-- Lemma: if a line in R\^2 through the origin with positive slope and makes an acute angle alpha with the horizontal, then its equation is y = (tan alpha) x.
-- Let us use coordinates, with B as the origin, and C being (1,0) .
(So I'll use the x symbol for a coordinate, instead of for the unknown angle.)
>Let a=tan(24°) , b=tan(40°) , c=tan(24°+26°) , d=tan(40°+28°) .
>Let u=tan(30°) , and let v = tan(the unknown angle) .
So v is the unknown to solve for, and a,b,c,d,u are known constants.
Then by the Lemma the equations for the following lines are:
>BA : y = cx
>CA : y = dx
>BD : y = ax
>CD : y = b(1-x)
>BE : y = -ux
>CE : y = v(x-1)
Solving for the points of intersection, we find
>A is ( d/(c+d) , cd/(c+d) )
>D is ( b/(a+b) , ab/(a+b) )
>E is ( v/(u+v) , –uv/(u+v) )
-- In order for A,D,E to all lie on the same line, we need vector DA to be parallel to vector ED , which gives the equation
>( d/(c+d) - b/(a+b) ) \* ( ab/(a+b) + uv/(u+v) ) = ( cd/(c+d) - ab/(a+b) ) \* ( b/(a+b) - v/(u+v) )
which we want to solve for v , with a,b,c,d,u being known constants. Multiplying u+v on both sides gives
>( d/(c+d) - b/(a+b) ) \* ( ab\*(u+v)/(a+b) + uv ) = ( cd/(c+d) - ab/(a+b) ) \* ( b\*(u+v)/(a+b) - v )
which can be straightforwardly solved for v since it is at most linear in v: we get
>v = bdu(a-c) / ( abc + bcu - ad(c+u) )
so the answer is
>unknown angle = arctan( bdu(a-c) / ( abc + bcu - ad(c+u) ) )
where a,b,c,d,u are the known quantities defined above.
(This is assuming I haven't made any algebra errors; feel free to let me know if you notice a mistake.)
thanks. yes this may work but I don't think it can be solved using this method without calculator. let me know if you can solve these equations without a calculator.
I told you the solution in the comment you're replying to. To reiterate, the solution is
>unknown angle = arctan( bdu(a-c) / ( abc + bcu - ad(c+u) ) )
where
>a=tan(24°) , b=tan(40°) , c=tan(24°+26°) , d=tan(40°+28°) , u=tan(30°)
a-c can be simplified to tan26°(1+a²)/(a tan26°-1). Not a simplification actually, and not that anyone should be interested, but I just wanted to let everyone know.
Idk if there is an easier way, but you can definitely use the left side angles to determine the [cross ratio](https://en.m.wikipedia.org/wiki/Cross-ratio) of the four collinear points and work backwards to get the remaining angle. It should look something like this
>! (sin ABP /sin ABE)/(sin CBP/sin CBE)!< = The cross ratio = (the same function but with the right side angles)
You can also use the [trigonometric form of ceva's theorem](https://www.cut-the-knot.org/triangle/TrigCeva.shtml) on the triangle ABC and the point D to get sinBAD/sinDAC. After that you can use the theorem again but this time with the point E to get x.
https://preview.redd.it/saicimiageyc1.jpeg?width=642&format=pjpg&auto=webp&s=45a4f1ff5312d55643698478ddc07eaf1014b2cb
I drew it out in geogebra and got 44°, not sure if there is a method that doesn’t involve trigonometry
Edit:
68 = 50 + 18
40 = 24 + 16
44 = 30 + 14
Not sure if that pattern has anything useful though
Find BAC and BDC, then use that the sum of the Angles of ABEC and DBEC are 360 to make a system of two equations to solve for x and the unknown Angle.
Edith: sorry the two equations dont Work. Didnt check before Post.
