#### Question

Let a, b, c be the three sides of a triangle △ABC, If the area S and the sides satisfy S = a^2-(b -c)^2 , and b+c = 8, find the maximum value of S

Collected in the board: Law of Cosines

Steven Zheng posted 4 months ago

Using the formula for the area of a triangle

S = a^2-(b -c)^2 = \dfrac{1}{2}bc\sin A

According to Law of cosines

a^2-(b -c)^2 = 2bc - 2bc\cos A

∴ 1 - \cos A = \dfrac{1}{4}\sin A

\dfrac{1 - \cos A}{\sin A} = \dfrac{1}{4}

\tan \dfrac{A}{2} = \dfrac{1}{4}

∴ \sin A = \dfrac{2\tan \dfrac{A}{2} }{1+\tan^2 \dfrac{A}{2} }

= \dfrac{\dfrac{1}{2} }{1+\dfrac{1}{16} } = \dfrac{8}{17}

S =\dfrac{1}{2}bc\sin A = \dfrac{4}{17}bc = \dfrac{4}{17}b(8-b)

= -\dfrac{4}{17}[(b-4)^2-16 ]

Therefore,

when b = 4, S_{max} = \dfrac{64}{17}

Steven Zheng posted 4 months ago

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