What is the actual answer?
#geometry-and-trigonometry
wheat belfry
Help pls, with the process
lone copper
Help pls, with the process
that’s an exam right
wanton portal
Hey, I have made a video on a "Secret Proof for Pythagoras Theorem". I think you all will throughly enjoy it. Should I share it here?
upper karma
sin^2x = 2sinxcosx
is this correct?
somber coyoteBOT
Mosh
humble pulsar
what?
upper karma
sinx . cosx = m
what's [sinx - conx]
the absolute value
I hope that's what it's called in eng
sinx . cosx = m
what's [sinx - conx](absolute value)
humble pulsar
?????
humble pulsar
I have no clue what you just typed
upper karma
still?
humble pulsar
nope
upper karma
what's wrong with it
somber coyoteBOT
peaceGiant
agile geyser
help me with something in geometry guys it is posted in question 3
please please
@wanton portal
timber tapir
can someone explain to me how tan(1) = is pi/2
like if tan = sinx/cosx
so then sin1/cos1, (pi/2)/0
undefined
dark sparrow
what are you talking about???
sin(1) isn't pi/2 and cos(1) isn't 0...
and tan(1) isn't pi/2
,calc tan(1)
somber coyoteBOT
Result:
1.5574077246549
,calc pi/2
somber coyoteBOT
Result:
1.5707963267949
dark sparrow
no.
timber tapir
is that not how that works or what
dark sparrow
it's sin(pi/2) = 1
timber tapir
oh ok
dark sparrow
the input of sin, cos, tan, etc. is the angle
timber tapir
so how am i supposed to find the value of sin(1)
timber tapir
yes i am
dark sparrow
what?
dark sparrow
arctan(1) is way different
timber tapir
how so
dark sparrow
well for starters this is an inverse trig function
arctan(1) is the solution of tan(θ) = 1
(subject to some restrictions on θ to make sure there's no ambiguity)
timber tapir
just learned the unit circle 2 weeks ago
halfway thru calc 1
lol
but that's the least of my worries
wintry tundra
arctan is not cot
It basically is
wintry tundra
Ur just biased
wintry tundra
I'm being 100% /srs
weary drift
I'm being 100% /srs
really?
weary drift
@wintry tundra pls don't shitpost here
wintry tundra
<@!608508888953782307> pls don't shitpost here
Fine
dark sparrow
clever fable
hey, I was just wondering, why is there two thetas, in the triangle here
shouldn't it be pi/2 - theta, (for the point M)
clever fable
(or is it just a mistake they made?)
signal swallow
your angle is always taken from the x axis, not the y
signal swallow
so is the diagram wrong then?
no, it is not wrong
pi/2 is the y axis, so if you did pi/2-theta, you'd be taking an angle from the y axis
angles are taken from the x axis
opaque ibex
hey, seems like c and d inversely proportional in the f(x)= a.cos(cx + d) + b function but i don't get why
opaque ibex
In the graph, the coefficient c is setting the amplitude and d is setting the translation, but the translations of d are affected by the coefficient c in the coordinate system. In a inversed proportial-->
cos(2x + 1) + 6 and cos(x + 1)
cos(2x + 2) + 6 and cos(x + 2)
cos(2x + 1) to cos(2x + 2) delayed 1/2 unit but why not 1 unit is what i'm asking
pure cape
tbh, d is not the translation in the x axis
the translation in the x-axis is d/c
which way?, that depends on the sign
and c and d are not inversely proportional to each other
pure cape
cos(2x + 1) to cos(2x + 2) delayed 1/2 unit but why not 1 unit is what i'm askin...
and that above, just answered why
pure cape
what have you tried?
you have used the wrong sides
you have to understand that the general volume formula for prisms is B*H where B represents the area of the base and H represents the height of prism extending from the base
now from that formula, we achieve the formula bhl/2 for triangular prism
where b is one of the side of the triangle base, h is the length of the altitude to that side aka the height of the triangle, and l is the length from the base
lemme just name some stuff first so its not vaguely describe
alright, so in this case, what you need is the area of the triangle
which is basically AB*BC/2, since this is a right triangle at B
well, one could argue that you can draw an altitude from B to AC
and yes, i agree, you can, but why make it hard when you have the easy way
so you just need BC, which is pretty straight-foward to find
so to conclude:
Base=AB (or BC)
Height=BC (or AB)
Length=BB'
opaque ibex
so finding BC using pythagoras theorem was a coincidence then I guess.
abstract cosmos
That's truee
pure cape
i never said it was false, and i have explained where and what the formula means
again i have listed out what should do
it seems like you are not reading any of my message
opaque ibex
how to find BC?
pure cape
pythagorean
or if you want to do it the hard way and find the altitude from B to AC then also pythag
opaque ibex
yeah that works but it says nothing about a right angle or B to AC will be a median
opaque ibex
so if it's not 90 degrees we can not use pythagoras
pure cape
yes
opaque ibex
so whats BC
pure cape
if the information right triangular prism is not given then not enough information is given
use pythag?
oh i just realised i read the question wrong
yeah, then not enough information is given then, we would need at least an angle or another side
void agate
How to find vector PQ in terms of vector AD and vector AB
i have found :
5/7 AD - 4/5 AB
Is this correct?
oblique hatch
So, your angle is theta, so that's the angle you'll be working off for each triangle. For example, on Triangle 1, opposite theta is 8 and adjacent is 6. What angle uses Opposite and Adjacent?
sand jolt
I'm having some trouble here with this first problem. I'm not the best at logic stuff like this. (Also the Coliseum window is just a room with a long window on one end of it.)
