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This question is based on the diagram

alt text

taken from this link. I don't understand why the areas of A6 and A5 adds up to $2bc\cos(A)$.

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Few days ago I noticed this discussion:meta.math.stackexchange.com/questions/132/… –  Quixotic Nov 17 '10 at 19:17
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Here's my version of that diagram: math.stackexchange.com/questions/803/… –  Blue Nov 18 '10 at 2:19
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2 Answers 2

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If you look at A5. It is a rectangle with side $b$ and, say, $x$. Now $\cos A= x/c$, so $x=c\cos A$. Hence the area of A5 is $bx=bc\cos A$.

A similar argument tells you that the area of A6 is $cb\cos A$.

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The other side of A6 has length $c \cos A$ and that of A5 length $b \cos A.$

So their total area is $b \times c \cos A + c \times b \cos A = 2bc \cos A.$

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