geometry-and-trigonometry

[stone bramble]

can someone explain this

[lapis moon]

in your identity, k = 0,1,..., n - 1. that means we would choose the left endpoints of each subinterval x₀, x₁, ..., xₙ₋₁

relate the varaibles in your identity into this problem:
α ← a, β ← Δx, n ← n

Δx = (b-a)/n

Δx is the width of each subinterval

we would stick to the 2nd image

with left endpoints.

apply your identity to get

,, \sum_{k=0}^{n-1} \sin (a+kΔx) = \frac{\sin\left[a+(n-1)\frac{\Delta x}2\right] \sin\left[n\frac{\Delta x}2\right]}{\sin{\frac{\Delta x}2}}

[somber coyote]

vin100

\begin{gather*}
\sum_{k=0}^{n-1} \sin (a+kΔx) = \frac{\sin\left[a+(n-1)\frac{\Delta x}2\right] \sin\left[n\frac{\Delta x}2\right]}{\sin{\frac{\Delta x}2}}
\end{gather*}

[lapis moon]

oh no it should be the step preceding this in the derivation

,, \sum_{k=0}^{n-1} \sin (a+k\Delta x)=-\frac{\cos\left[a+\left(n-\frac12\right)\Delta x\right] - \cos\left[a-\left(\frac12\right)\Delta x\right]}{2\sin{\frac{\Delta x}2}}

[somber coyote]

vin100

[lapis moon]

,, \sum_{k=0}^{n-1} \sin (a+k\Delta x)=-\frac{\cos\left[a+\left(n-\frac12\right)\Delta x\right] - \cos\left[a-\left(\frac12\right)\Delta x\right]}{2\sin{\frac{\Delta x}2}}

[somber coyote]

vin100

[lapis moon]

multiply both side by $\Delta x$ as we're going to find the left-handed Riemann sum

[somber coyote]

vin100

[lapis moon]

,, \sum_{k=0}^{n-1} \sin (a+k\Delta x) \Delta x=\atop -\left{\cos\left(b-\frac{\Delta x}2\right) - \cos\left(a-\frac{\Delta x}2\right)\right}\frac{\frac{\Delta x}2}{\sin{\frac{\Delta x}2}}

[somber coyote]

vin100

[lapis moon]

explain what?

i changed $a+\left(n-\frac12\right)\Delta x$ to $b-\frac{\Delta x}2$ because in our construction, the strip width is the length of interval $b-a$ divided by the number of subintervals $n$, i.e. $\Delta x = (b-a)/n$.

[somber coyote]

vin100

[lapis moon]

as the number of subintervals $n$ goes up, the subinterval width $\Delta x$ goes down. it's intuitive to see that the shape of these strips resemble more the curve

[somber coyote]

vin100

[lapis moon]

Geogebra activity used: https://www.geogebra.org/m/RCVce5W4

so it's intuitive to expect that total area of these strips $$\sum_{k=0}^{n-1} \sin (a+k\Delta x) \Delta x$$ get close to a particular number

[somber coyote]

vin100

[lapis moon]

the RHS of the above equality
$$-\left{\cos\left(b-\frac{\Delta x}2\right) - \cos\left(a-\frac{\Delta x}2\right)\right}\frac{\frac{\Delta x}2}{\sin{\frac{\Delta x}2}}$$
suggests the approximation $-(\cos(b) - \cos(a))$.

[somber coyote]

vin100

[fierce glacier]

Can anyone help

[lapis moon]

sorry $-(\cos(b) - \cos(a))$ isn't an approximation. it shld be the left-handed Riemann instead. how close are these two numbers? to simplify writing, let $h = \frac{\Delta x}{2}$. as $h$ is getting smaller and smaller, it's easy to see that $\sin h$ and $\cos h$ are getting closer and closer to 0 and 1 respectively, but how close?

this Riemann sum approximation has two factors. we'd express the approximation error for each factor in terms of $h$, before we consider the error when these two factors multiplied.

start with a simple observation

[somber coyote]

vin100

[lapis moon]

we see that $0 \le \sin h \le h$ when $h \ge 0$ is small.

[somber coyote]

vin100

[lapis moon]

from the Pythagorean identity, we expect $\cos h$ to be close to 1.

[somber coyote]

vin100

[fierce glacier]

Is this for me ?

[lapis moon]

i'm explaining the real meaning of the identity that DV Game posted: #geometry-and-trigonometry message

[fierce glacier]

Oh ok

[lapis moon]

that is, the left Riemann sum of the sine function, from which we define its area under the graph.

,, 1 - \cos h = 1 - \sqrt{1 - \sin^2 h} = \frac{\sin^2 h}{1 + \sqrt{1 - \sin^2 h}}

[somber coyote]

vin100

[lapis moon]

,, \frac{\sin^2 h}{\underbrace{1 + \sqrt{1 - \sin^2 h}}_{\ge 1}} \le h^2

[somber coyote]

vin100

[lapis moon]

,,\therefore 1 - h^2 \le \cos h \le 1

[somber coyote]

vin100

[lapis moon]

in the above screenshot,

,, \sin h \le h \le \tan h = \frac{\sin h}{\cos h}

[somber coyote]

vin100

[lapis moon]

,, 1 \le \frac{h}{\sin h} \le \frac{1}{\cos h}

[somber coyote]

vin100

[lapis moon]

,, \le \frac{1}{1-h^2} \le 1 + 2h^2

[somber coyote]

vin100

[lapis moon]

when $h^2 \le \frac12$. in fact, $h$ would be much small than that.

[neat moth]

I saw a mathematically provable claim in my field of study (not math.)
I translated the claim into the following mathematical model:
Let there be five three dimensional unit vectors a, b, c, d, and e
These vectors have two properties. One property is that the total sum of the five vectors is zero. And the other property is that adjacent pairs of vectors (a and b, b and c, … e and a) all have the same angle between them.
The claim is that given the above two statements, all five vectors must be coplanar.

I tried to go about proving or disproving this but I didn’t get very far

[somber coyote]

vin100

[lapis moon]

,,1 \le (1+2x)(1-x) = 1 + (1-2x)x \iff 0 \le 1 - 2x \iff x \le \frac12

[somber coyote]

vin100

[lapis moon]

the approximation error of the second vector is done

,, \frac{h}{\sin h} - 1 \le 2h^2

[somber coyote]

vin100

[lapis moon]

when $h^2 \le \frac12$

[somber coyote]

vin100

[lapis moon]

for the first factor, we'll first compare $\cos(b+h)$ with $\cos b$.

