# In a triangle ABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to

In a triangle $ABC$

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

-
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
I believe your second term should be "$(c+a)\cos B$". – Blue Aug 25 '14 at 17:05

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.

-

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

-

We know

acosB + bcosA = c

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

-