In triangle ABC with:
BC = 1;
angle B=45 degrees;
D is on AB such that DC =1;
E and H are on segment AC and EH = 1;
F is on segment AD such that EF=.5;
G is on segment CD;
EFGH is a rectangle.

Find angle A

(In reply to re: Solution by Charlie)

From (1)
             sin(A)
   x = 1 - ---------- = 1 - sqrt(2)*sin(A)           (3)
            sin(135)

Plugging (3) into (2) gives

    1
   --- = [1 - sqrt(2)*sin(A)]*sin(45-A)
    2

                                 sqrt(2)
       = [1 - sqrt(2)*sin(A)] * --------- * [cos(A) - sin(A)]
                                    2

       or

    1 = [sqrt(2) - 2*sin(A)]*[cos(A) - sin(A)]

      = sqrt(2)*cos(A) - sqrt(2)*sin(A) - 2*sin(A)*cos(A) + 2*sin(A)^2

      = sqrt(2)*cos(A) - sqrt(2)*sin(A) - 2*sin(A)*cos(A) + 2*[1 - cos(A)^2]

       or

    sin(A)*[sqrt(2) + 2*cos(A)] = 1 + sqrt(2)*cos(A) - 2*cos(A)^2

Squaring both sides and substituting 1 - cos(A)^2 for sin(A)^2 gives

    0 = 8*cos(A)^4 - 4*cos(A)^2 - 2*sqrt(2)*cos(A) - 1

This equation has two real roots and two complex roots. One of the
real roots is negative (which would give A>90) and the other is
~= .99175566.

Taking the arccos of this gives the answer.

 

Posted by Bractals on 2006-01-19 11:37:43