geometry-and-trigonometry

[slender socket]

So for the first one

What triangle do you use that includes sin

[upper karma]

I heard of them, never said I remember them xD

[slender socket]

Pahaha okay 1 sec

So the O, A and H stand for Opposite Adjacent and hypotenuse

[upper karma]

ah

[slender socket]

And the s c and t sin cos and tan

So for the first question we have to look at the SOH triangle

[upper karma]

ok

[slender socket]

Now we already know the opposite length to the angle and the length of the hypotenuse

So it’s just about plugging in values

Give the first question a go from there

[upper karma]

so it would be 8 / 3 in this case?

[slender socket]

Not quite

It would be opposite over hypotenuse

[upper karma]

ups my bad

didnt read correctly

3/8 😛

[slender socket]

Nice one

And because they only want the sinØ that’s your answer

Okay for the second one

Which triangle do you need

[upper karma]

CAH?

[slender socket]

Perfect

But for the Cah you need the adjacent length which you’re not given

[upper karma]

phytagoras

I can do that much 😛

[slender socket]

You got it

So work out the adjacent and put the result into the triangle

[upper karma]

the answer would be 0.92625 if I am not a complete retard

aaand 0.4048583

now this one

I dont understand what it menas

[slender socket]

Okay so

Ø is the angle

You’ve found sinØ cosØ and tanØ

Let’s look at the first one first

SinØ
To isolate the Ø we need to cancel out the sin

sinØ = 3/8

What could we do to both sides of the equation that would cancel out the sin

[upper karma]

divide by sin

or am I a complete idiot?

[slender socket]

Kind offf

You apply sin^-1

[upper karma]

oh goddamn

[slender socket]

Which you could think of as dividing by sin

So
Sin-1(sin(Ø))= sin-1(3/4)
Ø=sin-1(3/4)

Make sense?

Sorry that 4 should be an 8

On both of them

[umbral rivet]

Ooh, trig 👀

[upper karma]

I was confused for a sec about that 4 lol

dont scare me like that

[slender socket]

Haha sorry I forgot what the triangle was and guessed

[upper karma]

22.024312837, right?

[slender socket]

I don’t have a calculator to hand 1 sec

Yep spot on

[tropic stirrup]

Have fun

[upper karma]

goddamit, I will have a test on trigonometry in 3 weeks and I am no good at it

[tropic stirrup]

Whoops missing condition: green line is tangential to the circle

[slender socket]

Do you need a solution for that I don’t have a pen and paper nor can be bothered to grab one

You’ve got this BreakTruth, keep practicing and it’ll become second nature

[tropic stirrup]

(I made the question btw)

[slender socket]

Ah right nice

[upper karma]

I did trignometry last year as well, didnt understand jackshit

Caught Another Hippie
Tripping On Acid```
How to remember SOH CAH TOA
this is the only thing I can remember from last year

[tropic stirrup]

Lool

[slender socket]

Amazing

[upper karma]

xD

lol found this one

Some Old Hags Can't Always Hide Their Old Age

how would I do this one?

[tropic stirrup]

Express the definition of tan 25°, given that the height is h

🤔

[upper karma]

wut

[tropic stirrup]

What's the definition of tan?

[slender socket]

(Look at your sohcahtoa triangles)

[upper karma]

I amanged it

just forgot to come back xD

[tropic stirrup]

😑 👍🏼

[feral fractal]

anyone here?

need an answer asap

[upper karma]

ask then xd

[feral fractal]

angle(b)=45 degrees, median to b=7cm and a:c=3:4

in a triangle

usually, I would break the ratio apart so a=3x and c=4x

[upper karma]

care to upload a picture

[feral fractal]

and use it in an equation but im kinda stuck

I just have this

the last value is the median of b if it's called that

I'm just stuck on the ratio part because i have no idea in what equation i could implement it

[feral fractal]

Sketch: https://imgur.com/a/vMhGE

[upper karma]

4150. Calculate the volume and the total area

I'm not even sure if I am doing this right

Use the formulas.

I am but this is freaking out

am I not supposed to do

π*48*48

/3

since π*r^2

/3

Volume of a Sphere $$ = \frac{4\pir^3}{3}$$

[charred spear]

Rendering failed. Check your code. You can edit your existing message if needed.

