Post AmTTQbrnSprnZpy5Ts by ai@cawfee.club
(DIR) More posts by ai@cawfee.club
(DIR) Post #AmPJtvkcpyfdTH7I12 by karna@poa.st
2024-09-26T21:14:25.082771Z
4 likes, 2 repeats
What is θ? BD and DC are equal in length
(DIR) Post #AmPP0EOMwWNenGWvVw by lss@freesoftwareextremist.com
2024-09-26T22:11:33.694107Z
0 likes, 0 repeats
@karna is this even solvable? We only get one angle for triangles ABC and ABD.
(DIR) Post #AmPPbqfhaDsYEKZhIG by karna@poa.st
2024-09-26T22:18:22.649410Z
1 likes, 1 repeats
@lss yep. you also know that two of side-lengths are equal which makes it solvable. lmk if you want a hint
(DIR) Post #AmPSd3R9uHvDyQ2Ta4 by lss@freesoftwareextremist.com
2024-09-26T22:52:11.701768Z
1 likes, 0 repeats
@karna sorry, I've been out of school for too long.
(DIR) Post #AmPSd6zegUIZ0HCZGa by halberd@poa.st
2024-09-26T22:52:13.204283Z
1 likes, 1 repeats
@karna I can get 30° by law of sines, but there's definitely a smarmy "just put point X _here_, then it's a one-line proof" thing waiting in the wings.
(DIR) Post #AmPTCtf9izOM5Hg1tA by karna@poa.st
2024-09-26T22:58:41.370763Z
0 likes, 1 repeats
@halberd 30 is right. as for smarmy, its not a one liner but if you draw a line from the top perpendicular to the base of the triangle and take advantage of the resulting right angled triangles that get created you can use various tangent formulas and ratios to find the angle
(DIR) Post #AmPTr4ELqz3kRVIIRk by halberd@poa.st
2024-09-26T23:05:56.818004Z
1 likes, 1 repeats
@karna Okay, I was on the right track, at least. I had observed that if you let that perpendicular be 1, then you'd get side lengths as 1, tan(α), 1/cos(α). Then the side length equality became tan(θ + x + 15°) - tan(θ + x) = tan(θ + x) - tan(x) (letting x be the angle formed at A by dropping the perpendicular. I was hoping that expanding outward would possibly kill off the x, but expanding the tangent addition identities was a tedious prospect.
(DIR) Post #AmPVXdJpsyX1NsiiYq by karna@poa.st
2024-09-26T23:24:50.822771Z
0 likes, 1 repeats
@halberd Once you drop the perpendicular theres two angles whose tangents works out very nicely in terms of the length making up the base of the triangle and the bit you need to extend to reach the perpendicular. Combined with the sum-angle formula for tangents and it just works ™️
(DIR) Post #AmPdCsHTrTEdL9EO9o by greenshoots@poa.st
2024-09-27T00:50:44.025350Z
1 likes, 1 repeats
@karna I'm getting that θ is equal to 30°, would that be correct?
(DIR) Post #AmPfymnvj5zMAANnRA by greenshoots@poa.st
2024-09-27T01:21:47.991168Z
1 likes, 1 repeats
@karna And to show I didn't just copy my answer from others
(DIR) Post #AmPmjjsJ2ENaPfCYzY by karna@poa.st
2024-09-27T02:37:31.116998Z
0 likes, 1 repeats
@greenshoots Thats the right answer alright. btw what was the justification for w/2x = x/w in your second pic (which I'm assuming is the first in terms of reading order)
(DIR) Post #AmQhIe6I5hTk4XxrMW by greenshoots@poa.st
2024-09-27T13:11:18.496305Z
1 likes, 1 repeats
@karna I was comparing the two triangles, ABC and ABD, and since they shared at least one angle and ABD was inscribed within ABC, I assumed these two triangles were proportional to each other. Which then allowed me to use the similar triangles rule to define w/2x = x/w.Hopefully I came up with the right justification for my answer, that part would always get me on math exams back when I was still in school 😅
(DIR) Post #AmQjVkZFYdC7not0b2 by SuperSnekFriend@poa.st
2024-09-27T13:36:05.175120Z
0 likes, 1 repeats
@karna Shoot, I almost got the right answer last night. I should realized that the sum of angles Θ and A° must always be 45°, no matter where point A is. I knew A° was 15°, but I struggled to find the answer because I was trying to overcome not knowing angle B.It's been too long since I've learned math.
(DIR) Post #AmQjz8MUAdaJzqvXeq by lunarised@whinge.town
2024-09-27T13:41:23.463036Z
0 likes, 1 repeats
@karna i got 30°
(DIR) Post #AmQvRqXibAcyct4hma by karna@poa.st
2024-09-27T15:49:50.339864Z
0 likes, 1 repeats
@greenshoots Well you got to the right answer in the end but I dont think those two triangles count as similar...
(DIR) Post #AmQvUija5TshHDzgzA by Frondeur@poa.st
2024-09-27T15:50:21.654663Z
1 likes, 1 repeats
@karna so I have to go through all that to prove ABD and ABC are similar...
(DIR) Post #AmQvcohplglO7B2NxQ by karna@poa.st
2024-09-27T15:51:49.533128Z
0 likes, 1 repeats
@lunarised Correct answer! Out of curiosity, what approach did you take, law of sines, drop a perpendicular and tangents, etc.?
(DIR) Post #AmQvz3pIER6jNe8JXM by lunarised@whinge.town
2024-09-27T15:55:50.116578Z
0 likes, 1 repeats
@karna law of sins but using arclengths and circles because I forgot what to do ,:)
(DIR) Post #AmQwiNzO76pqwgT3Oi by Frondeur@poa.st
2024-09-27T16:04:02.114879Z
1 likes, 1 repeats
@karna My last resort is usually to drop an orthogonal line because I get the pythagorean equation which is independent of all the other relations I have...So I have the right unknowns for equations...but that always makes too much algebra to feel like I did it in the best way possible...but I couldn't think of anything slick here...
