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/sci/ - Science & Math

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File: 27 KB, 863x453, abd8e669-fb0c-4fee-84bb-3a59e137d183[1].jpg [View same] [iqdb] [saucenao] [google]
1592545 No.1592545 [Reply] [Original]

any idea how i can get the values for A and B?

>> No.1592557

A and B are worthless letters, they do not have value. I, for one, am glad we enslave them in day to ay speech without thought to their welfare.

>> No.1592567

OP here

at least i dont get the damn they are right here jokes

>> No.1592572

They aren't values, they are labels for the points, so line AB is 403.

>> No.1592575

draw a position vector with a ruler

>> No.1592576

Is that some new-fangled thing denoting a right angle on C? When did they stop using the thing going the other way

>> No.1592580

>>1592576

since i did not make this

>> No.1592586

Small tip - could possibly be wrong.
All angles of triangle add up to 180.
C = 90. If you reply I suppose I could find A and B for you through example?

>> No.1592593

>>1592586

yes C = 90degrees, and yes all angles on a triangle should add up to 180 degrees

yet this still made me brainfart

>> No.1592594

Fucking tangents, how do they work

>> No.1592605

>>1592594

http://answers.yahoo.com/question/index?qid=20080901162632AARbp2x

any help?

>> No.1592612

>>1592593
Ok, bare with me. Been a while since I've done much with trig.

Next post will have info...

>> No.1592622

>>1592594

Aw hell naw. Now its obvious.

>> No.1592650
File: 81 KB, 863x453, 1281710553303.jpg [View same] [iqdb] [saucenao] [google]
1592650

Can't solve the equatoin for x, problem would be done if i could do that.

>> No.1592664

>>1592650

i have seen someone solve it this way once i just can't remember how he did it

>> No.1592666

>>1592664
I could use a graphing calc to look for zeroes but that's kinda cheating
Once you find x it's just basic trig from there.

>> No.1592680

>>1592666

OP here

i have no problems with cheating because i would have done it already if i could

>> No.1592685

A hur fucking dur, why didn't I think of it before.
Just 345? You have a right angled triangle and the distance of the hypotenuse...

>> No.1592697
File: 14 KB, 336x318, 345.jpg [View same] [iqdb] [saucenao] [google]
1592697

You do the trig, tell me if it works out.

>> No.1592703

>>1592697
>>1592685
How do you know it's a 345 triangle?

>> No.1592705

>>1592703
Every right angled triangle is a 345.

>> No.1592706

>>1592697

ima give it a try
stand by

>> No.1592707

>>1592685
>He thinks all right triangles are 345

OP isn't there a geometry theorem on circles in triangles? There are lots of those and everyone forgets them because they're only used for the problems after they're presented.

Euclid would be disappoint.

>> No.1592714

>>1592707
>He thinks they aren't.
OP should've listened better in class.

>> No.1592713
File: 39 KB, 469x428, trollface.jpg [View same] [iqdb] [saucenao] [google]
1592713

>>1592705
4/10

>> No.1592716

>>1592705
45-45-90 triangles exist moron

>> No.1592718

>>1592713
4/10? Come on... Did you account for their wasted time?

>> No.1592720

Law of cotangents

>> No.1592722

>>1592716
...Give this guy a 7.
Those are called equilaterals. this is clearly not.

>> No.1592724

http://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle
Here, you can thank me later.

>> No.1592725

Triangle's just mad he can't solve the problem like me

>> No.1592726
File: 145 KB, 330x403, 1277319999824.jpg [View same] [iqdb] [saucenao] [google]
1592726

>>1592718
>>1592722
Ok, you're 6/10 now. But you're only above 2/10 because you're not a religious troll.

>> No.1592741
File: 12 KB, 336x318, The_Game.jpg [View same] [iqdb] [saucenao] [google]
1592741

>>1592725
orly? check this shit.

<---- u mad?

>> No.1592743

Not OP, but:
I got Angle A as 26.73 degrees.

And I got Angle B is 65.27 degrees.

Anyone got a similar answer?

>> No.1592747

>>1592724
Gives two quadratic equations with two unknowns, so should be solvable. But FUCKING MESSY. I would just plot the quadratic forms in the xy-plane and find intersections.

>> No.1592752

>>1592726
Uhh... That's an or- ugh nevermind.

