Given that 三角形ABC satisfies acosA = bcosB.
a) Using cosine law, prove that either a^2 - b^2 = 0 OR a^2 + b^2 = c^2
b) Hence comment on the type of 三角形ABC if 0<a<b
Application of trigonometry
2008-04-01 12:23 am
回答 (1)
2008-04-01 1:01 am
✔ 最佳答案
a) a cos A = b cos Ba [ ( b2 + c2 - a2 ) / 2bc ] = b [ ( a2 + c2 - b2 ) / 2ac ]
a2 ( b2 + c2 - a2 ) = b2 ( a2 + c2 - b2 )
a2b2 + a2c2 - a4 = a2b2 + b2c2 - b4
c2 ( a2 - b2 ) - ( a4 - b4 ) = 0
c2 ( a2 - b2 ) - ( a2 + b2 )( a2 - b2 ) = 0
( a2 - b2 )( c2 - a2 - b2 ) = 0
Hence a2 - b2 = 0 or a2 + b2 = c2.
b) If 0 < a < b, then a is not equal to b and so it's impossible for a2 - b2 = 0.
Then if a2 + b2 = c2, triangle ABC is right - angled ( converse of Pyth. Theorem ).
參考: My Maths Knowledge
收錄日期: 2021-04-14 19:53:10
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