

im strugglin over here

i literally cant get the last fucking side

excuse my french but like

this shit is pissin me off

what the fuck is the last thing to prove cus i literally cant put ad=db

[grave pond]

What are you trying to prove?

[still pelican]

acd=bcd

[grave pond]

In which sense of "="?

[still pelican]

congruent?

uhhh

[dense moon]

Struggling a bit here, I solved for x by setting both D and B equal to AC but I’m not sure how to find DBA, ADB, DAB, and CAD

[dense moon]

Is this uhh side angle side or sidex3

[still pelican]

yea

sss sas ass hl

stuff like that

[dense moon]

Yeah I’m learning that same stuff, let me see if I have last weeks notes

[smoky jetty]

looks like aaa to me, hmm

interesing...

[grave pond]

I mean, which kind of things do acb and bcd even denote in that equality? Areas of the triangles? Triangles which must be congruent? Angles which must be equal?

[still pelican]

wut

brain hurting

[dense moon]

Triangles/angles that must be proven to be congruent

[still pelican]

in a nutshell

im tryna put the last proof

which would be

[dense moon]

The ultimate goal is to prove how the sides are congruent

[still pelican]

ad=~db

right

but

[grave pond]

Either triangles or angles, then. You'll need to start by making up your mind which of them it is.

[still pelican]

i cant

its

triangles

[smoky jetty]

i think wat he meant is, what other properties of the diagram are u trying to prove? not sure

[still pelican]

the thing is

idk

because

[grave pond]

Looks like they're congruent by SAS.

[still pelican]

i cant put ad=~db as a proof

[still pelican]

yea i figured that

but i literally cant put that until i figure out the last proof

and ive literally tried everything

still no dice

[grave pond]

I, being very confused, think I will disengage here.

[dense moon]

Is there options given to you already ? Like multiple choice

[still pelican]

well

theres a dropdown

[dense moon]

Okay so

CAD = CBD

Since

You can’t do db

Or the other side of that

I think

Sas sss confuses me but I get the rough idea

[still pelican]

like

[dense moon]

😭

[still pelican]

i literally cannot answer

because im missing a proof

and idk what other proof there could possibly be

the only other one i could think of is ad db bisected

and that didnt even work

[dense moon]

Hm

Trying to think

Well fuck

I’m Brain dead

Finals have me burnt out already and they didn’t even start

[still pelican]

tbh

i dont think its us

[dense moon]

Fr

[naive kayak]

yoo

i need help withj my hw lmaoo

i dont wanna do it

can i send a pic, someone dxo it for me, i print it.

ill pay?

im too lazyu

whatever guess ill do it myself

[frail field]

This is my test

[naive kayak]

Oh yea that’s mad easy

I did that a while ago

Also

That’s algebra

Anyway what question

[lament flume]

Why did I get a different number?typing this into my calculator.

[naive kayak]

calculator is on dif mode prob

[lament flume]

My calculator is on degrees though.

That's the mode it told me to put it on ,on my self guided notes.

[naive kayak]

ok

[lament flume]

Does it look like I typed something in wrong ?

[dense moon]

Wht hw u got lmao

[naive kayak]

Geometry

Freshman

[dense moon]

Dm me

[still pelican]

almost an hour later

i finally did it

mind you

that was 1 of 5

[deft loom]

I need help with this

[silver wolf]

this question isnt phrased right it can be literally infinite

SPL would just approach 180 degrees

i assume |sl| is the length of sl

[deft loom]

Yes it is

[deft loom]

I thought so but someone found 12

[midnight dock]

help me my friend

friends

[upper karma]

most normal school word problem

[onyx cloud]

looks like they were using radians

[tender fog]

The logarithm of a number is the exponent to which another fixed number (the base) must be raised to produce that number. In the case of logarithm base 10, the base is 10 and the logarithm of a number is the exponent to which 10 must be raised to produce that number. So yes, in general, the logarithm of a number tells you how many times you can divide the number by the base before you get a result that is less than or equal to 1

I honestly have no knowledge on computer science logarithm as the only logarithm I know from Humss is Logarithmic Functions lol

[midnight belfry]

You're misunderstanding, it sounds like you're referencing the complexity theory in CS which is using the definition of logs but at base 2 specifically.