[upper karma]

Volume of a Sphere $$ = \frac{4\pi r^3}{3}$$

[upper karma]

Sorry, which one are you exactly trying to solve?

sphere?

oh

a)

So surface area and volume of a cone?

yes

So surface area of a cone $$= \pi a \sqrt{h^2+r^2}

So surface area of a cone $$= \pi a \sqrt{h^2+r^2}$$

[upper karma]

wait not surface area

total area*

but I'm only struggling with the volume

Well surface area in English. But the volume

oh

well I have the formula for the volume

but I keep getting wrong answers

not quite sure what I am doing wrong

=tex V=\frac{\pi r^ h}{3}

[upper karma]

=tex V=\frac{\pi r^2 h}{3}

[charred spear]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[upper karma]

Sorry.

Second one is correct.

yes

so shouldn't it be

pi*48*48*90/3

Yes.

can you tell me exactly how to type it on the calculator

(pi4848*90)/3

==(pi4848*90)/3

[charred spear]

217146.88421613

[upper karma]

oh so I actually got it right

If that's the answer that you got, then yes.

then why does the answer sheet tell me that it is

v = 2,2*105cm^3 = 200dm^3 (217,14)

Do they want the answer in dm^3?

Our current solution is in cm^3.

it is not said

but I assume so

how do I make it into dm^3?

Then just divide our solution by 1000 to get dm^3.

by 1000?

1000 cm^3 = 1 dm^3

10 cm = 1 dm.

didn't know that

good to know

so I was just confused on that part

Gotta watch those units.

if I may ask

what does

π48*48

actually mean

isn't it pi*48

*48

Yep.

then how come when I do it like that on the calculator

I get different results

parenthesis I assume?

Calculator syntax isn't entirely equivalent to our own.

something like it doesn't calculate in the order I want

yeah

Might be that it recognizes 48pi and not pi48.

Convention is normally 48pi, but pi48 isn't wrong.

how would I type this on paper?

just like it says in the formula?

or do I do it the normal way

=tex V=\frac{\pi r^2 h}{3} \ =\frac{\pi (48)^2 (90)}{3}

[upper karma]

=tex = 217.146,884 cm^3

[upper karma]

=tex = 217,15 dm^3

[upper karma]

yeag

217146/1000

Yes.

oh wait my formula sheet only has the mantelarea

how do you do the surface area again?`

=tex S=\pi r\sqrt{h^2+r^2}

[upper karma]

Anyways, g2g. Hope this helped.

thanks

[upper karma]

ABCD is a square and ABE has equal sides

Calculate x

[dark sparrow]

okay so ABE is an equilateral aka equal-sided triangle

what do you know about the angles in an equilateral triangle?

[upper karma]

60

degrees

[dark sparrow]

aha

so angle EAB = 60°

[upper karma]

indeed

[dark sparrow]

and angle DAB = 90°

since

ABCD is a square

so angle DAE = ?

[upper karma]

oh yeah

360/4

but can I even know what DAE is

when I don't know what the angle of DEC is

[dark sparrow]

DAE + EAB = DAB

is that clear?

[upper karma]

isn't it DABC?

[dark sparrow]

i said angle

an angle that doesn't have its own name can be named using three points

like so: this angle is PQR

[upper karma]

wait where is the angle of DAB located then

at D?

[dark sparrow]

no!

the vertex of the angle is the middle letter!

[upper karma]

oh

[dark sparrow]

cause like

think of going up one arm of the angle to the vertex and then down the other arm

[upper karma]

hm

[dark sparrow]

that's how angles are named

[upper karma]

bringing it down again

DAE + EAB = DAB

[dark sparrow]

mark those angles on your diagram

it should become obvious

[upper karma]

like this?

wait yeah

that makes DEE into 30 degrees

since it's a square with

90 degrees angles

90-60

[dark sparrow]

something like that

[upper karma]

yes

then I did it correctly

[dark sparrow]

colors are really helpful yo

if you don't have access to colors you can do single vs double arc marks

[upper karma]

double?

like 2 arc's

[dark sparrow]

like that

[upper karma]

yeah

[dark sparrow]

they don't have to be perfect arcs ofc, since that's just an auxiliary mark you're making

[upper karma]

then how do I solve the rest?

yeah I know

[dark sparrow]

well ok

so DAE is isosceles

and you know that the angles at the base of an isosceles triangle are equal

right?

[upper karma]

yes

60

[dark sparrow]

no

no!