I have a graph too if it helps any.
Like you're not standing outside since you're at y<0, you're above ground since z>0 and you're just standing 2 meters to the right for the x. Right?
humble pulsar
It's a right triangle, so one side is the diameter
sand jolt
I'm having some trouble here with this first problem. I'm not the best at logic ...
<@&286206848099549185>
chrome trout
Hey ya'll 🙂
I'm in a trig class, I like trig, im happy to do the work. I have been struggling understanding Trig concepts and I think it is because it is all lacking human interaction. Anyone care to review some trig topics with me? I have a test in 10 days = /
void moat
I'm in a trig class, I like trig, im happy to do the work. I have been strugglin...
do you know how to solve trig equations
also, do you know how to solve trigonometric proofs
and finally trig word problems
know all your double angle formulas, and trig identities
cirumscribed means that the triangle is in the circle
pale onyx
hOW
neat stag
are the designs made on table or cloth? but the cloth is hexagonal
lone copper
qn is not clear enough but it seems like the design is the part on the table that the hexagon doesn’t cover
neat stag
qn is not clear enough but it seems like the design is the part on the table tha...
like the area of table under hexagonal cover isn't designed?
lone copper
i think no that’s why the hexagon is there to cover the rest of the table
neat stag
dude this question
lone copper
this qn is very unclear
neat stag
i Haven't seen such table tbh
hollow raft
@hoary rover what in the fuck
What math are you in?
Like is this a high school geo?
winter pumice
Wow, I'm not familiar with the [ABCD] notation. Is that an area or perimeter?
hoary rover
area
nocturne remnant
Just assume ABCD is a square 
winter pumice
there's so much here it's like intending to throw students off
it's an exercise in willpower
idk actually maybe they really want you to remember all the notation
stray schooner
Do you still need help? I got an answer 30.
void agate
Hello, could someone help me solve this problem:
In quadrilateral ABCD, P and Q, are the midpoints of opposite sides AB and CD. The point M is the midpoint of BCD and N is the midpoint of PQ. Prove that the points A M and N lie on the same line.
I think it should look something like this:
nocturne remnant
C does not always have to lie on the line ANM
but in any case, ||Menelaus’ on BPQ should work||
shut finch
Im late to the party ik
cerulean nest
I have a graph too if it helps any.
Ahaha you use geogebra
Great app, I use it all the time
sand jolt
Ahaha you use geogebra
Lol yeah
silent plank
the work is messy but you'd need to determine the area of the base triangle (which is implied to be a right triangle)
which can be achieved by first determining the length of the unmarked leg using pythagoras
shut finch
what d you find confusing i will try and help
shut finch
OK you have to work out the sides of the triangle and use them in your formula
silent plank
note that 10ft is the length of the hypotenuse and is NOT the height of the triangle
shut finch
Exactly Romanov
silent plank
the work is messy but you'd need to determine the area of the base triangle (which is implied to be a right triangle)
which can be achieved by first determining the length of the unmarked leg using pythagoras
neat stag
Four circles are drawn, with the centre on the verticies of square... Each circle touches the boundaries of two other circles out of four.
How are their areas equal??
silent plank
doesn't really matter than much
if you consider 6 to be your base, 8 will be the height
and vice versa (you can consider 8 to be the base and 6 to be the height)
silent plank
either way you'll have 1/2 * 6 * 8 or 1/2 * 8 * 6 which has the same value
(which is implied to be a right triangle)
which can be achieved by first determining the length of the unmarked leg using pythagoras
winter pumice
Four circles are drawn, with the centre on the verticies of square... Each circl...
I'm not sure that they have to have the same area to satisfy these conditions...
winter pumice
Neat. So blue circles (every other) can vary in s/2 <= r_blue <= s/sqrt(2), as s/2 <= r_red <= s(1-1/sqrt(2)) ?
Doesn't that disprove the premise that they are all the same?
*same area
dusky surge
Yea I guess, since the radius of the circles varies...
Varies from (√√2)/2 to 2-(√√2)/2
Although I just did (√√2)/2 to 1 as demonstration
And of course you can prove it algebraically
neat stag
And of course you can prove it algebraically
i mean can you put it in simple terms
can I make 4 circles, centred on verticies of a square and each touching the other two, which are not equal in radius and, hence, area?
so whats the justification or reason for these circles having equal area
winter pumice
i mean can you put it in simple terms
They don't all have equal area. Please state the problem exactly as it was given to you. If you have, the claim that all circles have equal area isn't always true under the required conditions (that all circles are tangent to two circles).
however, circles in opposite corners do seem to be guaranteed to have equal area.
Area depends entirely on the radius (A=pi*r^2).
neat stag
They don't all have equal area. Please state the problem exactly as it was given...
Question: the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region (area in the square not occupied by any circle)
dusky surge
Question: `the figure, ABCD is a square of side 14 cm. With centres A, B, C and ...
Is there a picture we can refer to?
neat stag
Is there a picture we can refer to?
blue
they reject that only opposite ones can be equal
and take the second one where all r equal
dusky surge
From my example, I used square of side length 2, then the "shaded area" will be
2²-(r²+(2-r)²)π/2
Where r in a real number between 2^(¼)/2 and 2-(2^(¼)/2)
dusky surge
they reject that only opposite ones can be equal
If it's just the case, then it's not hard.
