See complete Problem 199 at:
www.gogeometry.com/problem/p199_triangle_altitude_angles.htm
Triangle, Altitude, Angles. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Sunday, November 2, 2008
Elearn Geometry Problem 199: Triangle angles
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tan68=dc/bd
ReplyDeletetan38=ad/bd
tanx=ed/ad
tan8=ed/dc
so, tan68/tan38=tanx/tan8
x=24
You can also choose to use a geometric solution rather than a trigonometric solution.
ReplyDeleteAntonio, would you like to give me some tip to do a geometric solution? What the first step? Please...
ReplyDeleteTips: Think in auxiliary constructions with 30, 60 degrees, equilateral triangle, isosceles triangle, congruence, cyclic quadrilateral.
ReplyDeletedraw BF = AB
ReplyDeletedraw EG perpendicular to BC
ang EFD = x = BEG ( perpendicular sides )
=> x = 22 ( from tr BEG )
To c.t.e.o:
ReplyDeleteThe answer is wrong.
What is the correct answer?
ReplyDelete24°
ReplyDeleteThe correct answer is 24 Degrees; this case trigonometry solution is a lot quicker than geometric solution. I found only one geometric solution for this problem, but it's in Arabic. If anyone has a geometric solution; it's best to copy a link to a website with that solution, like "Geometri Problemleri".
ReplyDeleteWho has an imageshack.us image for solving the problem?
ReplyDeleteBefore coming up with a geometric solution, you should discover something:
ReplyDeleteWant to see a paradox within an auxiliary construction?
Extend CE, intersecting AB at F (which means Angle CFB=60 Degrees). Extend BD, make Angle ECG=38 Degrees; G is on the extension of BD. Connect FG and AG. (Thus FBCG is cyclic, Angle FGB=14 Degrees and Angle CFG=68 Degrees. Angle AFG=52 Degrees. Triangles AGD and CGD are both 30-60-90 Triangles.) Angle AGF=16 Degrees. Extend FG until AFCH is cyclic, that means Triangle AGH is isoceles; Angle GAH=Angle GHA=8 Degrees. Construct median GI of Rt Triangle AGC to side AC; then AG=GI=GH, which means G is the circumcenter of Triangle AHI. So Angle GHI=Angle GIH=22 Degrees. Angle GCH=Angle GIH=22 Degrees, subtended on segment GH, GICH should be cyclic, however, Angle GHI=22 Degrees, Angle GCI=30 Degrees, both angle subtended on segment GI; therefore, there's a paradox!
Who can help me get solve this paradox? And post some geometric solutions up!
Sorry, I take it back, there's no paradox.
ReplyDeleteAnother solution:
ReplyDeletehttp://triangles-geometry.blogspot.com/2010/11/problem-001-auxiliary-lines.html
To Speed 2001
ReplyDeletePorque θ=8 Grados, x=3θ=24 Grados.
Because θ=8 Degrees, x=3θ=24 Degrees.
Problem 199: Solution by Mixalis Tsourakakis from Greece.
ReplyDeleteThanks Mixalis.
Problem 199 solution by Mixalis Tsourakakis at:
Deletehttp://www.gogeometry.com/problem/p199-geometry-mixalis-greece.pdf
X=24, using Trig.
ReplyDelete