[upper karma]

wait no

[dark sparrow]

isosceles

[upper karma]

I'm dumb

yeah, language barrier

[dark sparrow]

they are equal to each other, not to some fixed value or anything

[upper karma]

thought you were speaking about the other triangle

the top one ye

[dark sparrow]

DAE

[upper karma]

yes

so they are x+x+y=180

[dark sparrow]

except that a) you know y = 30, and b) you don't even need to whip out a full-blown equation for this

cause like

you know these three angles have got 180 degrees to share between them

[upper karma]

yeah

[dark sparrow]

angle DAE has taken 30

so that leaves 150 for ADE and AED

[upper karma]

I got 150 left

[dark sparrow]

and they need to split it evenly

so each gets 75

[upper karma]

they do?

how so

[dark sparrow]

well they must be equal

since yknow

[upper karma]

ohj yes

[dark sparrow]

this is an isosceles triangle

[upper karma]

jeez I keep confusing myself by looking at the other triangles

yeah I see now

one angle has taken 30 degrees

since then they are isco I cannot spell that word

they need to be split up

[dark sparrow]

Borrowed from Latin īsoscelēs, from Ancient Greek ἰσοσκελής (isoskelḗs), from ἴσος (ísos, “equal”) + σκέλος (skélos, “leg”)

"skelos" is kind of like "skeleton"

except k becomes c because latin

[upper karma]

yeah but we have another term for it

so I probably won't remember it

[dark sparrow]

likbent

[upper karma]

equal legs

[dark sparrow]

yeah same in russian

etymologically anyway

[upper karma]

so

show that y = 180 - 2x

can I explain this with

CAB = CBA

CBA = x

[dark sparrow]

yes

yes you can

[upper karma]

180-2x=y

good

now I actually understand this

In the triangle ABC is AB = BC

CE is a bisector

[dark sparrow]

aha

[upper karma]

Show that the angle is x = 75

[dark sparrow]

okay so first off, you know that BAC and BCA are both 70°

since again

B takes 40° and BAC and BCA have to split it evenly

[upper karma]

yeah

I just want to make sure

that makes it 110 left for x + ACE

BEC is 40 if I understand correctly?

wait no

[dark sparrow]

nnnope

[upper karma]

they never said that

[dark sparrow]

CE is a bisector though

so you know ACB gets cut in half

so ACE and BCE are both 35°

so now consider AEC

EAC took 70

[upper karma]

oh

[dark sparrow]

ACE took 35

[upper karma]

bisector makes it cut in half?

that clarifies it for me

[dark sparrow]

well yes that's what a bisector is

[upper karma]

honestly I didn't know you guys are easier to understand than my teacher

[upper karma]

def of bisector is a line/line seg/ arrray that splits a line segment into two equi distant parts

its can only split a ling seg

bc array and line have infin length

[dark sparrow]

...you're missing bisectors of angles

[upper karma]

can someone help me with this stuff? I dont get it

I never understand trigonometry

[thorn talon]

which one?

[upper karma]

I guess start from top to bottom xD

[thorn talon]

draw a diagram

[upper karma]

no paer nearby

¯_(ツ)_/¯

paper*

[tropic stirrup]

✋🏼 😩 👌🏼

[thorn talon]

ok let's do 6

[upper karma]

sure...

[thorn talon]

i assume you can use calculator?

[rustic talon]

SOHCAHTOA

Will help

[thorn talon]

^

@upper karma how familiar are you with the three basic trig functions?

[upper karma]

not much

[thorn talon]

ok

[topaz valley]

this shit's about to blow your mind

[thorn talon]

so let's use the above nmenomic

i wish i can spell

[tropic stirrup]

👀

[thorn talon]

so anyway: SOHCAHTOA

sine: opposite/hypotenuse

cosine: adjacent/hyptenuse

tangent: opposite/adjacent

i should have used equals

but meh

so looking at that picture of 6

which one would you use?

[upper karma]

SOH

[thorn talon]

good

now

since you are trying to find out angles

you need to use inverse sine

on your calculator it probably looks like sin^-1

[upper karma]

ye

tho it doesnt look like that 😛

[thorn talon]

arcsin?

idk

[upper karma]

I got a ti nspire cx, there is just a trignonometry button with all trig functions

[thorn talon]

interesting

anyway, that's how you find the angle for this

also, even though it's ^-1

it's not 1/sin

it's the inverse aka arcsine

[upper karma]

so arcsin(0.6/5)

[thorn talon]

yep

[upper karma]

4237

Show and prove why the area A can be calculated with the formula A = h(a+b)/2

[surreal bolt]

you take the trapezoid, copy it, rotate the copy 180, connect it back to the original and you get a parallelogram.

