Saturday, November 28, 2009

Problem 395: Square, 15 Degree, Equilateral triangle

Proposed Problem
Click the figure below to see the complete problem 395 about Square, 15 Degree, Equilateral triangle, Congruence.

 Problem 395: Square, 15 Degree, Equilateral triangle.
See also:
Complete Problem 395

Level: High School, SAT Prep, College geometry

7 comments:

  1. http://www.wolframalpha.com/input/?i=%281-1%2F2*tg%2815%29%29%5E2%2B1%2F4

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  2. http://geometri-problemleri.blogspot.com/2009/11/problem-52-ve-cozumu.html

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  3. solution plz some1

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  4. in triangle ABE ang(ABE)=75;BE=AB/(2cos15)
    with the law of cosine
    AE²=AB²+BE²-2AB.BE cos75=
    AB²(1+(1/(2cos15)²-cos75/cos15)= AB² then
    cos(75)=cos(45+30)=(sqr6-sqr2)/4
    cos(15)=cos(45-30)=(sqr6+sqr2)/4
    same result for CE
    CE=AE=AD
    ADE is equilateral

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  5. build on each side of the square (from the inside) isosceles triangle with 15 degrees..from there its easy..

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  6. Consider the equilateral triangle BCF at the exterior of the square ABCD say of side a. The quandrilateral FEDC and FEAB are rhombuses, therefore

    ED = FC = a and EA = FB = a

    (This holds because ang(FEC) = 75 = ang(ECF), so FE = FC = a and FE parallel to CD = a)

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  7. See
    http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=472125

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