Since it's some fixed graph and not moving ones.
winter pumice
Question: `the figure, ABCD is a square of side 14 cm. With centres A, B, C and ...
it's not animated I guess because there are two parameters...dumb
@dusky surge you know how to make the second slider animate?
does the side length need to be constant? that would be a shame
dusky surge
Haha I don't know how to make 2 slider animated
proven adder
yo can someone hop on call with me i got a problem solving arccsc(sec2pi/3)
winter pumice
ooh! me
dusky surge
Dunno lol, I don't really know how to use Desmos
winter pumice
Question: `the figure, ABCD is a square of side 14 cm. With centres A, B, C and ...
ok, well that's less interesting.
you have 4 quarters of equal-sized circles being removed from the square region.
The radius of each circle is half of the side length.
So the radius of each circle is 7.
The Area of each circle is what?
If you have 4 quarters, that's a whole circle being subtracted from the square's area, right?
storm portal
In 2D space, yes
In 3D space, not necessarily. It can be mapped onto itself. It depends on where the line is rotated about
molten delta
So we know that the formula for the circumference (the base of a cylinder is a circle, and the perimeter of a circle is the circumference) is (2)(pi)(r)
Yea?
Yes we can also use that formula as well
And since we are using pi = 3, let’s change the formula to (2)(3)(r),
So you got it?
Or…
Ok so
Right we are going to get there in a moment
So the formula in this case is C = (2)(3)(r)
C being the circumference
And since we’ve already been the circumference, which is 24, we plug it in
24 = (2)(3)(r)
Since the circumference is the perimeter of a circle
And a cylinder’s base is always a circle
Yep
24 = (2)(3)(r)
It’s a simple equation from there
Just solve for r
Yea for sure
It’s just we need to convert the d into r later on
So the formula now is C = (3)(d)
Yep
Then we just solve for d
Yep
Perfect
I mean we could use the process of elimination here
Or we can check mathematically
F would be 25/12
2.0833
Yep
Since we rounded down pi from before
So they rounded up the final answer
Hmm I mean that just depends on the question
If it asks you to round to the nearest tenth then that’s what you should do
I feel like the book your using shouldn’t have done that
It’ll likely be more specific in exams and tests
No problem have a good night/day
neat stag
ok, well that's less interesting.
you have 4 quarters of equal-sized circles bei...
the radius isn't always half the side length
neat stag
https://www.desmos.com/calculator/ojajaxehgd
as per this
dusky surge
the radius isn't always half the side length
Then the area is not fixed.
neat stag
Then the area is not fixed.
so then we can't find the shaded area?
dusky surge
We can.
dusky surge
From my example, I used square of side length 2, then the "shaded area" will be
...
This
Or for your case
14²-(r²+(14-r)²)π/2
Where r is the changing radius or one of the pair of circles.
winter pumice
the radius isn't always half the side length
His problem statement implies it is. We wrestled with that part already, they just want the case shown in the picture their teacher gave
neat stag
well i guess I'll have to stick to the radius being equal case
it makes it easy and i get the answer
and marks lol
dusky surge
Lol, I think that's what they want
mystic karma
blue
shouldn't the answer be 196 - 4 [(Pi*7^2)/4] ?
so basically the answer should be 42 Cm^2
winter pumice
shouldn't the answer be 196 - 4 [(Pi*7^2)/4] ?
yes
unless the radii are allowed to vary
but your picture and description pointing to the picture implied they weren't supposed to
mystic karma
but your picture and description pointing to the picture implied they weren't su...
i didnt share any picture , you got me confused with someone else
winter pumice
yeah, I was just telling my boi @dusky surge about that
dusky surge
Kek
winter pumice
But I wasn't refering to a person
just the picture
was this your problem?
thought it was theirs
mystic karma
😂 you guys got confused a simple question with a complex one
winter pumice
@neat stag
mystic karma
but how do you solve the complex type ?
winter pumice
complex meaning....?
mystic karma
in which the radii may vary
dusky surge
You can scroll all the way up~
winter pumice
the complicated one?
mystic karma
the complicated one?
yes
mystic karma
You can scroll all the way up~
ok
neat stag
<@!753961119613321306>
yes
winter pumice
there are two pairs with radii that have one value, and two pairs with a complimentary value
the sum of the radii must equal the side length of the square
so you can just subtract half-times the area for both cases from the area of the square whose area is (2s)^2 (if you're going by how I put it in Desmos)
mystic karma
ok thanks
winter pumice
let $s$ be half the side length of a square and choose r such that $$r \ni [s(2-\sqrt{2}),s\sqrt{2}]$$
un-simplified:
$$A = (2s)^2 - \frac{1}{2}\pi r^2 - \frac{1}{2}\pi(2s-r)^2$$
somber coyoteBOT
Disorganized
winter pumice
$$A(s,r) = 4s^2 -2s\pi(r - s)$$
upper karma
sinx/1+cosx + 1/tan
how do I solve this
somber coyoteBOT
Disorganized
mystic karma
how do I solve this
you rationalise
winter pumice
**Disorganized**
...maybe, don't trust me this early, or that late
upper karma
how
mystic karma
so basically multiply sinx/cosx+1 with cosx-1
upper karma
wait
winter pumice
@upper karma please confirm what the expression in the denominator is
upper karma
I don't get what you said
mystic karma
i also have a doubt lemme share it
2sec^2 theta - sec^4 theta - 2cos e * c ^ 2 * theta + cos e * c ^ 4 * theta = cot^4 theta - tan^4 i
winter pumice
is your expression
mystic karma
2sec^2 theta - sec^4 theta - 2cos e * c ^ 2 * theta + cos e * c ^ 4 * theta = co...
