If in a triangle ABC, a, b, c, and angle A is given and \[c\sin A < a < c\], then which of the following is true?
A. \[{b_1} + {b_2} = 2c\cos A\]
B. \[{b_1} + {b_2} = c\cos A\]
C. \[{b_1} + {b_2} = 3c\cos A\]
D. \[{b_1} + {b_2} = 4c\cos A\]

Hint: We will use cosine law to find a quadratic equation of b. Then we will apply the sum of roots formula to get the desired result.

Formula used:
The sum of roots of a quadratic equation \[A{x^2} + Bx + C = 0\] is \[ - \dfrac{B}{A}\].
The cosine law is:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
Taking cosine law \[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
Now rewrite the above equation
\[{b^2} - 2bc\cos A + {c^2} - {a^2} = 0\]
\[ \Rightarrow {b^2} - \left( {2c\cos A} \right)b + \left( {{c^2} - {a^2}} \right) = 0\] ….(i)
The above equation is a quadratic equation of b.
Compare equation (i) with \[A{x^2} + Bx + C = 0\]
A = 1, \[B = - 2c\cos A\], and \[C = {c^2} - {a^2}\]
Assume that \[{b_1}\] and \[{b_2}\] are the roots of the above equation.
Apply the formula sum of roots for equation (i)
\[{b_1} + {b_2} = \dfrac{{2c\cos A}}{1}\]
\[ \Rightarrow {b_1} + {b_2} = 2c\cos A\]
Hence option A is the correct option.

Note: Some students are confused with the formula of the sum of roots of a quadratic equation and the product of roots of a quadratic equation. They used \[\dfrac{C}{A}\] as a sum of the roots of the equation \[A{x^2} + Bx + C = 0\] which is an incorrect formula. The correct formula is the sum of the roots of the quadratic equation \[A{x^2} + Bx + C = 0\] is \[ - \dfrac{B}{A}\].