(inc picture)

[upper karma]

yeah but I'm supposed to show it with a formula

I mean he left me with this

[surreal bolt]

That is valid also.

[upper karma]

I don't understand it

[surreal bolt]

Oh the maim trick here is to think of the trapezoid as a rectangle and the leftover area as a triangle

[upper karma]

I mean I do understand it just not how to explain it

[surreal bolt]

so if you cut out the rectangle you have two triangles. If you take the left triangle and shift it to the right until it is next to the right triangle, then you make 1 triangle out of two.

[upper karma]

I know thaat it's practically a rectangle

just not how to explain it

[surreal bolt]

😃 thanks (downloading Geogebra)

[upper karma]

so how do I explain that

in my book

[surreal bolt]

um, not sure. There are usually a lot of ways to prove things. If your teacher wants you to know a specific proof, you're going to have to ask him both what method and how descriptive your proofs need to be.

[upper karma]

he wants me to prove that it works

I don't think he cares how I do it

since he says that if I find a method that suits me, it's valid

[surreal bolt]

again, there's differing opinions as to how much detail is required to show it "works"

So do you know the formula for the area of a parallelogram?

[upper karma]

yes

height * a + height * b

/2

I have a sheet for it

[surreal bolt]

Sorry for the handwriting.

I dunno what detail you need. Some teachers may require you to prove that connecting the two congruent trapezoids is, in fact, possible.

As in , the angles really are supplementary

[upper karma]

can you explain what he wrote?

[surreal bolt]

Um, the picture above from root(2) pretty much says it all.

2 is a rectangle with height h and width b. That means it has area bh
1 and 3 form a triangle if you slide them together. The height of the triangle is h. The base is (a - b). The total area is h(a - b)/2.

So the area of the rectangle and the post-slide triangle is bh + h(a - b)/2

Then there are some "substitutions" an simplifications.

[upper karma]

oh yeah

1+3

is a-b

since you take out the parts you use for the rectangle

[surreal bolt]

yessir.

bh = 2bh/2 (a substitution used)

We have:
bh + h(a - b)/2 ... starting point
==> 2bh/2 + (a-b)h/2 ... bh = 2bh/2
==> (2b + (a - b))h/2 ... factor out a h/2
==> (2b + a - b)h/2 ... simplify 2b + (a - b)
==> (a + b)h/2 ... simplify 2b + a - b

[upper karma]

now you are making me confused

I know I have h(a-b)+bh

[surreal bolt]

um h(a -b)/2 because the last part was a triangle.

i.e. area of rectangle + area of triangle --> bh + h(a-b)/2

[upper karma]

yeaah

[surreal bolt]

um yeah? h(a-b) needs a divided by 2

[upper karma]

yes

I know

since it's a triangle

[surreal bolt]

lol okay just making sure cause you wrote ... h(a-b)+bh

[upper karma]

ye I forgot

[surreal bolt]

kk well I gotta go. Hopefully that was sufficient to answer the question

[upper karma]

I'm still kinda hanging on, if anyone could help

[upper karma]

if so, @ me

[rare talon]

@upper karma divide a trapezoid by one of the diagonals

And count the area of the two triangles you made by it

You'll have two triangles, one with base a and height h, and the other with base b and height h

[upper karma]

@rare talon I thought they both had the base of a-b

[rare talon]

@upper karma I use different techniques

[upper karma]

hm

[rare talon]

Wait I'll upload a pic

This is how I divide

[upper karma]

is he looking for the value of the red segment ?

[rare talon]

@upper karma

He is trying to prove the area of a trapezoid

[upper karma]

that the formula is h(a+b)/2

[rare talon]

@upper karma

[upper karma]

yes

[rare talon]

Do you see my pic?

[upper karma]

I do

[rare talon]

Then try to calculate the area of each triangle

   A-----B
   /     \
 D _______C```

[upper karma]

oh you make it into 2 different?