this
upper karma
yes
winter pumice
$\frac{\sin{x}}{1+\cos{x}} + \frac{1}{\tan{x}}$?
somber coyoteBOT
Disorganized
upper karma
they want me to make it the simpliest
mystic karma
2sec^2 theta - sec^4 theta - 2cos e * c ^ 2 * theta + cos e * c ^ 4 * theta = co...
solve this also , its a solved example in my textbook yet i dont understand the final steps
winter pumice
just write $\tan{x}$ as $\frac{\sin{x}}{\cos{x}}$ to start
somber coyoteBOT
Disorganized
upper karma
did we do
1/tan > 1/sin/cos or cos/sin ?
winter pumice
(cosx)(sinx)/[(cosx)(1+cosx)] + (1+cosx)/[(cosx)(1+cosx)]
winter pumice
did we do
1/tan > 1/sin/cos or cos/sin ?
tanx = sinx/cosx
winter pumice
so if it's on the bottom of 1, that is, 1/tanx, that's 1/(sinx/cosx) = cosx/sinx
upper karma
1/tan > 1/sin/cos or cos/sin ?
winter pumice
2sec^2 theta - sec^4 theta - 2cos e * c ^ 2 * theta + cos e * c ^ 4 * theta = co...
I dont have a memory for this
dusky surge
Sexy
winter pumice
Are you sure you typed that as it was given to you?
mystic karma
I dont have a memory for this
meaning ?
mystic karma
Are you sure you typed that as it was given to you?
yea , i used google lens and then pasted here
winter pumice
meaning ?
it has some typos and it's not clear what goes with what
and I know typos very well
maybe take a screenshot
or use LaTeX
mystic karma
or use LaTeX
how do i use it
radiant girderBOT
dusky surge
Hmmm... Can it be shown like... A picture? All I can see is a downloadable file
mystic karma
how do i do that
dusky surge
Use the camera?
mystic karma
i was on my laptop
dusky surge
Lol the Google lens got cos(e*c)
mystic karma
yea XD
dusky surge
Okay, on my first thought
I think you can group
First 2 terms together and of course, the last 2 terms together
mystic karma
@dusky surge if you were bri'sh would you be known as crumpet
mystic karma
Okay, on my first thought
I think you can group
First 2 terms together and of co...
how do i do that ?
mystic karma
how do i do that ?
ok i get it now
dusky surge
Like since they have a factor
sec²θ in common
mystic karma
Okay, on my first thought
I think you can group
First 2 terms together and of co...
that step is given in the solved example but then they add 1 to both the groups for some reason
mystic karma
Like since they have a factor
sec²θ in common
yea
dusky surge
that step is given in the solved example but then they add 1 to both the groups ...
Wow that's a nice move
upper karma
What did I do wrong?
mystic karma
wait
dusky surge
this is the solution
This is so painful
upper karma
skipping steps, making mistakes
like whaat
dusky surge
I'll come back later
winter pumice
like whaat
RHS in step 4 is wrong
upper karma
Ok
winter pumice
also your notation for multiplying top and bottom by something, I've never seen that before. You followed through with it on the left but messed it up on the right in step 4. Maybe just write it out
"say what you mean"
pretend you are teaching someone else what you are doing
mystic karma
@winter pumice xplain this please
winter pumice
this solution is also correct right ?
that was brilliant until the last step
winter pumice
they are separating the sec(x) and csc(x) parts and anticipating a polynomial perfect square trinomial form
both polys are of the form x^2 - 2x + 1
they were just missing the 1's
mystic karma
oh now i see
winter pumice
the 1's they appended will cancel
mystic karma
so basically of the form ax^2 + bx + c
winter pumice
more specific then that, but yeah
mystic karma
this was very helpful thanks
winter pumice
they were going for a perfect square trinomial
analogous to x^2 - 2x + 1 = (x - 1)^2
if you let x=t^2, you can see it still works
t^4 - 2(t^2) + 1 = (t^2 - 1)^2
winter pumice
you can also do it with trig function substitutions. They had x= [csc(theta)]^2 in the first one and x = [sec(theta)]^2 in the second
winter pumice
u r a genius or what]
no, I learned this in highschool just like you
but I've been doing this for about 15 years longer
mystic karma
oh much appreciated man
obsidian marsh
Is this correct? It feels wrong for the area to go to infinity when the angle goes to zero.
quiet blade
Is this correct? It feels wrong for the area to go to infinity when the angle go...
How did you derive A = br/2? That isn't the area of the arc sector
Oh, got it. It can be obtained by multiplying the very first equality with r/2.
When alpha is really small, you have this really long radius*. As alpha tends to zero, r will tend to infinity. So the area might be pretty large.
light wagon
what's trigonometry?
light solar
mystic karma
what's trigonometry?
if sin0= 1 then its not trignometry
winter pumice
- what are the identities you wrote on the RHS
stray schooner
cotangent 2a
2 cotangent a
winter pumice
cotangent 2a
2 cotangent a
did your teacher tell you to write cotangent like that?