(DIR) Post #AmQwiS6ympHhjO3Amm by karna@poa.st
2024-09-27T16:04:02.937936Z
1 likes, 1 repeats
@lunarised gotcha, nice. Oh one more thing thats bothering me. Ive been meaning to ask for a while, why did you change your profile pic to tucker carlson? /s
(DIR) Post #AmQwoRZVhrY3nJFhGS by karna@poa.st
2024-09-27T16:05:07.913935Z
1 likes, 1 repeats
@Frondeur imo the dropping the perpendicular (eventually) leads to the slickest solution. but there is a little bit of algebra involved
(DIR) Post #AmQx6z43v5hcsnNouG by Frondeur@poa.st
2024-09-27T16:08:28.938761Z
1 likes, 1 repeats
@karna Ok, I just read through your answer...dont quite get the slick aspect but Ill try looking at the tangent addition identity like you said and see if that saves me a page of algebra lol...
(DIR) Post #AmQxFETwTx2112QCgq by karna@poa.st
2024-09-27T16:09:58.389449Z
1 likes, 1 repeats
@Frondeur The slick part is that you can sidestep a lot of algebra if you pick the right angles to take the tangents of
(DIR) Post #AmQzBKg7hZ3ez3O4RM by lunarised@whinge.town
2024-09-27T16:31:40.038660Z
1 likes, 1 repeats
heres some more (i hope solid) working just to see if i can do it using pure trig and algebra. pls be kind since i havent done this type of math for 10ish yearsAlso i like fly fishing!
(DIR) Post #AmQzQpF4LhQW7R7XBw by lunarised@whinge.town
2024-09-27T16:34:28.101952Z
0 likes, 1 repeats
@karna this is wrong. im dumb
(DIR) Post #AmQzSiedoQOwGcz6nI by eemmaa@soc0.outrnat.nl
2024-09-27T16:34:48.779Z
0 likes, 0 repeats
@lunarised@whinge.town witch craft, burn
(DIR) Post #AmR0J1plkTEioG8Hse by karna@poa.st
2024-09-27T16:44:16.399645Z
0 likes, 1 repeats
@lunarised I think there's a way to use the law of sines but its a little more involved. The hint I'll give is that whatever method you use, you need to use the fact that the two bottom lengths are equal in some way > fly fishinglmao perfect response to my joke question. made me laugh irl
(DIR) Post #AmR0PN3VlAk6MgaJMW by lunarised@whinge.town
2024-09-27T16:45:23.496751Z
1 likes, 1 repeats
@karna Yeah, i get that, (thats where sin(2) came from, but i misremembered rule of sines
(DIR) Post #AmRDYOdUYJ9xHD33PE by billiam@shitposter.world
2024-09-27T19:12:42.880978Z
3 likes, 2 repeats
@karna don't know nothing about no law of signs
(DIR) Post #AmRFC4EztrIDYm4VRQ by karna@poa.st
2024-09-27T19:31:05.519721Z
0 likes, 1 repeats
@billiam Nice use of 30-60-90 triangles
(DIR) Post #AmRR5ScU3ouoamfoIa by greenshoots@poa.st
2024-09-27T21:44:21.315607Z
0 likes, 1 repeats
@karna Ah, I didn't know that. Thanks for the reply though, these problems you post occasionally are helping me remember a lot of the math I have forgotten over the years.
(DIR) Post #AmTTQbrnSprnZpy5Ts by ai@cawfee.club
2024-09-28T21:19:59.678673Z
0 likes, 0 repeats
@billiam @karna Epic solution. Did you get AY = (2 + 3^0.5)/2 by doing AY = DY tan(ADY) and computing tan(75) or am I missing an easier way? Similar reasoning inspired by the same diagram: Draw DZTriangle BDZ is equilateral -> DZ = BZAngle ZDB = 60 Angle ZDA = ZDB - ADB = 15 Angle ZAD = ADB - ACD = 15 Triangle AZD is isosceles -> AZ = DZAZ = BZ AZB is an isosceles right triangle Angle ZAB = 45 Angle Theta = ZAB - ZAD = 15
(DIR) Post #AmTUwY9YuEb1HT4VkG by billiam@shitposter.world
2024-09-28T21:36:58.661818Z
2 likes, 1 repeats
@ai @karna thank you :frierensmile: yep, AY = 0.5*tan(75)that's a good idea with BDZ being equilateral. Although it takes a little effort to work it out, unless I'm missing something.
(DIR) Post #AmTVyCLj2iyCBt6wXw by karna@poa.st
2024-09-28T21:48:29.498780Z
1 likes, 1 repeats
@ai @billiam Minor typo at the end but nice use of equilateral triangles there
(DIR) Post #AmTWbd7lQdZuWcNojg by ai@cawfee.club
2024-09-28T21:55:35.654571Z
2 likes, 0 repeats
@billiam @karna I was thinking like this: in any right triangle, the median to the hypotenuse is equal to half the length of the hypotenuse (labeled "=x by theorem" in my picture). So the two little triangles in the picture are always isosceles. Since we also know that it's a 30-60-90 triangle in this problem, the bottom side is also "=x", so the bottom triangle in the picture is isosceles
(DIR) Post #AmTXCLHMjtcNtUNcgq by billiam@shitposter.world
2024-09-28T22:02:15.162435Z
2 likes, 1 repeats
@ai @karna oh right, of courseor just see that ZBC = 60BD = BZBDZ is isoscelesBZD = 60 = BDZyeah, my answer could have been a bit more succinct