標題:
我想問下呢條數點計?
發問:
若三角形的內角A,B所對的邊分別為a,b,己知bcosA+acosB==2,a-b=1,且角C=60度,則a=?
最佳解答:
By Cosine Rule, cos A = ( b2 + c2 - a2 ) / 2bc cos B = ( a2 + c2 - b2 ) / 2ac As given, b cos A + a cos B = 2 b ( b2 + c2 - a2 ) / 2bc + a ( a2 + c2 - b2 ) / 2ac = 2 ( b2 + c2 - a2 ) / 2c + ( a2 + c2 - b2 ) / 2c = 2 2c2 / 2c = 2 c = 2 By Cosine rule again, cos C = ( a2 + b2 - c2 ) / 2ab cos 60* = [ a2 + ( a - 1 )2 - 22 ] / 2a( a - 1 ) 1 / 2 = ( 2a2 - 2a - 3 ) / ( 2a2 - 2a ) a2 - a = 2a2 - 2a - 3 a2 - a - 3 = 0 By quadratic formula, a = {1 √[(-1)2-4(1)(-3)]} / 2(1) = (1√13)/2 For a is positive, a = (1+√13)/2
其他解答:
我想問下呢條數點計?
發問:
若三角形的內角A,B所對的邊分別為a,b,己知bcosA+acosB==2,a-b=1,且角C=60度,則a=?
最佳解答:
By Cosine Rule, cos A = ( b2 + c2 - a2 ) / 2bc cos B = ( a2 + c2 - b2 ) / 2ac As given, b cos A + a cos B = 2 b ( b2 + c2 - a2 ) / 2bc + a ( a2 + c2 - b2 ) / 2ac = 2 ( b2 + c2 - a2 ) / 2c + ( a2 + c2 - b2 ) / 2c = 2 2c2 / 2c = 2 c = 2 By Cosine rule again, cos C = ( a2 + b2 - c2 ) / 2ab cos 60* = [ a2 + ( a - 1 )2 - 22 ] / 2a( a - 1 ) 1 / 2 = ( 2a2 - 2a - 3 ) / ( 2a2 - 2a ) a2 - a = 2a2 - 2a - 3 a2 - a - 3 = 0 By quadratic formula, a = {1 √[(-1)2-4(1)(-3)]} / 2(1) = (1√13)/2 For a is positive, a = (1+√13)/2
其他解答:
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