Which is fairly common.

[limpid onyx]

Can anyone help me with a clean solution?

I’ve made a system of 4 ugly equations but there must be a nicer way

[grave pond]

Both ways of viewing the logarithm are valid as long as you're only interested in the integer part of the result.

[midnight belfry]

@shadow flax I'm just saying that they aren't two different ways of thinking about what a logarithm is doing. It's dividing by the base. Now what those things are that it's dividing is subjective.

[grave pond]

I think you might be misunderstanding what R- was saying. He wasn't speaking about the difference between base 2 or base 10, but about whether one thinks about logarithm as "how many times can I divide by the base" or as "which power must I raise the base to".

[smoky vector]

Hi, these aren't my answers. i was wondering if someone could explain each step of this proof im confused about it. The second statement is DA ≅ DA, it kinda looks like an SA

[limpid onyx]

Is setting de=0 a valid way to solve it?

this isn’t geometry

then why is it here

Do you know what trig is?

[grand pagoda]

13?

[smoky jetty]

Imagine you are Rob and are using the can opener. If I'm not mistaken, to open a can with it, you will make the can opener travel till it completes the circumference of the can. So, again, imagine you'd only let it travel halfway (i.e half of the circumference). From here, I think you should use the formula to find the circumference of a circle (2 x pi x r), then divide the product by 2, as the problem states that the can opener only needs to travel half-way. Thus, the quotient will be the arc length

I think there are more ways to find it than that formula, but I hope it made sense

[wooden mango]

what is there to solve

isn't every problem like that a guessing game

[fallow grove]

tan x is sin x / cos x

right?

[coral tusk]

Yes

This isnt trig, the numbers arent talking about the angles

[brave moss]

Seems to me like an exam

It looks awfully like Moodle

[smoky jetty]

Is there a proper way to visualize pi other than the ratio of diameter over the circumference? I've often encountered problems in trigonometry which includes the symbol pi, but until now, even after rigorously searching online, I can't get the reason why it's used in many formulas, nor can I even visualize it. Any response will be appreciated!

[coral valley]

help

[kindred girder]

Hi I have a test tomorrow could someone help me understand this type of problem?

[worn sun]

Can somebody pls help asap

[stray wigeon]

#help-3

[valid crypt]

help. How to memorize thrms?

[opaque panther]

If you have time, try developing an understanding of what they mean

@valid crypt If you have a fundamental understanding of what it is, there isn't much to memorize aside from the name of the theorem I imagine.

[valid crypt]

Oh ok

[lament flume]

Oh you are right they were I got the right number when i changed function to radians idk why the teacher didn't explain that they used radian for that promblem and not degrees but now thinking about it it makes sense though because we were finding the outside of the triangle and not a internal angle

Thanks

Actually no that still doesn't make sense because another promblem that is the same type actually comes up right with degrees. Maybe the teacher just messed that promblem up for the second time

Yeah I think their was a error on the paper sense radians are only supposed to be used when measuring circular objects

Oh it's not the same type of promblem because one is the length of the right angle but the other is for the hypotenuse lenght

I wonder if that is it

[lament flume]

Yeah he's just wrong then

[bleak blade]

for what?

nvm

i see

$AB^2=10^2+6^2-9^2$

[somber coyote]

BuboBlakistoni

[bleak blade]

$AB^2=55$

[somber coyote]

BuboBlakistoni

[bleak blade]

this is done with pythagorean

noticing $BD^2+45=CD^2$

[somber coyote]

BuboBlakistoni

[bleak blade]

due to same height

[limpid onyx]

This isn’t right

Nvm I solved it

Sorry my lengths given were not right, CE and EB were switched in my drawing. 55 would be correct for the given drawing though.

[stuck hound]

hello can someone help me

[bleak blade]

With what?

[stuck hound]

with an easy ass question 😭😭😭 but im so tired i cant eben do it

[fierce shadow]

https://youtu.be/OXrYNPJQoTA

[bleak blade]

send it

[errant lake]

Hey can someone help me

[fluid oxide]

Hello? I need help with construction! Can someone help?

