4

In a triangle $ABC$

$$(b + c)\cos A + (c + a)\cos B + (a + b)\cos C=?$$

burm1
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  • law of cosine c^2 = a^2 + b^2 - 2ab cos(C) a^2=b^2+c^2-2bc cos(A) b^2=c^2+a^2-2ac cos(B) – burm1 Aug 25 '14 at 16:58
  • Since you know the Law of Cosines, you can replace $\cos A$, $\cos B$, $\cos C$ in the expression with fractions involving $a$, $b$, $c$, then expand like crazy, and finally find that the result is pretty simple. (There's also far simpler approach.) – Blue Aug 25 '14 at 17:03
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    I believe your second term should be "$(c+a)\cos B$". – Blue Aug 25 '14 at 17:05

4 Answers4

7

we know

$$a = b \cos C + c \cos B$$ $$b= a \cos C + c \cos A$$ $$c= b \cos A + a \cos B$$ adding together we get $$(a+b) \cos C + (b+c) \cos A + (c+a) \cos B =a +b+c $$

Petite Etincelle
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5

Hint - draw a diagram of a triangle and see if you can mark a segment of length $b\cos A$ on side $c$ to get a geometric intuition about what is going on here.

Mark Bennet
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2

We know

acosB + bcosA = c

SO after opening brackets and rearranging them you'll see the answer that is a+b+c

DSinghvi
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-1

If we write:

  1. $b+c=2S-a$
  2. $c+a = 2S-b$

  3. $a+b= 2S -c$

    Then by expanding the bracket and taking $2S$ as common factor we can write $2S(\cos a +\cos b +\cos c)-(a\cos a+b\cos b+c\cos c)$
    Now since $a+b+c=180 \cos a +\cos b +\cos c = 1+$ 4\sin a/2.sinb/2.sinc/2 and acosa+bcosb +ccosc is $S$.
    Therefore solving further we get $2S(1+4 \sin a/2\sin b/2\sin c/2)-S = S + 4 r/R$ where $r$ is inradius and $R$ is circumradius as $4 \sin a/2 \sin b/2\ sin c/2 is r/R$ .

I'm stuck up after here.

Help me if I'm doing something wrong

TheHolyJoker
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dhruv
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