Sunday, September 28, 2008

Elearn Geometry Problem 185

Triangle and angles

See complete Problem 185 at:

Trapezoid, Triangle and angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Evrenin, the answer x = 30 is not correct, try again.

  2. Dear Romeo, the answer x = 30 is not correct, try again.

  3. x=10

    [img] 00.jpg[/img]
    [img] 002.jpg[/img]
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  4. Solution of problem 185.
    Let BC = u and CD = v. The extensions of AB and DC meet at E.
    We have ang(CBE) = ang(DAB) = 4x. Since ang(ACD) = 5x and ang(CAE) = x, then ang(AEC) = 4x, so triangles ADE and BEC are isosceles, with CE = BC = u and DE = AD = AB = u+v.
    Take F on AB so that AF = v. Then BF = BC = u, and triangle CBF is isosceles. As ang(CBE) = 4x, then ang(BFC) = ang(BCF) = 2x. Therefore, in triangle AFC, ang(ACF) = x = ang(CAF), so CF = AF = v.
    Triangle ABD is isosceles so ang(ABD) = 90º - 2x. If G is the intersection of BD and DF we have ang(BGF) = 90º. So BD is the angle bisector of CBF and DF = CD = AF = v. This means that CDF is equilateral.
    Hence x + 5x = 60º and x = 10º.

    1. Good work Nilton

      Did u have another solution Antonio?

  5. The answer is x=10 , the solution of the equation
    sin(4x)/sin(3x) == sin(8x)/sin(5x)

  6. The angle x satisfies the equation sin4x/sin3x = sin8x/sin5x whose solution in (0,pi/4) is x = 0.17453292519943298 rad = 10º