[bleak blade]

DA is parralel to EF so we can shift EF up

then solve equation

[errant lake]

Right thanks

32

[smoky jetty]

hope this one's correct, lemme try

[smoky jetty]

welp i could not. but, hey, can u share how you got the solution? Diving into that problem for 15 mins, yet I couldn't see how.

[steel breach]

how do i make at line segment for the radius equation that starts at (0,-4) and ends at (0,0) that is within y=0?

[silent plank]

the line connecting those points won't be a radius though

[steel breach]

ok can you just tell me how

[silent plank]

if you want a segment on the y-axis you can do something like

x= 0 {-4 <= y <=0}

make the necessary adjustments for all your equations to be consistent

[upper karma]

Hello does anyone know how I can calculate a-d in a faster way? Im pretty slow when it comes to those algebra ^^ thank you!

[neon trellis]

who decides the order for how the angle names are listed

DEA or ADE or EDA

does it matter

if it does, how do you come up with it

[wooden mango]

ADE and EDA are the same but DEA is different

the middle point in the name is the one where the angle is

or here’s a way to look at it

angle ABC is the angle where AB and BC meet

lmk if you want a picture 🙂

[neon trellis]

i see… that makes sense! then when i was trying to point at E being 90 degrees, what should i have done? i made it in the middle like i was supposed to

or is it correct

[wooden mango]

AED or DEA would work for that one

if you mean the one you marked 90 degrees

[neon trellis]

yes the like cube thing

[wooden mango]

cube thing?

[floral shale]

Cube

[grave pond]

Cube.

[still pendant]

hey guys, I'm having a problem with a homework question. Our teacher hasn't been here for 2 1/2 weeks now and she said a question similar to this will be on a test next week. She didn't give us an answer key for these packages though. Our supply has just been useless no offense to her just doesn't know any of the material. In particular the bearing and angle of depression... With the questions that have 1 or the other I understand but the one that has both in it is really confusing and the vids online don't have a question like this as a practice. Can someone help me with how this diagram is supposed to look? It'd be much appreciated

[dark sparrow]

cropped that for you

@tall ridge what stage are you on?

1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

i don't know, but you shouldn't give out answers like that.

[cunning yarrow]

Alright

[grave pond]

"Bearing" means "direction in a horizontal plane as measured by a magnetic compass, counted as degrees clockwise from due north".
"Angle of depression" means "direction in a vertical plane, counted as degrees below a horizontal line".
Does that help?

[grave pond]

You'll need several diagrams, one for computing the horizontal distance from D to each of N and M, and then one to solve the triangle between their ground positions.

[dark sparrow]

in principle you could make do with one 3D diagram but it may take some skill to draw it properly

as well as some good spatial imagination

[thick storm]

this may sound stupid but how do we know that the ratio of the sides of a right angle triangle are connected to the angles of said triangle? or basically why sin / cos / whatever actually works

ik its what defines sin and cos but its like how we dont have a function for the ratio of two sides in a regular triangle and we need to use the law of cosines or sine rule

[grave pond]

That's a better question that it looks like at first!

Basically the geometric definition of sine and cosine only works because we know that similar triangles are a thing, and two triangles with all angles equal have the same proportions between the lengths in them.
But this itself is a somewhat advanced fact about plane geometry; it requires the parallel postulate to prove. Euclid himself only proves it in book VI of the the Elements.

[thick storm]

ooh we basically use the fact that because similar triangles have the same ratio of sides and same angles then we can define a function which will give us the ratio thats pretty smart

[grave pond]

Exactly.

[dark sparrow]

BAD definitely isn't 42 degrees, also i am not the person who originally posted the question, also you should not give out answers especially wrong ones

[sly dawn]

I think CAD = ACD = 45 cuz AD and CD are equal and D = 90.

[lusty girder]

since triangle ADC has equal lengths on two sides, AD and DC, then the triangle itself is an isosceles triangle

[lusty girder]

this one's correct because an isosceles triangle has two same angles while one is different

[lusty girder]

this also applies to the other triangle in the question too, which is triangle ABC

[sly dawn]

its not mentioned that ABC is an isosceles triangle.

[lusty girder]

however they won't specify what the triangle is in these questions