標題:
A. Maths(proving)
發問:
In △ABC, if cosA=sinBsinC, prove that △ABC is an isosceles right-angled triangle.
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最佳解答:
Given cosA = sinB sinC cos A = ( 1 / 2 )[ cos ( B - C ) - cos ( B + C ) ] 2 cos A = cos ( B - C ) + cos A cos A = cos ( B - C ) cos ( 180* - B - C ) = cos ( B - C ) So 180* - B - C = B - C 2B = 180* B = 90* Put it back into cos A = sin B sin C, cos A = sin 90* sin C cos A = sin C If A = C = 45*, △ABC will be isos, so L.H.S. = cos 45* = 1 / sqr 2 R.H.S. = sin 45* = 1 / sqr 2 = L.H.S. Therefore △ABC is an isosceles right-angled triangle. 2008-03-01 11:20:19 補充： I've checked the question again and you should have missed out another given condition: 2 sin A cos C = sin B sin ( A + C ) + sin ( A - C ) = sin ( A + C ) sin ( A - C ) = 0 A = C and so △ABC is an isosceles right-angled triangle. 2008-03-01 11:21:14 補充： With only cosA = sinB sinC, you can only prove △ABC is right - angled since for cos A = sin C sin ( 90* - A ) = sin C A + C = 90* In this case, A = 30*, C = 60* is also possible.
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