[rare talon]

Yeah

Divide it based on line AC or line BD

Based on my illustration above

[upper karma]

oh

wait

https://i.imgur.com/K66IWz2.png

like this? @rare talon

[rare talon]

Yeah, I guess , I don't know how you really divide it

But actually I hope you can imagine

What is the base of triangle 1 , and the height

The base of triangle 2 , and the height

^if you're doing it that way, then it's fine and you can imagine what will happen

I actually just assume you separate those 2 triangles

ABD and BCD

Ping me when you have a problem!

[upper karma]

oh wait yeah

so I'm just supposed to calculate the 2 triangles? @rare talon

[rare talon]

Yeah

Try it

[rare talon]

@upper karma do you manage to get the proof?

[upper karma]

I never tried @rare talon

[rare talon]

Okay....

[upper karma]

got caught up in other stuff

[rare talon]

Don't worry about that 😄

[upper karma]

so I'm like supposed to do a*h?@rare talon

oh yeah

if I do a*h two times

[rare talon]

Yeah

Actually it's a*h/2

[upper karma]

yeah

but for two triangles @rare talon

it will be +-0

or am I just lost

[rare talon]

You’re lost

Remember the triangle I made for you

What is the area of triangle BCD ?

[upper karma]

b*h/2

?

oh

that basically just proves it

doesn't

since I have to get DAB too

which is a*h/2

which means that h(a+b)/2

I forgot how to do finite differences

Anyone know how

[rare talon]

Yeah

[upper karma]

@rare talon wait did I solve it?

[rare talon]

You should’ve

You got the area of BCD and ABD

Then just add it

To get the area of a trapezoid

[upper karma]

so ah + bh

[rare talon]

/2

[upper karma]

yes

= area of a trapezoid

[rare talon]

Yeah

Anyway the area of triangle is ah/2

Just making sure you’re not forgetting it

[upper karma]

yeah but so is the trapezoid

[rare talon]

Yeah because you prove it

😄

[upper karma]

a*h + b*h /2 triangle = trapezoid h(a+b)/2

[rare talon]

ah/2 + bh/2 = (ah+bh)/2 = h(a+b)/2

[upper karma]

ohh yeah

but isn't b and a

the same

if you split the trapezoid

like that

[rare talon]

No?

The shape of a trapezoid won’t change

It’s just a matter how you see it

[upper karma]

oh

[upper karma]

@rare talon hey

could you help me with this one?

[rare talon]

Problem?

[upper karma]

4234

[dark sparrow]

are you asked to find the area of the shaded region?

[upper karma]

I was typing

Show that the formula for the colored area is right

[dark sparrow]

half-circle minus the triangle

the half-circle has area πr^2/2

[upper karma]

yes

[dark sparrow]

the triangle has base 2r and height r, so its area is r^2

[upper karma]

r^2?

oh cause of the /2

[dark sparrow]

2r * r * 1/2

yes

=tex A = \frac{\pi r^2}{2} - r^2 = r^2\left(\frac{\pi}{2}-1\right)

[dark sparrow]

there we go

[upper karma]

I just don't understand the -2

[dark sparrow]

=tex \frac{\pi}{2} - 1 = \frac{\pi}{2} - \frac{2}{2}

[upper karma]

could you explain?

[dark sparrow]

i just did :v

[upper karma]

yeah but I don't really understand

[rare talon]

You can rewrite r^2 as 2r^2 / 2 if you feel comfortable with it

[upper karma]

I mean I know that I have πr^2/2 and r^2

[rare talon]

Yeah then to get the shaded area, subtract the area of half circle to the area of triangle

Then it is πr^2/2 - r^2

=tex \frac{\pi r^2}{2} - \frac{2r^2}{2}

[dark sparrow]

=tex \frac{\pi}{2} - 1 = \frac{\pi}{2} - \frac{2}{2}

[dark sparrow]

do you understand this?^

@upper karma

[upper karma]

not really

[dark sparrow]

2/2 = 1 😐

[upper karma]

oh

wait

oh yeah

it just confuses me when there are a lot of numbers

that's why I like individual calculations even though they are not as efficient

[upper karma]

Calculate the circumference and area of a square with the diagonal 70.7m

[dark sparrow]

circumference? of a square?

perimeter, surely?

and is that exactly 70.7 m? or

[upper karma]

yes

perimeter

and it is 70.7m

nothing else said

[dark sparrow]

k so

pythagorean theorem

[upper karma]

yeah, that's the part of the book I am in

I just don't know how

I don't know how u do it on a rectangle

[dark sparrow]

draw a square

draw one of its diagonals

you get two (identical) right triangles

[upper karma]

yeah

[dark sparrow]

and you know that all sides of your square are equal

because it's a square

[upper karma]

yeah

[dark sparrow]

so denoting your side with x

you know that x^2 + x^2 = 70.7^2

you can now find the values of 4x and x^2

[upper karma]

oh yeah

sqrt(70.2)?

honestly idk

[dark sparrow]

what is that meant to be

[upper karma]

I have no idea

[dark sparrow]

...

