Question

The difference between the two values of c is:
(A) $2\sqrt {{a^2} - {b^2}}$
(B) $\sqrt {{a^2} - {b^2}}$
(C) $2\sqrt {{a^2} - {b^2}{{\sin }^2}A}$
(D) $\sqrt {{a^2} - {b^2}{{\sin }^2}A}$

Answer 1

Hint:- Proceed by using the formula of cosine rule to form a quadratic in c and then use the properties of discriminant for real and distinct roots. Use the sum and product of the roots to find out the difference of the roots of this quadratic equation.

Complete step-by-step answer:
We know from the cosine rule of a triangle that $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
From the above equation, we get
${c^2} - 2bc\cos A + \left( {{b^2} - {a^2}} \right) = 0....\left( 1 \right)$
The above equation (1) is a quadratic in $c$
For real and distinct values of $c$ , the discriminant of the quadratic should be greater than zero.
$\therefore D > 0 \Rightarrow 4{b^2}{\cos ^2}A - 4 \cdot 1 \cdot \left( {{b^2} - {a^2}} \right) > 0$
$\Rightarrow 4{b^2}\left( {{{\cos }^2}A - 1} \right) + 4{a^2} > 0$
Using the trigonometric identity ${\sin ^2}A + {\cos ^2}A = 1$ , we get
$\begin{array}{l} \- 4{b^2}{\sin ^2}A + 4{a^2} > 0\\\ \Rightarrow {a^2} > {b^2}{\sin ^2}A\\\ \Rightarrow a > b\sin A \end{array}$
Let the roots of the quadratic equation (1) be ${c_1}$ and ${c_2}$
This implies that the two values of $c$ are ${c_1}$ and ${c_2}$
Using the sum and product of the roots of the quadratic equation (1), we get
$\begin{array}{l} {c_1} + {c_2} = 2b\cos A....\left( 2 \right)\\\ {c_1}{c_2} = {b^2} - {a^2}....\left( 3 \right) \end{array}$
In order to find out the difference between the two values of $c$ , we will use the algebraic identity ${\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab$
Thus, we get
${\left( {{c_1} - {c_2}} \right)^2} = {\left( {{c_1} + {c_2}} \right)^2} - 4{c_1}{c_2}$
Taking square root, we get
$\left| {{c_1} - {c_2}} \right| = \sqrt {{{\left( {{c_1} + {c_2}} \right)}^2} - 4{c_1}{c_2}}$
Replacing the values from equation (2) and (3), we get
$\left| {{c_1} - {c_2}} \right| = \sqrt {4{b^2}{{\cos }^2}A - 4\left( {{b^2} - {a^2}} \right)}$
$\begin{array}{l} \Rightarrow \left| {{c_1} - {c_2}} \right| = \sqrt {4{b^2}\left( {{{\cos }^2}A - 1} \right) + 4{a^2}} \\\ \Rightarrow \left| {{c_1} - {c_2}} \right| = \sqrt { - 4{b^2}{{\sin }^2}A + 4{a^2}} \\\ \Rightarrow \left| {{c_1} - {c_2}} \right| = 2\sqrt { - {b^2}{{\sin }^2}A + {a^2}} \\\ \Rightarrow \left| {{c_1} - {c_2}} \right| = 2\sqrt {{a^2} - {b^2}{{\sin }^2}A} \end{array}$
Since we have already found out that $a > b\sin A$ , the value of $\left| {{c_1} - {c_2}} \right|$ comes out to be positive, which is true.

Hence, $\left| {{c_1} - {c_2}} \right| = 2\sqrt {{a^2} - {b^2}{{\sin }^2}A}$ is the correct answer.

Note:- In these types of questions wherever there are two values of a variable is asked, it is advisable to proceed by finding out the quadratic in that variable. The properties of sum and product of the roots of a quadratic equation along with algebraic identities is used simultaneously to find out the difference between the roots of the quadratic equation.