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March 21st, 2013, 08:18 PM   #1
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Find Angle A

[attachment=0:76gf9of6]Find Angle A.JPG[/attachment:76gf9of6]
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March 22nd, 2013, 02:43 AM   #2
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Re: Find Angle A

area of triangle ADE = (1/2)(AD)(AE)(sin A) and area of triangle ABC = (1/2)(AB)(AC)(sin A)

Therefore (1/2)(AD)(AE)(sin A) = (1/4)(AB)(AC)(sin A)

This simplifies to 2(AD/AC) = AB/AE which is

2 cos A = 1/cos A

Thus cos A = 1/sqrt(2) and A = 45 degrees
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March 22nd, 2013, 05:29 AM   #3
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Re: Find Angle A

very good , you got it
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March 23rd, 2013, 12:47 AM   #4
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Let EF be an altitude of triangle ADE.
Triangles BEC, EFD are similar and in ratio BE/EF.
Triangles BEA, EFA are similar and in ratio BE/EF = AE/AF.
?EFD + ?EFA = (?BEC + ?BEA)/2 (given), so AF = AE/2.
By Pythagoras, AE = AF + EF, so AF = EF.
Hence angle EAF = 45.
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March 23rd, 2013, 08:41 PM   #5
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Re: Find Angle A

skipjack :you always have a very good and unusual solution
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