== 70.2^2

[charred spear]

4928.04

[dark sparrow]

2x^2 = 4928.04

x = ?

[upper karma]

4928.04/2

sqrt(Ans)?

[dark sparrow]

yes that will get you x

[upper karma]

x*4?

is the perimeter then

that means that x^2 is the area

== 70.7^2

[charred spear]

4998.49

[upper karma]

== 4998.49/2

[charred spear]

2499.245

[upper karma]

area

[charred spear]

49.99244943

[upper karma]

== sqrt(2499.245)

x

x*4

== 49.99244943*4

[charred spear]

199.96979772

[upper karma]

perimeter

[dark sparrow]

yes

[upper karma]

Solve the hypotenuse

[drowsy pilot]

so for right triangles what can you apply

what equation relates the two legs of a triangle with the hypotenuse

[upper karma]

I mean I guess it would be sqrt(6a^2+8a^2/7)

but how the hell do I solve it when I don't know what A is

[drowsy pilot]

you just leave it in terms of a!

and don't forget to square your 2nd leg too

[upper karma]

2nd leg?

[drowsy pilot]

so pythagorean theorem states that:

=tex a^2+b^2=c^2

[upper karma]

yeah

but I wrote both

[dark sparrow]

sqrt(6a^2+8a^2/7)
no

[upper karma]

just forgot to add ^2 to 8

[drowsy pilot]

you have to square the entire fraction

not just the numerator

[dark sparrow]

=tex \sqrt{\left(\frac{6a}{7}\right)^2 + \left(\frac{8a}{7}\right)^2}

[drowsy pilot]

^

[upper karma]

yeah but how am I supposed to solve that

on a calculator

[drowsy pilot]

apply the exponent to each fraction and you should be able to add them together

since denominators are equivalent

[upper karma]

u mean

14a/7

or what

[drowsy pilot]

so do the first fraction first

square each term of it

[upper karma]

I don't have a on my calculator

so

[drowsy pilot]

a is just a variable

so a^2 is just a^2

the value of a isn't important here

[upper karma]

but I can't type it

on my calculator

[drowsy pilot]

then consider just the numbers

=tex 6^2/(7^2)

[upper karma]

7??

oh

[drowsy pilot]

that's the denominator of your fraction, no?

[upper karma]

I thought you changed the 8 to a 7

[drowsy pilot]

noo

we're just talking about the first fraction

[upper karma]

can't I just do sqrt (6/7)^2+/(8/7)^2

tho

[drowsy pilot]

yes, that'll give you the coefficient

now just remember you have an a^2 with that

inside the square root

[upper karma]

well isn't that just reverse

[drowsy pilot]

what do you have inside your square root

after applying the exponent to both fractions

[upper karma]

6/7 + 8/7

?

[drowsy pilot]

you have to square the numbers

so square both the top and bottom of both fractions

[upper karma]

didn't I just do that

am I not supposed to square all of it

[drowsy pilot]

the fraction we started with was 6/7

[upper karma]

yes

[drowsy pilot]

you're familiar with squaring right?

[upper karma]

yes

[drowsy pilot]

so (6/7)^2 is equal to 6^2/7^2

so apply that

[upper karma]

yes

[drowsy pilot]

what did you get?

[upper karma]

0.73

[drowsy pilot]

you're gonna want to leave it in fraction form

[upper karma]

yeah but the problem with this

the only diffuculties I have

is having with variables

that aren't numbers

• having it

[drowsy pilot]

that's why im having you ignore the a for now

[upper karma]

staying as a fraction

[drowsy pilot]

so let's apply the rule i said

6^2/7^2

what's 6^2

[upper karma]

12

and 7^2 is 14

no

no

wait

[drowsy pilot]

squaring isn't multiplication

[upper karma]

36

and

[drowsy pilot]

yes

[upper karma]

49

[drowsy pilot]

so what is your fraction now?