[somber coyote]

vin100

[lapis moon]

,, |\cos(b+h) - \cos(b)| \le |\cos b (\cos h - 1)| + |\sin h| \le h^2 + h

[somber coyote]

vin100

[upper karma]

is this correct? im not really sure

[lapis moon]

in each figure, you are asked to find two trigo quantities. there're two figures, so there should be four figures. you've only done half of the work

the above inequality is due to compound angle formula, $|cos x|, |sin x| \le 1$ and $|a+b| \le |a|+|b|$

[somber coyote]

vin100

[lapis moon]

so we expect that the first factor to have an approximation error of $2h + 2h^2$

[somber coyote]

vin100

[lapis moon]

$h^2$ would be much smaller when compared to $h$. say $h = 0.01$, then $h^2 = 0.0001$.

[somber coyote]

vin100

[lapis moon]

now proceed to the approximation error of the product of these two factors

[upper karma]

Hmm, I'm sure my teacher didn't teach us that yet because we're still on introduction to trigonometry.

[lapis moon]

instead of directly writing them it would be much more meaningful to use some simpler symbols to illustrate a much more general observation

[lapis moon]

wdym? you're asked to find $\tan J$ and $\tan K$ in each figure. there're two figures, so you have to find four numbers in total. in your screenshot, there're only two found.

[somber coyote]

vin100

[lapis moon]

,, \begin{aligned}
& |fg - L_1 L_2|\
&= | {\color{blue}{f(g - L_2)}} + {\color{red}{L_2(f - L_1)}}| \
&\le |f| |g-L_2| + |L_2| |f - L_1|
\end{aligned}

[upper karma]

for question 1
tan j = 24/32
tan k = 32/24

for question 2
tan j = 8/15
tan k = 15/8

is this correct now?

[somber coyote]

vin100

[lapis moon]

so when $f$ and $g$ are close to $L_1$ and $L_2$ respectively, their product $fg$ is close to $L_1 L_2$.

[somber coyote]

vin100

[lapis moon]

\begin{alignat*}{2}
f&=-[\cos(b-h)-\cos(a-h)] &\qquad g&=\frac{h}{\sin h} \
L_1&=-(\cos b - \cos a) & L_2&=1
\end{alignat*}

[somber coyote]

vin100

[lapis moon]

It's easy to see that $|f| \le 2$, so

[somber coyote]

vin100

[lapis moon]

the desired error is

,, \le (2)(2h^2) + (1)(2h+2h^2) = 2h + 6h^2,

[somber coyote]

vin100

[lapis moon]

the coefficients '2' and '6' are never an accurate estimate of the actual approximation error. they just give a simple range of it, much larger than the actual range, because I've given simplified arguments at several places, say $1+\sqrt{1-\sin^2 h} \ge 1$ while estimating $\cos h$.

[somber coyote]

vin100

[lapis moon]

Hence we see that

,, \left|\sum_{k=0}^{n-1} \sin (a+k\Delta x) \Delta x - [-(\cos(b) - \cos(a))]\right| \le \Delta x + \frac32 (\Delta x)^2.

[somber coyote]

vin100

[lapis moon]

|left Riemann sum - theoretical value| <= Δx + negligible terms

as Δx gets smaller and smaller, the left Riemann sum would get nearer to the theoretical value

a more rigorous way to say this is that for all 0 < Δx < 1 (any reasonable small number, not so important), we have the above inequality

by changing $\sum$ to $\int$, and $\Delta x$ to $\dd{x}$, we denote

[somber coyote]

vin100

[patent valley]

I thought I was going to look at trigonometry

[lapis moon]

,,\int_a^b \sin x ,\dd{x} = -(\cos(b) - \cos(a))

[somber coyote]

vin100

[patent valley]

wow

this stuff really goes right over my head

[lapis moon]

The story starts from a trigo identity previous posted about the sum of sines: #geometry-and-trigonometry message

[pliant roost]

nice

I will work both the sum and the errors by hand to get a better understanding

I already knew about Riemann sums

[patent valley]

what

and this is trigonometry?

[pliant roost]

I asked about an interesting trig problem

[lapis moon]

actually we've just proved the definite integral of sine without using differentiation

[pliant roost]

And it turned out to be that cool

[pliant roost]

Haha! Yes!

[lapis moon]

in fact, integration was historically a subject separate from differentiation

[pliant roost]

And then you can get the indefinite integral

[patent valley]

Man I just realize how little I really know math

that's crazy

[pliant roost]

well me neither

[pliant roost]

well there can be some reasons

just because of the integral and derivative meanings are area and speed respectively, even though they're inverse of each other

[patent valley]

See you know what these terms mean

I don't even know where to begin

God I have a lot to learn

[pliant roost]

me too...

I don't understand why are positive-definite matrices are cool for example

[lapis moon]

the most tricky part is to give bounds to trigo fct using polynomials

#geometry-and-trigonometry message

[pliant roost]

fct is factorization right?

[pliant roost]

what is alpha there

[lapis moon]

sorry for my laziness too tired after 9 hours of typing

function

[pliant roost]

oh

[lapis moon]

it's the digit '2' 🙂

[pliant roost]

😂 thanks

I genuinely tried to find some angle

[pliant roost]

Actually the first triangle is a right triangle with the right angle being at the bottom

basically that's geometry rather than algebra

it's important to be able to mix alg and geom when doing trig

[lapis moon]

sorry i've forgotten the details of linear algebra. i've just wiki a bit. one of the equiv cond. of P(S)D mat. is sym/herm + +ve eig val

sym/herm mat allows two nice prop: 1. diag 2. ortho

diagonalizability allows simple decomposition

[pliant roost]

yeah I remember, it is symmetric and it is doing some z*Mz > 0 for any vector (with corresponding size of course) where * is the complex conjugate of a vector

[lapis moon]

orthongonal column vector basis allows simple orthogonal projection

[pliant roost]

well symmetric is cool because A=A^T

and orthogonal is A^T = A^-1

[pliant roost]

yeah, in 2d it can be seen very nicely

ok I read wiki

positive eigenval is good

also there is decomposition A = B*B

Cholesky decomposition iirc

[lapis moon]

sorry for delayed response i'm very excited about explaining deep stuff. observe that $P$ lies on the unit circle, so $P=(\cos \theta, \sin \theta)$ for some $0 \le \theta \le 2\pi$.
(a) use the Pythagorean Identity to find $k$
(b) use symmetry of the unit circle
(c) that should be clear when you've the diagram in (b)

[somber coyote]

vin100

[pliant roost]

anyways thank you, vin100, very much for explaining everything, didn't think I'd get that!

now I just gotta go

I will observe the stuff you found a bit later

[lapis moon]

u r welcome have a gd day

[upper karma]

@lapis moon can i ask?

is this correct?

[summer nebula]

@upper karma i see that you wrote sin instead of cosine for the second part

be careful not to mix up the two

you're on the right track

[upper karma]

oh tnx for telling me

but its correct right?

