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標題:
Application of trigonometry
發問:
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
最佳解答:
a) a cos A = b cos B a [ ( 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 ).
Application of trigonometry
發問:
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
最佳解答:
a) a cos A = b cos B a [ ( 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 ).
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