[upper karma]

36/49

[drowsy pilot]

good

now lets look at the second fraction

[upper karma]

it is

64/49

[drowsy pilot]

yes

now add those together, since denominator is the same

[upper karma]

36/49 + 64/49

[drowsy pilot]

what do you get

[upper karma]

100/49

[drowsy pilot]

good

now we put our a^2 back in

so you should now have

=tex \sqrt{\left(\frac{100a^2}{49}\right)}

[drowsy pilot]

does that make sense

[upper karma]

kinda

but why do I make it a 2nd time

[drowsy pilot]

why do you make what a 2nd time

[upper karma]

^2

[drowsy pilot]

im not sure what you mean

[upper karma]

the answer is 10/7

for the whole calculation

thank you ti 84

[drowsy pilot]

right

but the answer will also be in terms of a

so you did sqrt(100/49)

[upper karma]

10*a

[drowsy pilot]

10a/7 but yes

nice!

[upper karma]

I am allowed to have my calculator for the whole test

and we were supposed to use it

so if I take out a

I can always reapply it later on

as long as a doesn't change

[drowsy pilot]

well when you're adding or subtracting fractions yes

that was to eliminate some confusion you were having

but in general "a" can be considered as a constant you just don't know

still apply the same operations to it

[upper karma]

so what is this

4260

Calculate the area of ABC

I mean I know how I would solve it but idk how to get the other base

[drowsy pilot]

hmm

[upper karma]

BC is 20

only thing I know

[drowsy pilot]

right

my objective would be to find the length of side AB

im blanking almost entirely wow this is embarassing LOL

[upper karma]

@rare talon by any chance you know?

[rare talon]

Sorry

What is the question?

@upper karma do you know similiarity of triangles?

[upper karma]

similarity? @rare talon

[rare talon]

Yeah

[upper karma]

what do you mean

[rare talon]

https://www.mathopenref.com/similartriangles.html

See the picture

2 triangles on the picture are proportional similiar

[upper karma]

yeah it's the same thing with different

lengths

[rare talon]

Same angle, with different length

[upper karma]

^

[rare talon]

Now

Your task is

To spot which triangle are similiar

[upper karma]

those 2 are similar

the whole triangle

and the one to the right

[rare talon]

Actually

The left triangle

Is similiar too

[upper karma]

yeah kind of

[rare talon]

Now name the point of the height of triangle ABC from B is D

To make me easier to explain

[upper karma]

ok

[rare talon]

Now you notice

Triangle ABD is similiar to triangle BCD

[upper karma]

yes

[rare talon]

With:
<BAD = <DBC
<ABD = <BCD
And <ACB = <CAB

Do you agree with this so far?

[upper karma]

what does < mean

[rare talon]

Angle

=tex \angle

[upper karma]

oh

[rare talon]

Does it work

Yeah

I intend to do that

[upper karma]

I mean

all of them have 90 degree angles

so they are all the same in that way

[rare talon]

Nah

Not all of them have 90 degree

[upper karma]

not the left one?

[rare talon]

Uhhhh

[eager saddle]

I was playing with my pen today and i noticed that the shadow of the spring makes a sine wave and rotating it 90 degrees makes a cosine.

[rare talon]

Sorry what do you mean?

[eager saddle]

Nifty

[rare talon]

Wait

I'll draw a pic for you

[upper karma]

if they are all the same angle wise

and one has a 90 degree angle

then all of them have one

correct?

[eager saddle]

So are sine and cosine 2d parts of a 3d helix?

[rare talon]

Well they are

If I don't misunderstand

But let's continue

You notice that:
<ABD = <BCD

And

[upper karma]

yes

[rare talon]

<BAD = <DBC

Okay?

They are not 90 degrees

Note that

But they have the same angle

So:
AD / BD = BD / CD , agree?

[upper karma]

I'm confused

[rare talon]

You notice that:
<ABD = <BCD
<BAD = <DBC

Do you understand this?

[upper karma]

yes

[rare talon]

Then usually for similiarity, I removed the second letter from the angle representating them , and make a fraction out of it (actually I take the side opposite of corresponding angle)

So AD / BD (from 1st condition) = BD / CD (from 2nd condition)

From triangle ABD and BCD

[upper karma]

I am so confused and I have no idea how this has anything to do with this

I know that they are similar

[rare talon]

Try to review similiarity

Actually

Do you have a book with explanation of similiarity?