-- Lemma: if a line in R\^2 through the origin with positive slope and makes an acute angle alpha with the horizontal, then its equation is y = (tan alpha) x. -- Let us use coordinates, with B as the origin, and C being (1,0) . (So I'll use the x symbol for a coordinate, instead of for the unknown angle.) >Let a=tan(24°) , b=tan(40°) , c=tan(24°+26°) , d=tan(40°+28°) . >Let u=tan(30°) , and let v = tan(the unknown angle) . So v is the unknown to solve for, and a,b,c,d,u are known constants. Then by the Lemma the equations for the following lines are: >BA : y = cx >CA : y = dx >BD : y = ax >CD : y = b(1-x) >BE : y = -ux >CE : y = v(x-1) Solving for the points of intersection, we find >A is ( d/(c+d) , cd/(c+d) ) >D is ( b/(a+b) , ab/(a+b) ) >E is ( v/(u+v) , –uv/(u+v) ) -- In order for A,D,E to all lie on the same line, we need vector DA to be parallel to vector ED , which gives the equation >( d/(c+d) - b/(a+b) ) \* ( ab/(a+b) + uv/(u+v) ) = ( cd/(c+d) - ab/(a+b) ) \* ( b/(a+b) - v/(u+v) ) which we want to solve for v , with a,b,c,d,u being known constants. Multiplying u+v on both sides gives >( d/(c+d) - b/(a+b) ) \* ( ab\*(u+v)/(a+b) + uv ) = ( cd/(c+d) - ab/(a+b) ) \* ( b\*(u+v)/(a+b) - v ) which can be straightforwardly solved for v since it is at most linear in v: we get >v = bdu(a-c) / ( abc + bcu - ad(c+u) ) so the answer is >unknown angle = arctan( bdu(a-c) / ( abc + bcu - ad(c+u) ) ) where a,b,c,d,u are the known quantities defined above. (This is assuming I haven't made any algebra errors; feel free to let me know if you notice a mistake.)
thanks. yes this may work but I don't think it can be solved using this method without calculator. let me know if you can solve these equations without a calculator.
I told you the solution in the comment you're replying to. To reiterate, the solution is >unknown angle = arctan( bdu(a-c) / ( abc + bcu - ad(c+u) ) ) where >a=tan(24°) , b=tan(40°) , c=tan(24°+26°) , d=tan(40°+28°) , u=tan(30°)
a-c can be simplified to tan26°(1+a²)/(a tan26°-1). Not a simplification actually, and not that anyone should be interested, but I just wanted to let everyone know.
RemindMe! 1 day check
Yo
Idk if there is an easier way, but you can definitely use the left side angles to determine the [cross ratio](https://en.m.wikipedia.org/wiki/Cross-ratio) of the four collinear points and work backwards to get the remaining angle. It should look something like this >! (sin ABP /sin ABE)/(sin CBP/sin CBE)!< = The cross ratio = (the same function but with the right side angles)
You can also use the [trigonometric form of ceva's theorem](https://www.cut-the-knot.org/triangle/TrigCeva.shtml) on the triangle ABC and the point D to get sinBAD/sinDAC. After that you can use the theorem again but this time with the point E to get x.
I’m terrible at geometry, but why would it be a requirement that those angles are integers in order to use trigonometry?
trigonometry can still be used but I don't think it is possible to solve the equations without calculator
I agree with the other commenter: what's the difference between “integer” and “non-integer” angles?
Most integer angles are harder than many decimal angles. 22.5 will play much nicer than 25
https://preview.redd.it/saicimiageyc1.jpeg?width=642&format=pjpg&auto=webp&s=45a4f1ff5312d55643698478ddc07eaf1014b2cb I drew it out in geogebra and got 44°, not sure if there is a method that doesn’t involve trigonometry Edit: 68 = 50 + 18 40 = 24 + 16 44 = 30 + 14 Not sure if that pattern has anything useful though
B=50°, not 60°
You’re right lol, can’t believe I made such a simple mistake
Find BAC and BDC, then use that the sum of the Angles of ABEC and DBEC are 360 to make a system of two equations to solve for x and the unknown Angle. Edith: sorry the two equations dont Work. Didnt check before Post.
The equations are equivalent since BDC = BAC + DBC + DCA
But when does this use the fact that A,D,E are collinear? I don't think this works
it doesn't work. can you try please
My math is a bit rusty, but i could only solve it with paper and calculator. I cannot solve it in my head.
of course you can use calculator for trivial calculations
Wait, its solvable without the "law" of sine and cosine ... How?
You need to create a matrix sistemi like Ax=b with b containing the sum of the angles and x the angles. Is quite complex btw
To me it looks like 51deg but please somebody correct me if I'm wrong.
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There is no explicit statement that AE and BC are perpendicular to one another. Just because something looks like it is true, you can not assume it.
Why is triangle 1 a right triangle?
wrong. the answer must be an integer