[summer nebula]

the idea is correct, just a minor error

[upper karma]

alright tnx

[summer nebula]

[redacted]

[redacted]

OH WAIT

i see what you did there

nvm

you're correct

im just going too fast

[upper karma]

idk i just followed the formula of my teacher

[summer nebula]

yes its correct

im just use to doing it super fast

just fix the sine cosine part

you'll be fine

[upper karma]

alright

oh wait my answer in cosine is wrong

i used sine in calcu lol

[lapis moon]

sorry i felt asleep after typing math for hours

nice try

I would rather write

\begin{align*}\sin 32^\circ &= \frac{x}{18} \ x &= 18\sin 32^\circ \ &= 9.5 :\text{ (cor. to 1 d.p.)} \ \cos 32^\circ &= \frac{y}{18} \ y &= 18\cos 32^\circ \ &= 15.3 :\text{ (cor. to 1 d.p.)} \end{align*}

[somber coyote]

vin100

[lapis moon]

p.s. when i was in secondary school, i would even skip the two lines containing fractions to save time

[upper karma]

yes i forgot to change sine to cosine

[upper karma]

The acute angle between the terminal arm of an angle in standard position and the x-axis when the terminal lies in quadrant 2,3, or 4 is called the related acute angle

Guys is this true?

[green elm]

• The point P (-15, -8) is on the terminal side of the angle ϴ in normal position. Determine the
  value of trigonometric ratios for ϴ

[narrow wyvern]

Yo I’ve been having trouble with these, I’m not that smart when it comes to math so I need help lol

Even with trying to look up the answers I could literally only get one right if that explains how not good at math I am

[atomic agate]

Bruh moment

[runic beacon]

thats a lot of different types of questions

you should familiarize yourself with all area formulas

[upper karma]

anyone

mind helping me?

Write the equation in standard form for the ellipse with center at the origin, vertex (0,11), and co-vertex (
–
8,0).

[magic lichen]

guys in trignometry can you get expressions like $sin^3 (theta)$

[somber coyote]

Alpha | Sync

[nocturne remnant]

Yeah, what’s interesting about it?

It just means the cube of sin(theta)

[dark sparrow]

sure you can

[astral summit]

The coordinates of a point lies on the x-axis and at a distance of 9 units from the y-axis is -

[upper karma]

Why is T1 not equal to 5000cos(30)?

[sudden anchor]

cause it's 105 degrees not 90

[craggy maple]

can anyone guide me in this?

or give a hint?

[pliant roost]

What do you need to do?

Show that M and N are midpoints?

[sudden anchor]

what do you need to do

[craggy maple]

Show that BMN is equilateral

[craggy maple]

Show that BMN is equilateral

[wanton edge]

What does that mean?

D, E are on the same side of line AB

[sudden anchor]

It means that that like they are not on the opposite side, like if you connect them it won't intersect AB

I created a digram

but not sure how to prove it

[wanton edge]

Thanks for showing the diagram. Now I can understand the question

[sudden anchor]

Just the point G is the point N

[craggy maple]

Yeah

kinda confusing because

too much information available 💀

[craggy maple]

hey is DAB congruent to MBN?

[sudden anchor]

No I dchecked it

saidly no

[wanton edge]

Hey so I just proved that ABE is congruent to DBC

[craggy maple]

hey yeah wait

imma take the problem from there

thank you

[sudden anchor]

yess

[wanton edge]

You're welcome but we're not done yet

[sudden anchor]

they are congurent I checked

[craggy maple]

too many triangles get confusing sometimes 💀

[sudden anchor]

Nice observation

[craggy maple]

yes i know

i wanna work on the further stuff on my own

[wanton edge]

Want a clue?

I already solve it but I don't want to spoil the answer yet

[sudden anchor]

can you dm me

[wanton edge]

Who are you talking about?

[sudden anchor]

to you I want to see a soulution

[pliant roost]

Wait, shouldn't it be?

looks like it is

because I solved it that way 😅

[sudden anchor]

well looks like it but it isn't cause you can also choose point C to be wherever you like on the line AB

[pliant roost]

wait

let me try to prove it

[sudden anchor]

ok but you are proving something which isn't true lol

[pliant roost]

wait I'll either spot a mistake or prove it somehow

though no

they are just similar

aaand I needed similarity

[sudden anchor]

yess

so if you can prove that they're simliar you're done

[pliant roost]

then I'll just solve

[sudden anchor]

good

good idea

[wanton edge]

You only have to find 3 pairs of congruent triangles and then doing some stuff with length and angles and you're done

And that's what I did

[pliant roost]

I mean I got MB = MN and angle NBC = angle MBE, and I need to prove that angle EBN = angle DBM and I am done

[pliant roost]

bruh I forgot about another pair (ΔAMB and ΔDBN) and the angles ABM and DBN share the same angle, and the rest is 60 degrees

I am actually stupid

yeah, it is congruent to ΔDBN

[wanton edge]

AME is a straight line

[pliant roost]

nvm AMB

I just got confused by E on your graph right below

[wanton edge]

No problem

[upper karma]

do you guy know about Euclid's line and Euclid's circle?

[sudden anchor]

What do you mean

can you show it

[unreal relic]

why does cos 0°=1

[runic beacon]

look at the unit circle

[unreal relic]

unit circle???

oh i didnt get to that yet

[sudden anchor]

well you can also say cos 0 is 1 cause cos = base/hypotenuse and when cos 0 then base and hypotenuse are the same like they're the same line so it's 1

you can read this https://socratic.org/questions/why-is-cos-0-1

[unreal relic]

Thx

[lapis moon]

ratios of sides of a right-angle triangle can define sin, cos & tan for acute and right angles (with the exception of tan 90°, of course). using the unit circle can define these trig funct of any angle.

🔗: https://www.spektrum.de/lexika/images/mathematik/Einheitskreis-I1.jpg

[upper karma]

Have been struggling with this question lately, and would really appreciate it if someone could help confirm my answers or help find a better solution. What also confuses me a lot is the last statement From lighthouse B, lighthouse A has a bearing of 45°

[hushed bridge]

fax this made trig so much easier

[grave pond]

The 45° information means that lighthouse A is directly northeast of lighthouse B.

So the angle you've marked 45° on your sketch is wrong.

The direction from B to the boat is 237°-180­° = 57°, so angle B in the triangle is 57°-45° = 12°.

(Which makes the triangle pretty obtuse, so it looks like for question (b) one needs to assume that the shore is a straight line extending northeast past lighthouse A).

[upper karma]

Wait wouldn't also change the bearing placement for lighthouse A as well tho?

[grave pond]

What does "bearing placement" mean there?

[upper karma]

I mean, like the terminal angle

For light house A

[grave pond]

Yah the angle at A will just be 180° minus the agles at B and the boat.

[upper karma]

I trying to draw this out

[grave pond]

The relevant background fact here is that "bearing" always mean a direction measured relative to due north_.

[upper karma]

I'm still trying to understand and visualize your solution. But may I ask why did you subtract 57° by 45° to find angle B? Also my concern from before was that when you said that 45° meant that lighthouse A would be NE of lighthouse B, wouldn't that change the 285° bearing of lighthouse A from the boat?