I will adapt to that if you want

[upper karma]

I mean I only know stuff like

20/15 * 1,25

[rare talon]

I don't understand what you're writing

[upper karma]

let me type

20/15 = 1,25 then I can use that on a bigger triangle

to get out the length of it

[rare talon]

Oh

I can do that if you want

[upper karma]

yeah

[rare talon]

See triangle ABD and BCD

Again

Triangle BCD has all the sides in it

[upper karma]

yes

[rare talon]

What do you think triangle BCD larger than ABD is?

[upper karma]

12, 16 and 20 = lengths for BCD

I have no idea

[rare talon]

Triangle BCD has sides of 12,16, and 20

[upper karma]

yes

[rare talon]

Triangle ABD only has side as 12 right?

[upper karma]

oh yeah

yes

[rare talon]

Which is BD = 12

BD is opposite to angle BAD in triangle ABD

Agree?

[upper karma]

yes

[rare talon]

Note that angle BAD in triangle ABD = angle DBC in triangle BCD

Do you agree until this?

And the side opposite to angle DBC is CD

[upper karma]

yes

[rare talon]

Now you can see

What is BD / CD ?

BD is from triangle ABD and CD is from triangle BCD

[upper karma]

16/12?

wait

[rare talon]

The other way

[upper karma]

all of those

[rare talon]

BD / CD

[upper karma]

are similar

right?

[rare talon]

They are

[upper karma]

left and right are similar

give me one second then

[rare talon]

Since we proved that all the angles are the same

Okay good

[upper karma]

12/16

[rare talon]

Good

Which is 3/4

So triangle ABD is 3/4 "larger"(which is actually smaller) than triangle BCD

Do you agree until this?

Save this to your brain

[upper karma]

yes

[rare talon]

Now

[upper karma]

can I * 0,75

to all of the angles?

[rare talon]

*sides?

You can

But make sure you do it right

[upper karma]

yes

[rare talon]

Tell me what do you get for AD and AB

[upper karma]

so

[rare talon]

(I actually don't need AB, but it's a good exercise for you)

[upper karma]

I have the lengths 9, 12 and 15

[rare talon]

Which length is 9?

AD / BD / AB ?

[upper karma]

BD

[rare talon]

No

BD is 12

The problem state it

Clearly

[upper karma]

the left triangle's equivalent to right's BD

is 9

[rare talon]

Which side?

Is you say equivalent?

AD / AB ?

Or whatever length you want

[upper karma]

let me just tkae a picture of my notebook

[rare talon]

Okay

[rare talon]

Can you actually

Label the sides?

Which one is A , B , C ?

Or D

[upper karma]

1 sec then

[rare talon]

And

Label the angle

If possible

To show me which angle has the same angle with which angle from the other triangle

[upper karma]

to make you understand what I think

[rare talon]

Okay

They are good

But wait

I want you to fully understand things about similiarity

Check the picture

Now can you see which angles are the same?

[upper karma]

yes

because if you rotate it

the angles are the same

[rare talon]

Yeah , you can take it that way

Okay now so

You get that AD = 9

And DC = 16

So AC = 25 , am I right?

[upper karma]

yes

25*9

[rare talon]

No

Not that

AC = AD + DC

AC = 9 + 16

[upper karma]

25*12 I mean

/2

[rare talon]

Oh

Yeah

[upper karma]

gets me the area

[rare talon]

If you want to find the area

Okay

Good

[upper karma]

which was what I was looking for

[rare talon]

c:

The other way

You get AB = 15

And BC = 20

And you can get the area from that too

[upper karma]

/2 on both?

[rare talon]

What do you mean by both?

Well

I just give you an alternative way

[upper karma]

15*20/2

[rare talon]

Yeah

[upper karma]

yeah

[rare talon]

You can do that too

[upper karma]

then correct

[rare talon]

Since ABC is right triangle

Then we can treat BC as height, and AB as the base

(you can imagine it by rotating the triangle ABC to be able to see that)

Whichever way is fine, don't worry!

[upper karma]

can you help me with this one?

I want you to tell me why you do it like this instead of helping me with the answer

[rare talon]

Okay

Good attitude

[upper karma]

I know the formula

[rare talon]

But the discussion will be lenghty