Congrats. Nice demonstration

Very unique question with unique solution . Very very niCe ❤❤❤

Hey I was actually kind of wondering if there is a way to tell when you should use the exterior angle theorem and also if you can most versions of some of the geometry challenges using the cotnagent theorem.

This one turned out to be surprisingly easy. At first I thought to myself that I would just watch the solution because I was feelin a bit lazy and it looked tricky, but I ended up solving it in less than a minute. Then went on to check the solution and it was the same as mine.

Thanks for watching

φ = 30°; ∆ ABC → ABC = θ = ? B = AT + BT; BC = BD + DM + EM + CE = a + b/2 + b/2 + a; ECA = AEC = 8φ/3 → CAE = 2φ/3; EAD = φ sin⁡(BMT) = 1 → BM = CM = a + b/2 → BCT = TBC = TCA = ABC = θ = 4φ/3

linda questão

Try to look Langley Triangel, then, it is very easy to see the search Angle is 10°

Parabéns

ctgθ=(sin20/sin100sin80)+ctg50...θ=40

I'm going with 40°. 1. ∠DEA = 100° 2. ∠AEC = 80° 3. Therefore, △AEC is isosceles, segment AE = segment AC, and ∠EAC = 20°. 4. But that results in △DCA being isosceles, with segment DC = segment AC 5. Segment BD = segment EC, therefore, segment BE = segment DC. 6. Segment BE = segment DC (#5), and segment DC = segment AC (#4), and segment AC = segment AE (#3), therefore segment BE = segment AE (transitivity). 7. Therefore, △BEA is isosceles, and ∠EBA (θ) = ∠EAB. 8. Two equations in 2 unknowns: a. θ = 30° + α (#7), and b. θ + (30° + α) + 100° = 180° (△BEA) Substitute a into b to get: θ + (θ) + 100° = 180°, simplify: 2θ = 80°, or θ = 40°. Now to watch and see how you did it, and whether I got it right. Cheers! - s.west

@skwest

Ай бұрын

Got it. Thanks for the challenge.

Let BD = EC = x. ∠AEC = 50°+30° = 80°. ∠DEA = 180°-80° = 100°. ∠CAE = 100°-80° = 20°. ∠BDA = 180°-50° = 130°. In ∆EAC, ∠AEC = ∠ECA = 80°, so ∆EAC is an isosceles triangle and AE = AC = y. In ∆DCA, ∠ADC = ∠CAD = 50°, so ∆DCA is an isosceles triangle, and AC = DC = y. DC = y and EC = x, so DE = y-x. As BD = x, BE = (y-x)+x = y. As AE also equals y, ∆BEA is an isosceles triangle and ∠ABE = ∠EAB = θ. ∠BEA = 100°, so: θ + θ + 100° = 180° 2θ = 80° θ = 80°/2 = 40° In ∆BDA angles are 40°+130°+10° = 180°, so θ = 40° confirmed.

What is the thinking about this problem? This is what Mathletes call "Angle chasing". The plan is more important than the geometry. Anyone able to follow the solution understands the geometry. Its the thinking that matters.