[grave pond]

How would that change the sailor's direct measurement that the problem explicitly gives us?

The lighthouse is where it is. The sailor saw it in the direction he saw it, and wrote down what that direction was.

But may I ask why did you subtract 57° by 45° to find angle B?
The boat sees lighthouse B i direction 237. Someone standing at that lighthouse can look back along that line and see the boat in direction 57. Since the lighthouse sees the boat in direction 57 and the other lighthouse in direction 45, the angle of the triangle at the corner is the difference between those two bearings.

[upper karma]

I understand the part where lighthouse B sees the boat in direction of 57°, however when you say the lighthouse B sees lighthouse A in direction of 45° is what confuses me, because where would that be? Is it angle A? Or B?

[grave pond]

45 is the angle between due north and the direction the lighthouse keeper at B sees the light from A in.

It is not one of the angles in the boat-A-B triangle.

[upper karma]

Wait wait

[grave pond]

Argh, I swapped the letters A and B in the drawing.

Corrected

[upper karma]

I was getting close to what you are showing me as well

[grave pond]

Right.

[upper karma]

From I how I see this now is that

Lighthouse B was had an alternative angle, and to get angle 12°, I had to subtract 45° (caused by the NE shift of Lighthouse A) from the direction of Lighthouse B to the Boat which as 57°

The angle of the boat was difference of 285° and 237° which as 48°

And now to find angle A, I have to subtract 180° by 48° by 12° = 120° is that correct @grave pond?
I know my diagram of angle A looks acute, but I understand the correct value

[grave pond]

Yes.

[upper karma]

Okay then, I think I have the peices now to find out the distances of the lighthouses for the boat. Using sine law.

Thank you so much for the needed support, @grave pond, I really appreciate you.

[grave pond]

yw

[thin sable]

anyone plz tell this ques

[grave pond]

That sounds bizarre. First is looks like ABC is an arbitrary triangle, but then immediately it redefines C to mean the midpoint of AB, and the rest of the exercise looks like it makes sense with this new definition.

It would make so much more sense if we cross out the words "of ΔABC".

[lapis moon]

draw some auxiliary lines passing through A and B perpendicular to CZ, then observe some similar right triangles.

[magic lichen]

guys wuts cos 45

[nocturne remnant]

Google it

[magic lichen]

guys

how do u draw tan on a unit circle

[lilac whale]

sin/cos

[lilac whale]

$\frac{\sqrt{2}}{2}$

[somber coyote]

D00M_Re1ated

[magic lichen]

how do u draw it on a unit circle

[lilac whale]

oh

im actually not sure

[magic lichen]

i didnt want the formula

[lilac whale]

do you have to do that?

[magic lichen]

no i just wanted to know

[lilac whale]

im not really sure if you can

[magic lichen]

the proof of how u draw a tan on unit circle

u can draw it but

like

how does it work

so basically u draw a tangent on point P where point P is where ur terminating ray touches the circumference

[lilac whale]

oh

[magic lichen]

and then

distance of point P till where tangent touches x axis is tan theta

[lilac whale]

huh

[magic lichen]

distance of point P till where tangent touches y axis is cotan theta

[lilac whale]

interesting

[magic lichen]

but i wanted to know proof of that

arya i need help

xd

[lilac whale]

i do not know where the proof comes from

I think you can google it but im not sure

[magic lichen]

like

._.

[nocturne remnant]

You mean this?

[magic lichen]

YE YE

[lilac whale]

oh

[magic lichen]

how do u proove that that is tangent

[nocturne remnant]

Proof: similar triangles
sin x / cos x = tan x / 1

[magic lichen]

ok i know that

how does it proof that line PF is tangent theta

[nocturne remnant]

OPE and OFP are similar

[lilac whale]

you can try an example with a specific point

like use a certain angle on the unit circle

and try it yourself

[magic lichen]

they are congruent

[nocturne remnant]

???

[magic lichen]

yes...

[nocturne remnant]

No???

Are we looking at the same pair of triangles…

[magic lichen]

sin theta in ope is

[lilac whale]

there is only one triangle

[nocturne remnant]

Also the angle is not given to be 45 degrees so you would have no idea

[magic lichen]

angles gonna be same dud

cuz tangent of a circle is perpendicular to point P

right?

[nocturne remnant]

Ye

Perpendicular to OP*

[magic lichen]

so if angle ope is x then angle fpe = 90-x

then ur other angle in both the triangles will be the same

and u have 90 degree angles so they are congruent

[nocturne remnant]

No bruh

[magic lichen]

then what

bro one side is equal in both triangles and all angles are equal

[nocturne remnant]

I already said it

[magic lichen]

they congruent

[nocturne remnant]

No obviously one triangle is much bigger duh???

[magic lichen]

how

[nocturne remnant]

Can you read carefully

[magic lichen]

wait lemme think

yeah ok so how do u use similar triangles to prove that the length PF = sine/cosine

[urban whale]

YO

so

tell me could i calculate the sleeve length of this using trig ?

[magic lichen]

not a trianlge

[magic lichen]

o ok

[magic lichen]

so u want to know hypotenuse

[urban whale]

yes

[magic lichen]

units on the x axid

axis

u forgot

how much is 1 unit

[urban whale]

coulnt you use tan to find it

5 like the y

[magic lichen]

no u cant use tan

[urban whale]

hmm

i d k how

[magic lichen]

its pythogoreas theorum not trig

its

[urban whale]

ah

[magic lichen]

$\sqrt15^2+15.5^2$

no wait nonononono

[somber coyote]

Alpha | Sync

[magic lichen]

wth

so basically

im lazy

[urban whale]

all good

[magic lichen]

u root the whole thing

not only 15^2

[dark sparrow]

{}

you need some of these

$\sqrt{15^2 + 15.5^2}$

[somber coyote]

Ann

[urban whale]

thanks

[nocturne remnant]

FP/OP=PE/OE

[magic lichen]

oh so sin/ cos = slope = tan

[magic lichen]

isnt it
$OF/OP = EF/OP =EO/PF$

how do u do this

[somber coyote]

Alpha | Sync

[nocturne remnant]

OF is irrelevant

[magic lichen]

ok

isnt it the hypotenuse of the bigger triangle

[nocturne remnant]

Yes but we’re not looking at hyp

[magic lichen]

then we are looking at sin and cos

right?

[nocturne remnant]

I already gave you what sides you should look at

[magic lichen]

but what u are giving is

fp/op

=

pe/oe

oh ur looking at the slope of both triangles?

[nocturne remnant]

I guess you could say that

[magic lichen]

then how does tangent relate to sin/cos

[nocturne remnant]

I used op as the shorter side of the big triangle, not the hyp of the smaller one

[magic lichen]

tan*

ye ye i got it

[nocturne remnant]

Just substitute the known values into this

[magic lichen]

op is 1

cuz its unit circle

[nocturne remnant]

Yep

[magic lichen]

the u get

fp=pe/oe

OH

FP IS TAN

[nocturne remnant]

Yea

[magic lichen]

pe/oe is sin/cos

TY

[magic lichen]

same thing for finding cotan right?

