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Question: In triangle \(A B C\) \(( b + c ) \cos A + ( c + a ) \cos B\) \(+ ( a + b ) \cos C =\)....

In triangle ABCA B C

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

A

0

B

1

C

a+b+ca + b + c

D

2(a+b+c)2 ( a + b + c )

Answer

a+b+ca + b + c

Explanation

Solution

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

From expanding and collecting terms using projection rule, a=bcos⁡C+ccos⁡Ba = b \cos C + c \cos Betc.