Given: △EAD with ∠A=90. B and C on segment AD with AB=1, BC=2, CD=3. ∠ADE≅∠BEC.
Find AE
Let x = length of AE
Let t = the two angles ∠ADE and ∠BEC
Let s = ∠AEB
tan(t) = x/6
tan(s) = 1/x
tan(s+t) = 3/x
tan(s+t) = [tan(s) + tan(t)] / [1 - tan(s) * tan(t)]
3/x = [1/x + x/6] / [1 - 1/x * x/6]
3/x = [1/x + x/6] / [1 - 1/6]
(3/x)*(5/6) = (1/x + x/6)
5/2 = (1 + x^2/6)
15 = 6 + x^2
x^2 = 9
x = 3
So the length of AE is 3
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Posted by Larry
on 2020-01-09 10:46:27 |
Using some trig