[nocturne remnant]

I think it’s similar

[steel ibex]

What is k in this identities?

[dark sparrow]

for any integer k.

says it right there

[dark sparrow]

you should not ping people out of the blue like that 🤨

[runic beacon]

try relating b to other angles then

why not?

see the

point at which angle b forms for example?

can you see it now?

theres a lot of ways (you can also consider that fact that line 3 is parallel to line 2)

corresponding angles will be equal

well

theres a lot of ways

but if you consider the fact the sum of angles on a straight line is 180 degrees

and look carefully at the figure

unknown angle has a relation with c

isnt it?

no?

[runic beacon]

??

you know what that small symbol means in c and the unknown angles?

Yes

so?

what equation do you get?

x?

or c?

yes

so, solve for c

Ok

can you find the other angles now?

give it a try

what do you get?

yes

yes

which one?

yes

so the triangle with b as its exterior angle is isosceles

yes

they are related?

also you have the third angle as 2c

vertically opposite

wdym by thats?

yes

you can

you can use anything

that works

did you get it?

[runic beacon]

No Problem

have a nice day

[terse breach]

how do I find the values of the angles x y z using trigonometry?

[serene dove]

du suger balle @terse breach

https://tenor.com/view/asian-baby-finger-cute-gif-13450306

[neon crystal]

KÄFT BOEL

du kan inte ens 1+1

[terse breach]

solved it

[trim breach]

What is your question? All of question 7? @vestal siren

[vestal siren]

yeah

[trim breach]

Starting with 7a, you are asking what percentage [of all students] surf. So you will need the grand total first.

Are you able to find the grand total?

[vestal siren]

for what

[trim breach]

The contingency table.

The one in question 5.

[vestal siren]

yea i have

166

[trim breach]

Yes. So the percentage of students that surf should be all students that answered “Yes” to surfing, divided by the grand total, multiplied by 100.

[vestal siren]

so 100/77?

[trim breach]

Not quite.

[vestal siren]

is it flipped

[trim breach]

What is the marginal total of surfers?

How many people answered “Yes” to surfing?

[vestal siren]

77

[trim breach]

That is for skateboarders, not surfers.

[vestal siren]

where do i find surfers what

the other one is pets

wait

32??

[trim breach]

That is for surfers that also skateboard.

You want to look at the marginal total, meaning everyone that surfs and skateboards and everyone that surfs but does not skateboard.

You already wrote it down in the margin.

[vestal siren]

i’m confused now

[trim breach]

If I tell you that the number of surfers is 97, do you know how I got it?

[vestal siren]

oh yeah okay 97

do i divide by 100

[trim breach]

You divide by the grand total first, which you said was 166 earlier.

Then, multiply by 100.

[vestal siren]

ok

next do i do 69 divide 166

[trim breach]

For 7b, no. That would be the proportion of students that don’t surf. You want the ones that don’t skateboard.

The 166 part is correct though.

Just need a different numerator.

[vestal siren]

so 24

[trim breach]

More than just those 24 answered no to skateboarding.

You want the entire “no” column for skateboarding.

[vestal siren]

45 or add more

[trim breach]

Underneath the “Skateboarding” label, there is “Yes” and “No”. Anyone in the “No” column cannot skateboard.

[vestal siren]

89

[trim breach]

Yes!

So divide that by 166, then multiply by 100 for the answer.

[vestal siren]

okay i got it

[pseudo crystal]

How do you find the third point in a triangle if you know the coordinates of the first and second point?

A(0, 1, 0) and B( 0, -1, 0)

[static jolt]

i think need more info

[pseudo crystal]

Well I guess you could assume that the triangle is equilateral

[lofty tangle]

what 0.5S^2sin(θ mean

like what is it used for

the * wont show up

[grave pond]

You'll probably need to show more context if you want an answer that's not just "half of the square of S times the sine of theta".

[lofty tangle]

i was asking a sub for help on a problem in geometry and he handed me a sticky note for finding the area of an equilateral triangle

[grave pond]

Hmm, it is indeed the area of an isosceles triangle if S is the length of the two equal legs and theta is the angle between them.

Apparently you're supposed to substitute theta=60° yourself?

[lofty tangle]

yeah

but does it still find the area of an equilateral triangle if you substitute s for side length and theta for 60?

[grave pond]

Yes, an equilateral triangle is a special case of an isosceles one.

You can view it as an instance of the general area of a triangle: area = ½ab·sin(C).

[lofty tangle]

alr

ty

[dark sparrow]

check whether the inequality -π/2 ≤ -π/3 ≤ π/2 is true

[steel ibex]

Is there any alternative methods?

[dark sparrow]

what alternative do you need to the definition of an interval?

i mean for this one in particular if you're so averse to what i just said you could instead verify that |-π/3| ≤ π/2 or something...

honestly. why do you need alternative methods for verifying whether a point lies in an interval

[dark sparrow]

the study of the relationship between lengths and angles, to put it broadly.

often in relation to triangles specifically, whence comes the name.

[royal portal]

Hello i got a geometry problem and id really need an idea to solve it out

AC distributes the trapeze in 2 similar trianglea

Triangles

Some ideas?

<@&286206848099549185>

[dark sparrow]

@royal portal do you still need help with this?

[royal portal]

@dark sparrow yes please

[dark sparrow]

i would begin by noting that angles BCA and CAD are equal

this should help pin down what pairs of sides correspond to each other

[unreal relic]

is it said that the thing is a uhhh

BC || AD

@royal portal

[royal portal]

Nvm

Solved it already

Thx anyways

[upper karma]

I need pink area

[upper karma]

a Circumference of two circles have a difference of 30cm.Find how much their radius differ(of the 2 circles)

I need all area

[strange umbra]

try using pick's theorem

||A = 8 + 4/2 - 1 = 9||

[tawdry bane]

can anyone send me some good practice problems

[lapis moon]

in geometry? i recommend the 400 propositions in chapter 6 of 📖 Machine Proofs in Geometry downloadable at http://www.mmrc.iss.ac.cn/~xgao/publ.html

[upper karma]

ABCD Square

[lapis moon]

this is another textbook problem that can be solved using the area method

https://media.discordapp.net/attachments/433980473367330826/972657786418782288/unknown.png?width=1180&height=669

source: Machine Proofs in Geometry

for secondary school students, just treat the areas as positive.

together with Zhang's Elimination Point method (消點法)

https://images-ext-1.discordapp.net/external/RwrVGCmlA2TPh4IZ6x_BAgxFS-od-QGVW7094I7gXYU/%3Fwidth%3D696%26height%3D670/https/media.discordapp.net/attachments/486841085210132490/968248935682551858/unknown.png

source: see bottom line of the above screenshot

free points: A,B
constructed points:

1. C,D: ABCD is a square; F: AF = (2/5) * AB
2. E = AD ∩ CF; K = BD ∩ CF

your desired ratio is |FK:EF|. we apply the Co-Side Theorem to eliminate E and K from this ratio. (sorry i made a careless mistake in the second ratio: the denominator shld be ADC minus ADF.)

\begin{align*}
\frac{FK}{CF} &= \frac{FK}{FK + KC} = \frac{\triangle BDF}{\triangle BDF + \triangle BDC} \
\frac{EF}{CF} &= \frac{EF}{CE - EF} = \frac{\triangle ADF}{\triangle ADC - \triangle ADF}
\end{align*}

[somber coyote]

vin100

[lapis moon]

here i'm too lazy to type $S_{\triangle ABC}$ for the area of $\triangle ABC$. I just write $\triangle ABC$ to represent its area instead.

