求解数学题

  • 主题发起人 主题发起人 jo jo
  • 开始时间 开始时间
答案 2.5
让 角 MCB = a, then MF=10sin(a), ME=10sin(a)cos(a), FE=10sin(a)sin(a), MC=10cos(a), EC=10cos(a)cos(a),
BE=ME * ctan(2a)=10sin(a)cos(a)cos(2a)/sin(2a)=5cos(2a), BC=BE+EC=5cos(2a)+10cos(a)cos(a),
最后, DE = EC - (1/2)BC = 10cos(a)cos(a) - (1/2)[5cos(2a)+10cos(a)cos(a)] = 10cos(a)cos(a) - (1/2){5[2cos(a)^2 - 1] + 10cos(a)^2}
=10cos(a)cos(a) - 5cos(a)^2 + 5/2 - 5cos(a)^2 = 5/2 (Tricky: 幸运全约掉!)
smart!
 
答案 2.5
让 角 MCB = a, then MF=10sin(a), ME=10sin(a)cos(a), FE=10sin(a)sin(a), MC=10cos(a), EC=10cos(a)cos(a),
BE=ME * ctan(2a)=10sin(a)cos(a)cos(2a)/sin(2a)=5cos(2a), BC=BE+EC=5cos(2a)+10cos(a)cos(a),
最后, DE = EC - (1/2)BC = 10cos(a)cos(a) - (1/2)[5cos(2a)+10cos(a)cos(a)] = 10cos(a)cos(a) - (1/2){5[2cos(a)^2 - 1] + 10cos(a)^2}
=10cos(a)cos(a) - 5cos(a)^2 + 5/2 - 5cos(a)^2 = 5/2 (Tricky: 幸运全约掉!)

让 角 MCB = a, (角 MBC = 2a);

MC=10cos(a),
CE=10 cos(a)*cos(a),
ME=10sin(a)cos(a), then BM=5 (Triangle BME);
BE=5cos(2a);

Next: BE+DE=CE-DE
2DE=CE-BE
DE=0.5(10cos(a)cos(a)-5cos(2a))
=0.5(5)=2.5
 
這是用三角计算出来,能用纯幾何得到答案嗎?
 
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