>> No.1592750
File: 189 KB, 1024x550, 1281613797407.jpg [View same] [iqdb] [saucenao] [google]
1592750

>>1592726
So kind <33
Have some interents

>> No.1592765

Oh btw... I think you'll find I'm only off by a couple decimal places... 8D

But thats why I invented the approximately equal to sign.

>> No.1592801

>>1592765

Its funny because you're quite stupid but think that you're quite smart.

>> No.1592811

>>1592801
orly? and you've tried my solution?

>> No.1592812

>>1592743
I got something really close to a 30-60-90 triangle actually.

>> No.1592821
File: 4 KB, 238x349, 2345313634.gif [View same] [iqdb] [saucenao] [google]
1592821

>Every right angled triangle is a 345

x = x
x = 3
x = 4
3 = 4?

Fuck off retard.

>> No.1592834

oh shit, this reminds me of trilinear coordinates. shit was so cash

>> No.1592848

>>1592812
To explain, first I'm going to rename the angles a,b,c. Sides A,B,C are opposite these angles. S = (A+B+C)/2

According to the law of cotangents Cot(c/2) = (S-C)/(radius of inscribed circle)
So Cot(pi/4)*74 = (A+B+C)/2 -C
74 = (A+B-C)/2
148 + 403 = 551 = A+B
A+B=551
A^2+B^2 = 403^2.
solve for A and B to get the lengths, then you can compute angles.
Might have messed something up though

>> No.1592857

>>1592812
A = 36.86
B = 53.13
Approximately.

>>1592821
...Go try 345 on an equilateral, then come back and apologise for being an idiot.

>> No.1592862

this kind of problem appears on the aime every fucking time.

learn trilinear coordinates for this stuff. it's loev

>> No.1592874

>>1592857

That isn't an equilateral triangle, you lose again faggot.

>> No.1592883
File: 74 KB, 479x435, 1280540068682.jpg [View same] [iqdb] [saucenao] [google]
1592883

>>1592874
...orly?

>> No.1592882

>>1592857
is ... is he trollin ... ? im so consfused ...

>> No.1592886

>>1592883

yarly.

Equilateral triangles aren't right triangles. Nigger.

>> No.1592891

>>1592882

Yeah he is either really stupid or trolling.

>> No.1592898

>>1592857
No, I get sides, B = ~348, A= ~203.
a = arcsin(203/403) = 30 degrees.
b = arcsin(348/403) = 60 degrees.

>> No.1592915

>>1592886
Yes, and?

I never said the triangle you posted was an equilateral.

>>1592898
That puts me at approx. 6.86 degrees off.

>> No.1592922

>>1592915
no you just said all right triangles are 345 which is the most moronic statement i have heard in a while. THOUGHT EXPERIMENT: imagine an actual 345 triangle in your mind. choose one side thats not the hypotenuse . make it longer. to prove the point make it 10000 units longer. now tell me its still a 345 so i can disregard you as a troll

>> No.1592924

>>1592922
Hey man, you could've disregarded me as a troll when I said it. Enjoy your wasted time and rage.

>> No.1592948

>>1592924

classic
>say something stupid
>someone refutes you
>tell them how you think theyre wrong
>they prove you wrong, easily
>cant come up with a valid defence, claim 'ITROLU'

moron

>> No.1592960

>>1592948
>claiming I claimed I was a troll.
Anyway. Does OP have an answer yet?

>> No.1592974

>>1592960

See, he isn't a troll, he is just really really stupid.

>> No.1592987
File: 15 KB, 480x360, ZJ.jpg [View same] [iqdb] [saucenao] [google]
1592987

>>1592974
orly?

>> No.1593029
File: 70 KB, 1663x496, what do i even.jpg [View same] [iqdb] [saucenao] [google]
1593029

>> No.1593063
File: 37 KB, 502x266, 4chan.jpg [View same] [iqdb] [saucenao] [google]
1593063

The first equation gives a quadratic in x. The picture really doesn't distinguish between A and B, so just use that their sum is 90.

>> No.1593073

>>1593063

I get approximately 59.81, 30.19

>> No.1593109
File: 1.80 MB, 2736x3648, P8130198.jpg [View same] [iqdb] [saucenao] [google]
1593109

Answers correspond to OP's pic I think.

>> No.1593225

>>1593109

/thread

>> No.1593333

If you believe that the sides of every right triangle are in the ratio of 3,4,5 (and I do) where 5 is the hypotenuse, then based on the picture

(a/3) = (403/5) and (b/4) = (403/5)

where a is the side opposite angle A and b is the side opposite angle B.