[somber coyote]

vin100

[lapis moon]

apply the definition of $F$ to further simplify the expressions

[somber coyote]

vin100

[lapis moon]

\begin{align*}
\triangle BDF &= \frac12 \left(1 - \frac25 \right) , AB^2 = \frac{3}{10} AB^2 \
\triangle ADC &= \triangle BDC = \frac12 AB^2 \
\triangle ADF &= \frac12 \cdot \frac25 AB^2 = \frac15 AB^2
\end{align*}

[somber coyote]

vin100

[lapis moon]

divide the first ratio by the second ratio, and the answer will come out.

i omit AB² to save time

,, \frac{FK}{EF} = \frac{\frac{\frac{3}{10}}{\frac{3}{10}+\frac12}}{\frac{\frac15}{\frac15+\frac12}} = \frac{\frac38}{\frac27} = \frac{21}{16}

[somber coyote]

vin100

[lapis moon]

,, \frac{FK}{EF} = \frac{\frac{\frac{3}{10}}{\frac{3}{10}+\frac12}}{\frac{\frac15}{\frac12-\frac15}} = \frac{\frac38}{\frac23} = \frac{9}{16}

[somber coyote]

vin100

[lapis moon]

i made a mistake at the first time. the second one is right.

[willow lynx]

i need help with this i can show you what i have so far but i doubt it’s right

[lapis moon]

the so-called "mapping notation" isn't the one used for elements.

,tex \verb|\mapsto| gives $\mapsto$ in \LaTeX

[somber coyote]

vin100

[teal kayak]

look at cool shape i made

does this thing have a name or am i cool and spesil

[nocturne remnant]

https://en.wikipedia.org/wiki/Spiral_of_Theodorus

[teal kayak]

That's cool

I am still cool for randomly discovering this spiral by playing around on geogebra

[upper karma]

https://youtu.be/yBw67Fb31Cs

How the fuck is trigonometry linked to frequency wave looking graphs

God

Math never fails to confuse the fuck out of me

[fallen sleet]

what is the standard way to find the number of triangles given SSA?

I managed to make this unholy formula as a discriminant (negative means zero triangles, zero means one triangle, positive means two triangles)(trig functions in degrees): $(\frac{sin(90-\angle A)}{sin(\angle A)})^2(4a^2-4c^2)+4a^2$

[somber coyote]

alshfik

[ember timber]

Hi, anyone to help me check my answer? I got 7.5, but the reference answer is 17.5 ! I don't know where I made a mistake.

[ember timber]

oh, I has found the mistake! I should use the perp not the proj~

[strange umbra]

ABC is an acute triangle, furthermore there are two X and Y such that ACX and CBY are isosceles right-angled triangles (with the right angles at X and Y). M is the midpoint of AB. How can I prove that XMY is also a isosceles right triangle?

[dark sparrow]

@strange umbra are you allowed to use complex numbers? i have a feeling there might be some very aethetically pleasing proof of this if you are

[strange umbra]

sure

[dark sparrow]

interpret the coordinate system you've plotted this on as the complex plane, so that A is at 0, M is at some real point a, B is at 2a, and C is at some point z in the first quadrant

i claim that, after some algebra and observing some 90- and 45-degree angles in the picture, you can get $X = \frac{1+i}{2} z$ and $Y = (1+i)a + \frac{1-i}{2} z$

[somber coyote]

Ann

[dark sparrow]

and then (identifying complex numbers with vectors) we get $\overrightarrow{MX} = -a + \frac{1+i}{2}z$ and $\overrightarrow{MY} = ia + \frac{1-i}{2}z$, and so $\overrightarrow{MX} = i\overrightarrow{MY}$ just as we want.

[somber coyote]

Ann

[dark sparrow]

though i do freely admit that i am skipping over a lot of details here.

[ember timber]

Exercise me.Is this latex？Where‘s can I find the format document

[upper karma]

I think it's easier if you let M be the origin

Let A be a complex number a, C is at c, et cetera

The info that AXB is an isoceles right triangle is equivalent to
(c-x)*i=a-x

Similarly for BYC

Just make sure you get orientation right since multiplication by i is a 90 degree rotation counterclockwise

We wanna prove YMX is isoceles right

It helps to place the origin at M

This way we just need to prove that y*i=x

Also since M, the origin, is the midpoint of AC we have a+c=0

[strange umbra]

thx a lot guys 👍

[upper karma]

What is 1/3 plus 9/9 ?

[upper karma]

Whoops should be a+b=0

[dark sparrow]

do you know how to add fractions?

[opaque gull]

Anybody familiar with normal distribution?

[loud shard]

can someone please help me with this problem?

[sly venture]

its find the area of a n-sided polygon with a perimeter of 242 and an apothem spanning 18 units

cans omeone help

[white matrix]

@velvet flicker are you able to use CAS?

if the angles are 28 and 49

the last angle would be 103

i find mf using the sine theorem

and then use the cosine theorem to find AF

since the length can't be negative, AF would be 15.49 and M would be 7.46 if im not mistaken with the angles

since the way that you write the angles are different from mine

@loud shard

[velvet flicker]

why was I pinged

in the geometry channel of all places

[white matrix]

my fault brodie 🤣🤣

[tall python]

bro i need somebody to break down n explain some stuff to me

😭

beggin y'all bro

[opaque gull]

What's up

[undone path]

why

why make ur life harder

[oak sandal]

Yo, having trouble going from 2/tan theta to -2csc(theta)

Anyone has a clue they can toss?

Went to 2cos(theta)/sin(theta but not sure what to do from there

[woeful tangle]

i need help with my ixl

[lunar grail]

So you're trying to find the coordinates of the new square

Do you have an idea of what C' might be?

[woeful tangle]

sorry i dont you think we can get in a call and i screen share bc i really need to get a 80 so i can boost my grade so i can pass

let me know if not its fine

@lunar grail

[lunar grail]

uh no call plz

[woeful tangle]

we cant get in a call and i screen record the ixl?

you dont have to talk

@lunar grail

[lunar grail]

no thanks, we can just type here

[woeful tangle]

ok

this is the promlem im on

what do i do?

