Question

In \[\Delta ABC\] find \[2\left( {bc\cos A + ca\cos B + ab\cos C} \right)\]:
A. 0
B. \[a + b + c\]
C. \[{a^2} + {b^2} + {c^2}\]
D. None of these

Answer 1

Hint: We will use the cosine law to solve the question. Then we will simplify the obtained expression to get the desired answer.

Formula used:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
Given expression is
\[2\left( {bc\cos A + ca\cos B + ab\cos C} \right)\]
Remove the parenthesis by multiplying 2
 \[ = 2bc\cos A + 2ca\cos B + 2ab\cos C\] ….(i)
Redefine cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[ \Rightarrow 2bc\cos A = {b^2} + {c^2} - {a^2}\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[ \Rightarrow 2ac\cos B = {a^2} + {c^2} - {b^2}\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
\[ \Rightarrow 2ab\cos C = {a^2} + {b^2} - {c^2}\]
Substitute the value of \[2bc\cos A,2ca\cos B,2ab\cos C\] in (i)
\[ = {b^2} + {c^2} - {a^2} + {a^2} + {c^2} - {b^2} + {a^2} + {b^2} - {c^2}\]
\[ = {a^2} + {b^2} + {c^2}\]
Hence option C is the correct option.

Additional information:
Cosine rule is applicable, if the length of two sides and angle between them is known.
There is another formula that sine law. It states that the ratio of one side to the sine of its opposite angle of a triangle is the same for the other two sides.
The mathematical representation is
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
To apply sine law, we need to know one side, the opposite angle of that side, and one of the side rests of them or two angles, and the length of one side which is opposite side of the two angles.

Note: The given question is based on the cosine law for triangles. Some students make the mistake of applying the cosine formula \[{a^2} = {b^2} + {c^2} - 2bc\cos B\] which is incorrect. The correct formula is \[{a^2} = {b^2} + {c^2} - 2bc\cos A\].