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amaths
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最佳解答:
By Sine rule, a/sinA=b/sinB=b/sinC=k a = k sin A Therefore, a/cosA = k sinA/ cos A = k tan A Similarly, b/cosB = k tan B and c/cosC = k tan C So, k tan A = k tan B = k tan C i.e., tan A = tan B = tan C Since 0< A, B, C < 180 degrees So, A = B = C and triangle ABC is a equilateral triangle.
其他解答:
Let a/cosA = m a/m = cosA b/m = cos B c/m = cos C By cosine rule: a^2 = b^2 + c^2 - 2bc cos A = b ^2 + c^2 - 2bc (a/m) a^2 = b^2 + c^2 - 2abc/m ---(1) b^2 = a^2 + c^2 - 2abc/m --- (2) c^2 = b^2 + a^2 - 2abc/m --- (3) (1) - (2) : a^2 - b^2 = b^2 - a^2 2a^2 = 2b^2 a = b Similarly b=c
amaths
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In triangle ABC , a/cosA = b/cosB = c/cosC . What kind of triangle is triangle ABC ?最佳解答:
By Sine rule, a/sinA=b/sinB=b/sinC=k a = k sin A Therefore, a/cosA = k sinA/ cos A = k tan A Similarly, b/cosB = k tan B and c/cosC = k tan C So, k tan A = k tan B = k tan C i.e., tan A = tan B = tan C Since 0< A, B, C < 180 degrees So, A = B = C and triangle ABC is a equilateral triangle.
其他解答:
Let a/cosA = m a/m = cosA b/m = cos B c/m = cos C By cosine rule: a^2 = b^2 + c^2 - 2bc cos A = b ^2 + c^2 - 2bc (a/m) a^2 = b^2 + c^2 - 2abc/m ---(1) b^2 = a^2 + c^2 - 2abc/m --- (2) c^2 = b^2 + a^2 - 2abc/m --- (3) (1) - (2) : a^2 - b^2 = b^2 - a^2 2a^2 = 2b^2 a = b Similarly b=c
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