Because that is the second really unusual writing of a trig identity I have seen in 24 hours
stray schooner
Yes, we have these in all textbooks
winter pumice
are you sure
somber coyoteBOT
Disorganized
stray schooner
Yes
somber coyoteBOT
Disorganized
stray schooner
We have Russian textbooks, we write tangent and cotangent - tg, ctg
winter pumice
We have Russian textbooks, we write tangent and cotangent - tg, ctg
Ah!
ok
Ok so you remember how there are laws of exponents and logarithms?
there's a really long version of that for trigonometric identities.
"trigonometric identities". Usually if you search that, you get a short list. But sometimes you get longer lists.
First one might require the product formulas for sine and cosine
let me look at it a min
*let me try it
winter pumice
Solved part A by reorganizing top and bottom, using sum-to-product identities.
Cosine is an even function so it didnt matter how I ordered the first 2 terms in top and bottom when I used the identities.
silent plank
how much trig have you learned?
have you learned about sine,cosine,tan yet?
ok
which trig function do you think would be most useful here?
no
relative to the 67° angle,
can you identify the position of the side x
no
shut finch
Romanov can I also help please
silent plank
the side x isn't touching the 67° angle
shut finch
Yes sir
silent plank
yes x is opposite the 67° angle
can you identify the position of the 2m side relative to the 67° angle
yes
so you have the angle, the opp and the adj
yes
shut finch
Yeah you would do tan(67) /2 ?
Yh
Yeah I’m pretty sure you do tan(67)/2
Since toa
You put it in brackets on the calculator
mystic karma
how do you calulate trignometric thetas for various angles
for now i just memorised trig ratios for 0, 30 , 60 and 90
for example how do i calculate the value of tan67 deg ?
signal swallow
put it in a calculator
mystic karma
i could but when i don't have calculator sitting around ?
signal swallow
you could do a taylor expansion if you really want
wouldn't be fun though
and it wouldn't be exact
signal swallow
yes
signal swallow
depends
but normally if i'm not allowed a calculator on an exam, the exam doesn't need a calculator
mystic karma
i see
signal swallow
they use easy numbers that i can work by hand, the exam is more of a test of concept
mystic karma
understandable
so what i understand up until this point , there are like 5 strategies to prove trignometric identities -
1direct identity use
2 lcm
3 convert into sin and cos
4 rationalise ( only works for sin and cos)
5 using polynomial identities
can anything be added or corrected ?
signal swallow
haven't proved trig identities in a bit, but seems about right
factoring is also useful
mystic karma
whats factoring ?
signal swallow
$x^2-4=(x+2)(x-2)$
somber coyoteBOT
a disappointing son
mystic karma
oh yes
signal swallow
$x^3-3x^2+2x=x(x^2-3x+2)$
somber coyoteBOT
a disappointing son
signal swallow
👍
silent plank
what's your issue with this?
split agate
i have serveral questions i need help with tho
worn sigil
Can someone explain to me why that is the period
<@&286206848099549185>
Isn't finding the period 2pi/b
and not pi/b
why is that saying pi/(1/2)
pls ping me
north cosmos
neon grotto
Hey! Can someone help me with this: A square ABCD is given. Points P and Q are taken on the sides AB and BC such that the BP is equal to the BQ.BH is the height of the triangle BPC. Find the angle DHQ
winter pumice
Hey! Can someone help me with this: A square ABCD is given. Points P and Q are t...
Unclear where H is.
neon grotto
h is on CP
winter pumice
Please screenshot the problem so we get the original statement and any figures it refers to
winter pumice
Put it up here so everyone can refer to it
...until it gets buried over the next few days
Since Discord 'forums' move quick
neon grotto
but the problem is not in english
neon grotto
Bulgarian
winter pumice
Can you copy-paste the text?
neon grotto
no
neon grotto
can you help me with the problem
winter pumice
I probably have to translate it first, but maybe I dont, depends
winter pumice
You can translate it. Replace every instance of a string of letters with the Bulgarian word for something else. Pie, maybe (the pastry, not the number), run it through Google Translate, and then bacl-substitute the strings of letters
Please put the screenshit up though so we can compare it
Ok, how about this: is H an intersection point?
Can you just draw the figure
Because the description sounds good until H comes out of nowhere
@neon grotto
neon grotto
give me a second
i have written on the paper sorry
@winter pumice is it good now?
probably, it makes the last two triangles make sense
winter pumice
@neon grotto I suspect that tri HBC is similar to tri HCD
neon grotto
i dont think so
By the way dont continue torturing yourself i think i solved the problem!
winter pumice
Oh good
neon grotto
Thank you for your time and help
winter pumice
What did you get?
neon grotto
angle dhq is 90 degrees
winter pumice
Wow, really?
winter pumice
Ok, bye
neon grotto
have a wonderful day
upper karma
fish
upper karma
Maybe try draw it i cant help you now cuz im in a car
nocturne remnant
can someone please help me with this question
Oops I mixed up B and A
But notice that the 2 similar triangles must both contain theta and 180-theta
Which is not normally possible (as the remaining angle would be 0)
Unless…UNLESS!
theta = 180-theta
So that gives theta = 90
copper jackal
I dont know where to post this but can somoene please tell me how to get the bisector of a line pls ping if you want to send a solution cause i have a maths test tomorrow
uncut summit
Thanks elon
upper merlin
upper karma
imma have to review this
dusky surge
cos means cosine.
winter pumice
The area of trapezium ACFD is the sum of the areas of tri ACD and tri CDF
We know that AC and DF are parallel because the planes they are in are parallel, and they are hypotenuses of similar triangles (tri ABC and tri DEF by SAS-similarity).