This means that a =241.8 and b=322.4. That can be confirmed by using the Pythagorean theorem (403^2 = a^2 + b^2) and substituting 241.8 for a and 322.4 for b.


Based on the law of sines and knowing that the sin of C is 1 since C is a right angle

(a/Sin A) =403 and (b/Sin B) = 403 or (241.8/Sin A) =403 and (322.4/Sin B) = 403 or A = 36.87 and B = 53.13

>> No.1593736 [DELETED] 
File: 61 KB, 252x221, 1281204942920.png [View same] [iqdb] [saucenao] [google]
1593736

>the sides of every right triangle are in the ratio of 3,4,5

wat

>> No.1593747
File: 61 KB, 252x221, 1281204942920.png [View same] [iqdb] [saucenao] [google]
1593747

>>1593333

>the sides of every right triangle are in the ratio of 3,4,5

wat

>> No.1593763

>>1593333

That nigger is trolling again, just ignore him

>> No.1593774

>>1593109

Is it just me, or does math look beautiful when using traditional pen & paper?

>> No.1593803

>>1593109

That would be good if the sum of the angles in a triangle added up to 210.6 degrees.

>> No.1593817

>>1593109
I like the part where the angles don't add up to 180°.

>> No.1593823
File: 25 KB, 921x606, picard-facepalm.jpg [View same] [iqdb] [saucenao] [google]
1593823

>>1593803

FUCK ASS HORSESHIT how could -- I cant even-- FUCK

>> No.1593825

>>1593803
Non-Euclidean geometry!

>> No.1593827
File: 37 KB, 1152x720, incircle.jpg [View same] [iqdb] [saucenao] [google]
1593827

>>1593333

It is very easy to check to see that your answer is incorrect using the formula to find the radius of an incircle (which was already given).

*ignore my mislabeling of the sides in mathematica, it doesn't matter anyway.

>> No.1593839
File: 98 KB, 600x450, 1277675077786.jpg [View same] [iqdb] [saucenao] [google]
1593839

>>1593823

>> No.1593876
File: 24 KB, 445x480, 1279472808902.jpg [View same] [iqdb] [saucenao] [google]
1593876

>>1593839

>> No.1593919
File: 16 KB, 305x340, 1263839013785.jpg [View same] [iqdb] [saucenao] [google]
1593919

Pic related, these problems are of equal difficulty

>> No.1593916
File: 11 KB, 254x251, s_624343.png [View same] [iqdb] [saucenao] [google]
1593916

>>1593876

>> No.1593949

>>1593333
not all right angle triangles have the ratio 3:4:5. the mistake you're making is thinking that if it has a 90 degree angle somewhere that ratio applies, but it doesn't.
you answer was close because the angles in the problem were not far off from being 3:4:5 but were obvious altered a bit so that that wasn't a feasible way of doing the question.

>> No.1594015

This shit is impossible

>> No.1594056
File: 64 KB, 863x453, 313.jpg [View same] [iqdb] [saucenao] [google]
1594056

>>1592545
Will this work?

>> No.1594157

A = 59.8028
B = 30.1972

Roughly. You don't even need to use any trig until the end, its easier to find the length of the sides (all algebra) and then trig to find the angles with that information.

>>1593073
>>1593073

This guy was correct.

>> No.1594178

P and B make C;
Circumcenter rules:
Equaverticies, Not a bunch of fools (ha ha ha)
Angle Bisector
makes the incenter
Equadistant from the sides, which we would prefer OH!
Median, median, median to centroid
Oh what fun it is to ride 2/3rd of the way! hey!
Altitude altitude what's the height of this!
Orthocenter can be found: What eternal bliss!

>> No.1594466

>>1594157
What did you do to get the sides?

>> No.1594506

>>1594466

r = A/(a + b + c) where A is the area and a b c are the sides of the triangle, and r is the radius of the incircle. You know how a b and c relate because of Pythagorean theorem, you know how to find the Area of a triangle, and you know what r and c are. Its just a matter of finding values for a and b that satisfy pythagorean and the incircle radius formula at the same time. I used a computer because they are good with that kind of stuff.

>> No.1594892

sure is high school in here

>> No.1595203

A = 30.1922
B = 59.8078

rounding to 4 decimals