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0|0|0|49ce805c7a481fae.jpg||||Another solution of 60-70 variant of 80-80-20 triangle|lutzdc||15:13:52|03/28/2009|Someone passed me this triangle and asked me to find the angle CED.  Of course I got sucked into the 4 equations%2C 4 unknowns path and figured out that this was going to get me nowhere.%0D%0A%0D%0AMy solution involved breaking it into right angle triangles as per the attached%2C then assigning an arbitrary length to the bottom side of one of the lower triangles in the attached picture.%0D%0A%0D%0A1%29 Determine missing angles to add up to 180 per triangle%3A EBD%3D20%2C ECD%3D10%2CCBD%3D40%2CBEC%3D30%2C intersection angles of 50 and 130. %0D%0A2%29 Create 2 right-angle triangles from the 60-70-50 lower triangle and solve for h%3A h%3Dxtan%2860%29%0D%0A3%29 Determine f%3A sin%2870%29%3Dh%2Ff%2C therefore by substitution of h%2C f%3Dxtan%2860%29%2Fsin%2870%29%0D%0A4%29 Determine c%3A sin%2810%29%3Dc%2Ff%2C therefore by substitution of f%2C c%3Dsin%2810%29.xtan%2860%29%2Fsin%2870%29%0D%0A5%29 Determine g%3A sin%2860%29%3Dh%2Fg%2C therefore by substitution of h%2C g%3Dxtan%2860%29%2Fsin%2860%29%0D%0A6%29 Determine b%3A sin%2820%29%3Db%2Fg%2C therefore by substitution of g%2C b%3Dsin%2820%29.xtan%2860%29%2Fsin%2860%29%0D%0A%0D%0ANow solve for lengths e and d%0D%0A7%29 Determine e%3A sin%2840%29%3Dc%2Fe%2C therefore by substitution of c%2C e%3Dsin%2810%29.xtan%2860%29%2Fsin%2870%29%2Fsin%2840%29%0D%0A8%29 Determine d%3A sin%2830%29%3Db%2Fd%2C therefore by substitution of b%2C d%3Dsin%2820%29.xtan%2860%29%2Fsin%2860%29%2Fsin%2830%29%0D%0A%0D%0A9%29 Determine a%3A Using the law of cosines%3A a%5E2 %3D e%5E2   d%5E2 - 2e.d.cos%2850%29%2C therefore by substitution of e and d%2C a %3D sqrt%5B %28sin%2810%29.xtan%2860%29%2Fsin%2870%29%2Fsin%2840%29%29%5E2   %28sin%2820%29.xtan%2860%29%2Fsin%2860%29%2Fsin%2830%29%29%5E2 - 2.cos%2850%29.%28sin%2810%29.xtan%2860%29%2Fsin%2870%29%2Fsin%2840%29%29.%28sin%2820%29.xtan%2860%29%2Fsin%2860%29%2Fsin%2830%29%29 %5D%0D%0A%0D%0A10%29Determine angle CED%3A Using the law of sines%3A sin%28CED%29%2Fe %3D sin%2850%29%2Fa%2C therefore by substitution of e and a%2C%0D%0A%0D%0Aangle CED %3D arcsin %7Bsin%2850%29.sin%2810%29.xtan%2860%29%2Fsin%2870%29%2Fsin%2840%29 %2F sqrt %5B%28sin%2810%29.xtan%2860%29%2Fsin%2870%29%2Fsin%2840%29%29%5E2   %28sin%2820%29.xtan%2860%29%2Fsin%2860%29%2Fsin%2830%29%29%5E2 - 2.cos%2850%29.%28sin%2810%29.xtan%2860%29%2Fsin%2870%29%2Fsin%2840%29%29.%28sin%2820%29.xtan%2860%29%2Fsin%2860%29%2Fsin%2830%29%29 %5D%7D%0D%0A%0D%0Aangle CED %3D 20 degrees