This means trapezium ACFD is a plane figure
This is why we know we can add the triangle areas we picked: it wont matter how we cut the trapezium up because all its points are coplanar
(Somebody pop in here and tell me every "trapezium" is a plane figure)
Anyway. Should be easy for you to take it from there, calculating the areas of the triangles in the trapezium, because they are right triangles as well (justify this!) and you can get their side lengths easily. Let me know if you get stuck.
winter pumice
I have not written anything down to figure out this will work, try again following the steps I described
winter pumice
Ok, what are the side lengths of ABC and DEF
upper karma
I cant figure it out
winter pumice
^ you dont know how to solve a right triangle given 2 of the 3 sides?
upper karma
½bxh?
winter pumice
Thats an area formula
upper karma
oh
winter pumice
What formula gives you the sides of a right triangle
upper karma
no clue
winter pumice
Get a clue then: Google it
upper karma
pythagoureous ?
winter pumice
Yes
winter pumice
Yes
You only need the 2 formulas you listed to solve the problem
Plus a little logic
winter pumice
Ok
(DE)^2 + (EF)^2 = (DF)^2
So then
(6)^2 + (6)^2 = (DF)^2
36 + 36 = (DF)^2
72 = (DF)^2
what next?
upper karma
find out Df?
winter pumice
How
upper karma
idk
winter pumice
What is the inverse of squaring an unknown?
*what is the inverse of squaring something
upper karma
I not even onto pythagorous yet
somber coyoteBOT
Shen
storm portal
Just be careful with signs.
winter pumice
Thanks Shen, for confusing me with OP
But I guess its good the definition is here now
winter pumice
Nah, its good, it's pretty
storm portal
Thanks 😅
Wasn't sure if I should delete it because you're trying to get him to think about it
Respect 👊
winter pumice
I not even onto pythagorous yet
Did your teacher give you this problem?
winter pumice
Did you cover the distance formula yet?
upper karma
yes
winter pumice
That's a form of Pythagorus. You guys have covered this
It would have come up by now
upper karma
oh then no
winter pumice
You already listed the formula for Pythagoras, are you saying you don't know how and have never taken a square root?
winter pumice
Taking a square root is an inverse operation to squaring
If you take the square root of x^2, you get x again
upper karma
yes ik
winter pumice
...because they are inverse-operations
winter pumice
This is the way I like to explain it:
x | x^2
1 | 1
2 | 4
3 | 9
4 | 16
5 | 25
6 | 36
When we square the input x, we are doing x*x
So we go from the left column entry to the right column entry
$\sqrt{x}$
somber coyoteBOT
Disorganized
winter pumice
is a function that can take us from an entry in the right column BACK to the entry in the left
It tells you what number got squared
So $\sqrt{36}=\sqrt{6^2}=6$
somber coyoteBOT
Disorganized
winter pumice
@upper karma understand?
storm portal
so sqr of 81 = 9
Correct
winter pumice
The list above is easy b.c. everything on the right is a PERFECT SQUARE: some number times itself.
But getting back to above...
winter pumice
72 = (DF)^2
72 is not a perfect square because it isn't a number times itself
But 72 = 36*2
and 36 IS a perfect square
So we do
$\sqrt{72}=\sqrt{36\cdot2}=\sqrt{36}\sqrt{2}=6\sqrt{2}$
somber coyoteBOT
Disorganized
winter pumice
and that's just fine for us.
BTW, we know we can do this from Laws of Roots (Exponents), which you might not have seen in detail yet
Don't worry about it
Splitting the root like this under multiplication is allowed.
You need to get the hypotenuse of the triangle on the front face now. Try it
AC
vital fractal
upper karma
do you still need help on this?
rare sierra
dense belfry
if i have the relation $4\sin\left(x\right)<3\cos\left(x\right)$ is there a way that i can rewrite this as a specific inequality for $x$?
somber coyoteBOT
aquaticape
upper merlin
if i have the relation $4\sin\left(x\right)<3\cos\left(x\right)$ is there a way ...
represent in tan?
dense belfry
represent in tan?
that what I thought but somehow that messes up all the intervals where the inequality should be true
for example, 2 radians works for tan but not for the original inequality
upper merlin
that what I thought but somehow that messes up all the intervals where the inequ...
the problem occurs when cos(x) is negative
upper merlin
was anyone able to solve this?
i ran a script which generated random points just to make sure there wasn't any printing mistake in the question
it seems to show that the ques is correct
mystic karma
was anyone able to solve this?
use the google lens homework section it will scan the problrm and give you a solution online , it works for most questions might have a doubt in
upper merlin
use the google lens homework section it will scan the problrm and give you a sol...
