 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A cute little angle to find (Posted on 2006-01-18) 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

 See The Solution Submitted by Jer Rating: 3.6667 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re(2): Solution - Derivation | Comment 6 of 7 | (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 thereal 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 Please log in:
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