[lunar grail]

count 2 units to the right, and 8 units up from C

[woeful tangle]

-4,-5 and -4, 3?

[lunar grail]

-4, 3 would be C'

[woeful tangle]

oh

[lunar grail]

so then find the other points using that same method

and then tell me your full answer and I'll see if you missed it

[woeful tangle]

i got it right yay

[lunar grail]

-4, 5 would be D'

nice!

[woeful tangle]

these ones always get me

[lunar grail]

this one is easier! Because instead of going to the right you only need to count 10 units down

so try counting from R down

[woeful tangle]

-7,-9?

[lunar grail]

Yes

[woeful tangle]

would U be -7,0

[lunar grail]

yes, U' would be

[woeful tangle]

would T be 2,0

and would u actullay be 2, -9

sorry got it mixed up

would S be -7,0

[lunar grail]

yes

[woeful tangle]

https://tenor.com/view/sad-baby-frown-cry-tantrums-gif-4649018

i got it wrong

[lunar grail]

no this was wrong sorry I was in another chat

[woeful tangle]

heres my new problem

[lunar grail]

So use the same methods as before

[woeful tangle]

is N 7, -8

is M 8, -2

i got it right

[lunar grail]

nice, going to bed, have a good night!

[vagrant ore]

I need some help with this problem

[dark sparrow]

angles ABE and EBC are two equal halves of angle B

use this to write down an equation in x, solve it for x, then find the measure of angle ABE

[unreal shuttle]

(english is not my primary language, i hope all my english math terms are correct)

Hello
I want to find a point P(x(t),y(t),z(t)) where the point is always on the intersection of a sphere and a plane.
Let's say the sphere is: x²+y²+z²=2²

ex 1: the plane is: z=0
For this example it's pretty easy, I found P(2cos(t), 2sin(t), 0)

ex 2: the plane is x-z=0
For this I found P(√(2)cos(t), 2sin(t), √(2)cos(t))

Ex 3: the plane is x- 0.2y - 0.4z = 0
For this I did not find a solution and is why i am here.
This is how i tried it.
z = (x-0.2y)/0.4
x²+y²+((x-0.2y)/0.4)² = 4
x²+y²+6.25x²-2.5xy+0.25y² = 4
7.25x²+1.25y²-2.5xy = 4
In the previous example i did not have the xy term so I could easily make an ellipse out of this and convert it to cos and sin. Then I'd have x(t) and y(t) and could find z(t) with the equation of the plane. I think this example is also an ellipse but not aligned properly around the x and y axis. So i think this wil get me something of the form P(a*cos(t), b*cos(t-c), d*cos(t-e)) I am not sure of this.

[steel ibex]

Does LHS=RHS?

[static jolt]

Yes

1/3/4 = 4/3

[steel ibex]

How 4 goes at the numerator?

[static jolt]

(1/3) ÷ (1/4)

@steel ibex I was wrong sorry they arnt equal

They called "complex fractions"

1/3/4 = 1/12 not 4/3

[steel ibex]

Thank you, for your clarification 👍

[static jolt]

The formula to calculate the area of a regular polygon is, Area = (number of sides × length of one side × apothem)/2, where the value of apothem can be calculated using the formula, Apothem = [(length of one side)/{2 ×(tan(180/number of sides))}].

[torpid mountain]

helppppppppppppppppppppppppppppppppppppppppppppppppppp

math is so math

or im just dumb

im probably just dyumb

dumb

[opaque gull]

I don't understand the "English" in it like I'm not sure if away is the hypotenuse distance or whatevs

i think its like this

then use pyth theorem

[pliant roost]

It does because the 1 / 3 / 4, not 1 / 3 / 4, since the equality sign is on the first division. So the LHS and RHS are indeed equal

[dark sparrow]

y'all need parentheses

1/(3/4) vs. (1/3)/4

[pliant roost]

yes...

[vagrant ore]

I need help with this

[azure crown]

make a right triangle with an altitude from <x then use trig

[vagrant ore]

How?

[lapis moon]

use the law of cosines

[summer salmon]

I have this question on an assignment of mine, super confused as to where to begin, I have a unit circle and a diagram I drew of it but idk where to go from there

[astral hull]

in the figure, cd is a line segment of the diameter, and bc is perpendicular to cd, prove that abcd is concyclic. Can someone give me some hints to deal with it?

[wanton edge]

How I do it
Step 1. Prove that ∠BAD = 90°
Hint: this picture

[wanton edge]

@astral hull From the above picture, draw the line segment AD. You can see that ∆OAD is isosceles

[magic lichen]

guys i didnt understand equations for sinusoidal graphs

and how u use them for finding amplitude, period and range

[summer salmon]

amplitude is just half the difference between the highest point of the graph and the lowest point of the graph

(max-min)/2

[upper karma]

$\sin(\pi/2 - x ) = \cos x$. right if i change sin into $\sin^2 $ and cos into $\cos^2$. will the eq be alright??

[somber coyote]

moltres109

[astral hull]

thanks for giving me a new perspective. Is APO always collinear?

[wanton edge]

Yes. Because those two circles have the same tangent point

[astral hull]

Oh i see it the common tangent point. Thanks a lot

[wanton edge]

No problem

[magic lichen]

no im talking about the equations

like $ \frac -1/2 \cos 3x $

uh...

$y=-1/2 \cos 3x$

ok

wh

what does the equation mean

[somber coyote]

Alpha | Sync

[vagrant ore]

Is the answer just 6?

[magic lichen]

7 right?

wait nvm

[vagrant ore]

Nah it was 6

[magic lichen]

ye im pretty sure

ye ye

[magic lichen]

guys in a sinusoidal graph

$y=a \sin(bx+c)+d$

[somber coyote]

Alpha | Sync

[magic lichen]

what value does c change

like a changes amplitude, 2pie/mag b changes period, and midline changes d

what does c change

[dark sparrow]

pi, not 🥧

c is the phase shift

[magic lichen]

ye i got it

what i meant was

how do u calculae if theres a phase shift or not

[dark sparrow]

there is always a phase shift

it might be zero or not zero

[magic lichen]

bruh

ok u didnt get the quesiton

i was asking how u calculate phase shift if ur given only 1 graph

[dark sparrow]

look at where the sinusoid crosses its midline

i don't have the energy to describe this in full or with any sense of formality right now

[magic lichen]

ye i got it

its basically the offset of the first up going point on the x xis where y=0

rigt?

right*?

[dark sparrow]

where y=d

cause that's your midline

[magic lichen]

ye ye

:V

[vagrant ore]

Would the answer be 3 sqrt(3)?

[dark sparrow]

no

[vagrant ore]

Oh. How would I figure out the answer?