i tried searching on google, found only 1 solution which assumed M to be on the median/altitude
mystic karma
mystic karma
https://youtu.be/3ET7YGAR26I
@upper merlin see if this works
upper merlin
https://youtu.be/3ET7YGAR26I
i was talking about this itself, it assumes M to be on the altitude but its not mentioned in the question
mystic karma
oh ok
chrome laurel
hey
If x is an angle in the third quadrant and csc(x)= -13/5, find the value for 3sin^2(x)-2cos(x) as a fraction with no decimals.
can anyone help me?
dark sparrow
have you made any progress so far?
chrome laurel
well i know how to solve for x kind of, but how does it being in the third quadrant affect it
dark sparrow
you don't need the value of x itself here
chrome laurel
huh?
dark sparrow
the fact that x is in Q3 tells you what signs sin(x) and cos(x) are
hopefully you are able to calculate sin(x) directly from what's given to you, and then cos(x) from that and the Pythagorean identity?
lime crownBOT
dark sparrow
...
upper karma
Same angle, equal the terms and voila
dark sparrow
||sounds like someone is here in bad faith||
light arch
Bro this dude be cheating in his 3rd grade homework
silk charm
light arch
Who cares lol
dark sparrow
is "rock solid logic (sarcastic)" actually an expression in english 
dark sparrow
in russian we say железная логика which means "iron logic"
upper karma
Rock solid is a thing I think
light arch
@upper karma not in a sarcastic way
upper karma
Wdym? Just say it sarcastically duh
light arch
dark sparrow
light arch
@dark sparrow what's your major if I may ask, i see you a lot here, you know your stuff
dark sparrow
some flavor of applied math
i'm doing masters rn
i guess less applied than what i did in bachelors
...hello
do i know you?
<@&268886789983436800>
silk charm
ah someone beat me to it
dark sparrow
what even was this
viral jolt
silk charm
it might be an alt of the person i just banned for cheating
given what they said
these people
dark sparrow
im not even like. a Furry furry
worthy igloo
it's the internet ma its not that deep
nocturne remnant
They are required to use cosine law
And the similarity is probably a side effect of trying to make everything whole numbers
silent plank
are they actually required?
sonic nova
you would use it to find an angle C
silent plank
or did they assume that law would be most useful here
sonic nova
then sub it into cosine rule again to figure out the side
upper karma
cosine law is complete overkill for this
Is there like a geneva convention on the brutality of problem solving methods?
weary drift
assuming so is pedagogically beneficial
winter pumice
Can someone explain to me what i did wrong? Im confused
If you DILATE a point M(a,b) from the origin by a factor of 3, then the distance from (0,0) to (a',b') is scaling by a factor of 3
The ray from O to M is stretched out by 3. Where is this new point?
Draw it
stuck iris
Is it possible to express $\cos(a - b) = f(a) \times g(b)$ ?
somber coyoteBOT
J J
stuck iris
I've tried factorising it in numerous ways but i feel like its not possible? is this a provable thing?
dark sparrow
it most definitely isn't possible.
if it were, then the solution set of cos(a-b)=0 would consist purely of horizontal and vertical straight lines on the coordinate plane
which it does not
@stuck iris
upper karma
also keeping a=b implies f(a)g(a)=1 or g(a)=1/f(a) which is identically true
which means we could write cos(a-b)=f(a)/f(b) and since cosine can be zero it implies either f can be zero or infinite
both of which implies cosine should not be bounded but we know cosine is very well bounded so hence this assumption is false
though yea this is a pretty weak argument but pretty good insight regarding the problem
stuck iris
if it were, then the solution set of cos(a-b)=0 would consist purely of horizont...
RIGHT i didnt think of this, thanks
nocturne remnant
Try to find the exact positions of the vertices of the triangle by using the info already given
echo solstice
AB=5 DE=5x2 l think (not sure)
dusky surge
Does it help if you prove that triangle ABC is similar to triangle A'B'C'?
Since you will find some angles in same segment if the circumcircle of A'B'C' is constructed
kindred coral
How would I be able to determine what single shape corresponds to the requirement if several shapes fit the description?
loud shard
can someone please help me with this?
bleak fox
can someone please help me with this?
Take the given equation and square both sides
upper merlin
Does it help if you prove that triangle ABC is similar to triangle A'B'C'?
ye thanks got it.
cursive vapor
hi
i have an exercise
given the are of circular sector= 16pi find the perimeter of this sector
just tag me please
hollow plume
- Since angles 4 and 7 make a straight line (180 degrees), the angles are supplementary.
- Angle 12 is 90 degrees, you can assume that the 2 other supplements are equal to 90 degrees. Hence they’re complementary. (Angles 10, 12, and 3) make a 180 degree straight line.
- Angles 4 and 5 are not adjacent to each other to make a straight line. They’re not a linear pair.
- Angles 2 and 10 are not directly opposite from each other. Also, angle 10 is less than 90 degrees and angle 2 is greater than 90 degrees.
- Angles 1 and 2 are adjacent to each other and make a straight line. They are a linear pair.
- Angles 7 and 9 are directly opposite from each other. Since angles 4 and 7 make 180 degrees and angles 9 and 8 also make 180 degrees, all 4 angles make 360 degrees. Again, angles 7 and 9 are directly opposite. Hence, they’re vertical angles
hollow plume
@stray pollen ^^
stray pollen
Thank you so much
signal swallow
15. Since angles 4 and 7 make a straight line (180 degrees), the angles are supp...
don't give people answers
pastel pewter
hi , can somebody help me please, how to get sine and cosine value of an angle with unit circle? for example sine 1 degrees, sine 27,7 degrees, or sine 89 degrees
lapis pine
quiet blade
Not enough information to determine
raw tiger
hi , can somebody help me please, how to get sine and cosine value of an angle w...
