My Math Forum Find Angle A

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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 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 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
 March 22nd, 2013, 05:29 AM #3 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: Find Angle A very good , you got it
 March 23rd, 2013, 12:47 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2269 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 AF² = AE²/2. By Pythagoras, AE² = AF² + EF², so AF = EF. Hence angle EAF = 45°.
 March 23rd, 2013, 08:41 PM #5 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: Find Angle A skipjack :you always have a very good and unusual solution

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