[nocturne remnant]

Easiest way would be to use Pythagorean theorem on all 3 right triangles to obtain 3 equations

And then solve for z

Well no actually

I think

By similar triangles 3/z = z/30
And then solve for z directly

[nocturne remnant]

This is much faster

[vagrant ore]

Would you cross multiply?

[nocturne remnant]

Yeah you can do that

[vagrant ore]

So 45 is the answer?

[nocturne remnant]

How did you get 45

z times z is z^2 not 2z lol

[vagrant ore]

Oh :P

3 sqrt(10) should be right

[nocturne remnant]

Yeah

[upper karma]

@vagrant ore I think it's 9

[vagrant ore]

Nope

[dark sparrow]

yes

[fallen patio]

feel free to dm me :)

😭 💀

if ur willing too ofc

[ember oasis]

can anyone tell me how pi/6 equates to sqrt3/3?

[dark sparrow]

pi/6 itself doesn't

tan(pi/6) does

[ember oasis]

ah yeah but how'd it get to sqrt3/3?

[dark sparrow]

draw a 30°-60°-90° triangle

and observe that the side opposite 30° (a.k.a. pi/6) is 1 and the side adjacent to it is sqrt(3)

[steel ibex]

One way to think about it is that if you take a right triangle with one angle being pi/6, then the side opposite of the pi/6 angle will be 1, the side adjacent to the pi/6 angle will be sqrt(3), and the hypotenuse will be 2. So, another way to think about pi/6 is that it is the angle that gives a right triangle with sides of 1, sqrt(3), and 2.

[ember oasis]

gotcha thanks!

why is sin(3.45098) = 0.060... on a calculator while sin(3.45098) on symbolab is -0.30447...

[steel ibex]

The two values are actually the same; it's just that your calculator is in radian mode while Symbolab is in degree mode. To convert from radian to degree, multiply by 180/π. Therefore, sin(3.45098) in radian is actually sin(3.45098 * 180/π), which is equal to -0.3044737347536959.

[ember oasis]

OH THANK YOU

[ember oasis]

what about inverse signs?

[dark sparrow]

other way around, the calculator is doing it in degree mode while symbolab is in radian mode

[ember oasis]

what do they mean when they say "find the measure" ?

is it the angle of the acute central angle?

[dark sparrow]

"the angle of the angle" is kind of weird

but yes you're asked to find the size of the angle

ie how many degrees it has

[ember oasis]

quite lost on this one, are the given values enough?

because from my previous examples, it would normally have pi on the arc length

[dark sparrow]

yes, your angle will be 34/39 radians

[ember oasis]

ohh and then convert that to degrees?

[dark sparrow]

yes

[ember oasis]

gotcha ty !

[robust tinsel]

Yes, this is for a test. No, I don’t want the answer directly. I need someone to define some of these

I need definitions for Congruent and perpendicular.

[wanton edge]

Congruent* is basically another way to say "identical". Perpendicular is when two lines meet at right angle (90°)
(*In geometry)

[robust tinsel]

That’s… actually more simple and helpful than anything I’ve seen on Google. Thanks!

[wanton edge]

You're welcome

[robust tinsel]

How do I dilate a figure?

[sleek star]

(1;-3)

[robust tinsel]

Uh…

What?

[sleek star]

oh wait figure

xD

dint read that

sorry

[robust tinsel]

You’re fine, im kinda stupid and don’t know half of math lol

And I graduate in 2 weeks…

[sleek star]

how many units dilate

[vagrant ore]

I need help

[grave pond]

You need to know that sin(30°) = ½.

Or possibly remember that cos(30°) = sqrt(3)/2.

[errant salmon]

this should help

memorizing the trig ratios will help a lot in the future

[near trail]

I have a question or two. How can I calculate the green line length if I know the placements of two points (A and B) on the circle boundaries? What is this problem called (so I can research it more)?

[vagrant ore]

What if the angle is 35 instead of 30?

[grave pond]

Then the result cannot written exactly using root signs; you have to resort to trig tables or a calculator anyway.

[vagrant ore]

So for this problem would I do 15sin(35)?

[chilly raven]

I really need help on this one

[errant salmon]

i’m not sure i follow

[grave pond]

Yes.

[errant salmon]

but to find x u can use SOHCAHTOA i think

[grave pond]

A whole bunch of angles in the diagram are the same, and the length of each of HE and GE becomes the side length of the square, divided by the sine of one of the identical angles.

[chilly raven]

I don't understand

[vagrant ore]

How do I do this?

[static jolt]

find area of the circle so 3.14*6^2

[vagrant ore]

So 113.1

[static jolt]

the portion of the shaded region plus the triangle is: 113.1*(5/12)

5/12 is the 150/360

[vagrant ore]

So the answer should be 47.125 then

[static jolt]

now have to find the area of the triangle and subtract it from 47.125

[vagrant ore]

What would the height be?

[static jolt]

so you know all three angles

its 150 + 15 + 15

[vagrant ore]

Oh ok

180

[static jolt]

yea but to find the height will have to split the triangle into 2 triangles

[vagrant ore]

Ok

[static jolt]

so then will have 2 right triangles

[vagrant ore]

Yes

[static jolt]

so each triangle angle is 90 + 75 + 15

[errant salmon]

just find the area of the sector

then find area of triangle

and do area of sector minus area of triangle

[vagrant ore]

How would I do those?

[static jolt]

the area of the sector is 47.125

we already found that

[vagrant ore]

Oh ok

[errant salmon]

fax

[vagrant ore]

So area of a triangle

[errant salmon]

yea

1/2ab sinC

is area of triangle

[vagrant ore]

Huh?

[errant salmon]

sorry

$0.5 \cdot a \cdot b \cdot \sinC$

that’s the equation for area of triangle

[vagrant ore]

But for my problem what would the variables be

[errant salmon]

ok

so

label the sides on ur triangle

btw calculator is allowed right

but yea so you can technically make any side a and b

in this case you’d want them to be sides that you know the length of

[vagrant ore]

Yes

[errant salmon]

and you’d want sinC to be an angle that you know right

so you have all the information needed

[vagrant ore]

Yes

[errant salmon]

what do u think a and b will be

[vagrant ore]

So like this?

[errant salmon]

yes

sure

that’ll work

[vagrant ore]

Ok

[vagrant ore]

So now what?

[errant salmon]

what’s the area of the triangle

[vagrant ore]

I don't know

[errant salmon]

calculate it

you can’t more forward without it

[vagrant ore]

How? It's only variables

[errant salmon]

?

i don’t get what u mean

you have

the variables

a and b are equal to 6

you labeled it correctly

sinC = sin(105)

make sure your calculator is in degrees

[vagrant ore]

Sin(105) is 0.9659

[errant salmon]

g

calculate the area of the triangle using the formula i gave

[vagrant ore]

Which one?

[errant salmon]

$0.5 \cdot a \cdot b \cdot sinC$