Just use the inverse trig functions 🙂
pastel pewter
Just use the inverse trig functions 🙂
can u give some example, i can't imagine how it works
upper karma
Hi how can I solve this
I know the answer by the way... just want to know why
Ignore what is there
winged falcon
i don't even know how to being with this one tbh
upper karma
I know the answer by the way... just want to know why
<@&286206848099549185>
quiet blade
Hi how can I solve this
What did you get for RQ?
upper karma
10 cm
quiet blade
Great, and the hypotenuse
upper karma
What did you get for `RQ`?
PR is 10 root 5
quiet blade
What is the area of the triangle?
Actually don't calculate it
just tell me with words, what times what
quiet blade
Aha, so you have PQ*QR/2 = PR*(height you are looking for)/2
and you have all other values except the height
upper karma
Aha, so you have `PQ*QR/2 = PR*(height you are looking for)/2`
No I am looking for the altitude of hypotenuse... height can be said as 20 if the baseline is QR
quiet blade
H right, so the area of the triangle is also base = PR times height H, divided by 2
upper karma
Yes
quiet blade
Then use this PQ*QR/2 = PR*(height you are looking for)/2
upper karma
Wait u can look at it from diffrence angles and get it ?
quiet blade
Yes, the formula is the base of a triangle, multiplied with the altitude that falls on that base, divided by 2
There are three bases, therefore 3 instances of the same formula
upper karma
I think the question is more directed toward the altitude of hypotenuse theorm that I know nothing about
But... u like hacked the question
That's so cool
quiet blade
I don't exactly know what that theorem is, but most people would solve it as I did
And that theorem is most likely a consequence of that we went over
It probably is PQ*QR / PR is the altitude
quiet blade
Then use this `PQ*QR/2 = PR*(height you are looking for)/2`
which is immediate from ^
delicate rampart
livid moss
Are you asking if that's true? Then yes
clear sun
urban bolt
How to solve that 2 problems
quiet marlin
urban bolt
@quiet marlin
urban bolt
Ohh, I get how to solve that, I was just need to take arccosine of both sides, and there is ultraeasy inequality $$sin(x-\pi/6) < 5\pi/18$$
somber coyoteBOT
qcanser
thorny forge
what does it mean by (b)
silent plank
start by drawing two tangent lines from the person to the tank
thorny forge
yes oki i just dont get how ur meant to find the cicumference visible to person
silent plank
the part that's visible is the arc between the two points of tangency
thorny forge
hmm
silent plank
consider properties of radii, tangents and right triangle trig
thorny forge
i get it abit but how do i find the angle
silent plank
right triangle trig
thorny forge
i dont get that
molten rivet
what does it mean by (b)
Red portion is visible
thorny forge
Red portion is visible
aaaa okai thank you,, how do i find that angle from the centre
to the edge of the red portion
narrow sleet
If you draw a line from the centre to the edge, you form a right angled triangle
Ur hypothenuse is 80
One of your side is 20
so....
thorny forge
mm im having trouble visualising it do u mind drawing it its okai if not T-T
ohhh wait
im supposed to find this angle using that and use that angle to find the length seen
thorny forge
<@!569352801940013078>
ooooo thank you!! i got the answer
upper karma
please help in this ques..
mossy pike
please help in this ques..
Well, A=60 degrees, B=80 degrees, C =120 degrees, D=100 degrees; A,B,C,D belong to a circle. Maybe these facts will help.
Try to think about angles which are 60 degrees. Perhaps their sine or cosine will come useful.
Also you can use the intersection of AC and BD, call it O, and find AB = AC = AO + OC.
It seems like the answer is 6/cos(80)
approx 35
upper karma
Question:
If three angles of a quadrilateral are equal then it is a parallelogram. (true/false)
silent plank
wdym by 3 angles of a parallel
upper karma
It seems like the answer is 6/cos(80)
yes but not in options
winter pumice
Please help in this problems
You didn't draw L and your figure is too small to read
mossy pike
yes but not in options
Perhaps the problem is incorrect
delicate lintel
any tips for proofs?
muted sparrow
any tips for proofs?
study analytic geometry
upper karma
Just started studying this, the first method I learned, wondering if I did I correctly because I’m trying to understand it.
weary scaffold
can someone help me with a few trigonometry questions?
upper karma
Just started studying this, the first method I learned, wondering if I did I cor...
Also the variables changed because of substitution, that’s what my book said anyways, just put into one big method to see if I understood what it was trying to say.
thorny forge
i understand i have to do 6400 x sin(15) but what do i do after that
dark sparrow
@thorny forge it's 6400 cos(15°), not 6400 sin(15°). also you don't "have to" do anything. there's no such thing as "having to" do something in math.
once you have written that down, it's time to put 6400 * cos(15) into a calculator, then round to the nearest kilometer
also don't use the letter x to denote multiplication.
thorny forge
ohhhh sorry i meant cos aaa but the answer is less than what the calculator gives me, am i meant to take away something after it
dark sparrow
what does the calculator give you?
thorny forge
calculator gives me 6223 but answer is 6182, its same with the rest of the questions asking for radius,, i dont know what i take away after
