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Question: In triangle ABC, (b + c) cosA + (c + a) cosB + (a + b) cosC is equal to: A. 0 B. 1 C. a + b + ...

In triangle ABC, (b + c) cosA + (c + a) cosB + (a + b) cosC is equal to:
A. 0
B. 1
C. a + b + c
D. 2(a + b + c)

Explanation

Solution

We will first mention the cosine formula and find the value of b cosC + c cosB to be equal to a and so on. Now, after that, we will put in these values in the given expression and thus we have the answer.

Complete step by step answer:
We know that the cosine formula that is the Projection formula is given by the following expression:-
⇒cos⁡A=b2+c2−a22bc\Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} ……………(1)
Similarly we have that:
⇒cos⁡B=a2+c2−b22ac\Rightarrow \cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} …………...(2)
⇒cos⁡C=a2+b2−c22ab\Rightarrow \cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} ………...…(3)
Now, we will find the value of b cosA + a cosB by using the equations (1) and (2):
⇒acos⁡B+bcos⁡A=a(a2+c2−b22ac)+b(b2+c2−a22bc)\Rightarrow a\cos B + b\cos A = a\left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right) + b\left( {\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)
Cancelling off a and b from the denominator, we will get:-
⇒acos⁡B+bcos⁡A=a2+c2−b22c+b2+c2−a22c\Rightarrow a\cos B + b\cos A = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2c}} + \dfrac{{{b^2} + {c^2} - {a^2}}}{{2c}}
Since, the denominator is same, we will now add the numerator:-
⇒acos⁡B+bcos⁡A=a2+c2−b2+b2+c2−a22c\Rightarrow a\cos B + b\cos A = \dfrac{{{a^2} + {c^2} - {b^2} + {b^2} + {c^2} - {a^2}}}{{2c}}
Clubbing the like terms to get the following expression:-
⇒acos⁡B+bcos⁡A=2c22c=c\Rightarrow a\cos B + b\cos A = \dfrac{{2{c^2}}}{{2c}} = c …………….(4)
Similarly, if we calculate a cosC + c cosA and b cosC + c cosB, we will obtain:-
⇒ccos⁡B+bcos⁡C=a\Rightarrow c\cos B + b\cos C = a ……………….(5)
⇒ccos⁡A+acos⁡C=b\Rightarrow c\cos A + a\cos C = b ……………….(6)
Now, if we look the expression whose value we were required to find, we will see that:-
⇒\Rightarrow (b + c) cosA + (c + a) cosB + (a + b) cosC = b cosA + c cosA + c cosB + a cosB + a cosC + b cosC
⇒\Rightarrow (b + c) cosA + (c + a) cosB + (a + b) cosC = [b cosA + a cosB] + [c cosB + b cosC] + [a cosC + c cosA]
Now, using the equations (4), (5) and (6), we will obtain:-
⇒\Rightarrow (b + c) cosA + (c + a) cosB + (a + b) cosC = a + b + c

Hence, the correct option is (C) a + b + c.

Note: The students must note that if they directly remember the equations (4), (5) and (6), it will be an extremely easy task for us. But, you must know how we derive that using the Projection formula.
The students must commit to memory the following formula:-
⇒cos⁡A=b2+c2−a22bc\Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}