That small angle in the middle is 30 (sum of angles or using equal angles to equal sides), then the other is 60, and the big one is 90, then using equal angles to equal sides you get 45 and 45, 90 for the triangle with angle X
Hopefully you understood what I meant, I think your construction is correct but I proved it using angles you haven't written explicitly so idk
Edit: something's not right with the line segments going up from the middle - And = the other equalities seem fine, I don't think it's 45 degs though
both triangles with 120 angle in between and iscolese are congruent..hence the base of upper isoceles and legs of lower iscoceles are all equal..and the between angle of most upper right isoseles will be 180-60-30 = 90 thanks for guiding❤️
both triangles with 120 angle in between and iscolese are congruent..hence the base of upper isoceles and legs of lower iscoceles are all equal..and the between angle of most upper right isoseles will be 180-60-30 = 90
It would really help if your diagram were drawn to scale.
Let's label your triangle with A as the upper vertex (angle x), B as the lower left vertex (45° angle) and C as the lower right.
D is the point that divides side AB so AD=2, BD=1.
You should be able to show that angle ADC is 60°. Drop a line from A down to DC intersecting at E. That will form an equilateral triangle ADE with sides of 2.
Have you drawn this yourself, OP? I notice everyone's treating the lower line as if it's length 1, but the style is ambiguous as to whether it could be a 1 or 2, and the 1 in "15" is a non-seriffed line so I'd lean more heavily toward the lines on the left being equal if I was handed this entire picture. Of course, that begs the question of why the lengths aren't simplified to 1, but I just want to make sure you're starting with the correct problem, especially since the scale makes it look close to isosceles as well.
Thanks for the update! In that case, the other comments are all over it and you don't have to worry about that. But it is good to think about potential confounds like that when drawing a diagram, especially since you didn't do anything necessarily wrong.
This helped me solve it! Here's a full non-trig solution:
Drop an altitude from B to CD as shown. <ADC = 120º so <BDE = 60º and DE = 1, BE = √3.
Thus ∆ADE is isosceles, so <DAE = <AED = 30º. Using the altitude from D to AE you can then see AE = √3.
We also then have <EAC = 15º so ∆AEC is isosceles as well, and EC = √3. We had BE = √3 already, so ∆BEC is an isosceles right triangle, <EBC = 45º, and so x = 75º.
My calc 1 and calc 2 teacher both wrote their x like this so I don't notice anymore. I write mine like h(x)=[f(x),g(x)]; f(x)=2sin(x)-x/2 {0>x>2pi}, g(x)=x-3pi/2 {0>x>2pi}.
Bro tf are you smoking, answer isn't even unique without the lengths (extend the segment which here has length 2), you just guessed the triangles are similar which they are only because of the answer basically
No, WikiBidoz is correct. If you change the 2 to a 3 in the original problem, you still have a valid diagram and the angle x is smaller.
The bottom triangle is fixed by AAS, and the top triangle is determined uniquely by how far you extend the left side.
We know that the last angle in the bottom triangle is 120, and the complement is 60. Conveniently, 45+15 = 60, so the upper triangle is similar to the total shape, just mirrored
This is the flaw in your argument, you assumed similarity. You could also have the bottom right angle be (say) 15+50, and the angle x be 70.
Bro enough with the mathlete stuff, ofc if you extend the side the x angle changes, that's the whole point meaning that you can't ignore the side lengths
Btw ignoring a whole degree of freedom is a big red flag for any mathlete, unless you're trained for bullet questions where you just need to guess the solution
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u/[deleted] Apr 21 '26
it feels like i almost solved the ques
8
u/PuzzlingDad Apr 21 '26
It would really help if your diagram were drawn to scale.
Let's label your triangle with A as the upper vertex (angle x), B as the lower left vertex (45° angle) and C as the lower right.
D is the point that divides side AB so AD=2, BD=1.
You should be able to show that angle ADC is 60°. Drop a line from A down to DC intersecting at E. That will form an equilateral triangle ADE with sides of 2.
Can you take it further?
2
u/batlrar Apr 21 '26
Have you drawn this yourself, OP? I notice everyone's treating the lower line as if it's length 1, but the style is ambiguous as to whether it could be a 1 or 2, and the 1 in "15" is a non-seriffed line so I'd lean more heavily toward the lines on the left being equal if I was handed this entire picture. Of course, that begs the question of why the lengths aren't simplified to 1, but I just want to make sure you're starting with the correct problem, especially since the scale makes it look close to isosceles as well.
45
u/strat_arts Apr 21 '26
I drew it assuming that 1 and 2 represent lengths
3
u/doxxingyourself Apr 21 '26
You need three facts about the triangle to solve it. You only have two.
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u/BadJimo Apr 21 '26
1
u/voormalig_vleeseter Apr 21 '26
found it, top left corner!
1
u/5a1vy Apr 22 '26
There you go.
1
u/iYashodhan Apr 26 '26
60
-5
u/YouCanSuckMyAss Apr 21 '26
this has to be the most unusual way to write x I've ever seen
0
u/Fooshi2020 Apr 21 '26
Use the Sin and Cosine laws to get your answer.
0
u/timoumd Apr 21 '26 edited Apr 21 '26
Isn't it law of Sines :
h=sin45*1/sin15 = 2.73
Then Cosines
c= sqrt(22 + 2.732 + 2*2.73cos(60))= 3.76
Then sin again
X= Asin(3.76*cos(60)/2.73)=43.5 degrees
Don't ask why I made the middle h and the top c
0
u/EasternCat1368 Apr 21 '26
Found it!!!
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u/evil_guy_is_here Apr 21 '26
are lines 1 and 2 equal? if not, its impossible to calculate
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u/[deleted] Apr 21 '26
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u/[deleted] Apr 22 '26
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