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Distance to Diagonal 2 (Posted on 2013-06-06) |
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Let ABCD be a parallelogram with ∠BAD = 45° and ∠ABD = 30°.
What is the distance from B to diagonal AC in terms of just |AB|?
What is the distance from B to diagonal AC in terms of just |AD|?
Solution
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Comment 1 of 1
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It feels as if there should be a simple geometry proof here. I can’t find one - so here’s some trig:
Denote the lengths of AB, AD by x and y, and the angle CAD by Z. Let p be the length of the perpendicular from B to AC.
Using the Sine Rule in triangles ABC and ABD:
y/x = sin(45o – Z)/sin Z = sin 30o/sin 105o
Solving for Z: sin Z = sin(45o – Z) sin(45o + 60o)/sin 30o
2 sin Z = (cos Z – sin Z)(1 + sqrt3)
tan Z = (1 + sqrt3) / (sqrt3 + 3) = 1 / sqrt3
Z = 30o
So p = y sin 30o = |AD|/2
and p = x sin 15o = |AB|sqrt2(sqrt3 - 1)/4
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Posted by Harry
on 2013-06-14 16:34:52 |
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