Sunday, November 2, 2008

Elearn Geometry Problem 199: Triangle angles

Triangle, Altitude, Angles

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.

20 comments:

  1. tan68=dc/bd
    tan38=ad/bd
    tanx=ed/ad
    tan8=ed/dc
    so, tan68/tan38=tanx/tan8
    x=24

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  2. You can also choose to use a geometric solution rather than a trigonometric solution.

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  3. Antonio, would you like to give me some tip to do a geometric solution? What the first step? Please...

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  4. Tips: Think in auxiliary constructions with 30, 60 degrees, equilateral triangle, isosceles triangle, congruence, cyclic quadrilateral.

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  5. draw BF = AB
    draw EG perpendicular to BC

    ang EFD = x = BEG ( perpendicular sides )

    => x = 22 ( from tr BEG )

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  6. What is the correct answer?

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  7. Another Hint: this site has the solutions to this problem, but it's in Arabic. You need someone to help translate the solutions.
    http://www.mathmontada.net/vb/showthread.php?p=5496

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  8. The 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".

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  9. Who has an imageshack.us image for solving the problem?

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  10. Before coming up with a geometric solution, you should discover something:
    Want 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!

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  11. Sorry, I take it back, there's no paradox.

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  12. Another solution:
    http://triangles-geometry.blogspot.com/2010/11/problem-001-auxiliary-lines.html

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  13. To Speed 2001

    Porque θ=8 Grados, x=3θ=24 Grados.
    Because θ=8 Degrees, x=3θ=24 Degrees.

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  14. Problem 199: Solution by Mixalis Tsourakakis from Greece.
    Thanks Mixalis.

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  15. The answer is 24.
    http://geometri-problemleri.blogspot.com.es/2013/02/go-geometry-problem-199.html

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  16. There is a generalization:
    A:B = (x,60+x,60-4x):(30-2x,30+2x,3x)
    where A:B denotes the angles that can be interchanged according to Trigonometric Form of Ceva Theorem.

    x=8,
    30-2x=14,
    60+x=68,
    30+x=38,
    3x=24,
    60-4x=28,

    Here, http://geomania.org/forum/fantezi-geometri/ucgen-icerisinde-p-noktasi-model-3-6/msg14451/#msg14451 , there is a solution in Turkish, you can trace from the picture.

    Also this problem is related with another type of classical problem.
    I classified them here: http://geomania.org/forum/fantezi-geometri/model-ucgen-p-noktasi/msg5609/#msg5609

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