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(45A)
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 20060119 11:37:43 