Question

In ambiguous if $a$, $b$ and ${\text{A}}$ are given and if there are two possible values of third side, are ${c_1}$ and ${c_2}$, then:-
A) ${c_1} - {c_2} = 2\sqrt {\left( {{a^2} - {b^2}{{\sin }^2}{\text{A}}} \right)} $
B) ${c_1} - {c_2} = 3\sqrt {\left( {{a^2} - {b^2}{{\sin }^2}{\text{A}}} \right)} $
C) ${c_1} - {c_2} = 4\sqrt {\left( {{a^2} + {b^2}{{\sin }^2}{\text{A}}} \right)} $
D) ${c_1} - {c_2} = 2\sqrt {\left( {{a^2} + {b^2}{{\sin }^2}{\text{A}}} \right)} $

Answer 1

Hint: First use the cosine rule to form a quadratic equation in terms of $c$, and then find the sum and the product of roots using that quadratic equation and after that, use these values to find the required difference of roots.

Complete step-by-step answer:
We have given that in ambiguous, $a$,$b$ and ${\text{A}}$ are given and there are two possible roots of third side which are${c_1}$ and ${c_2}$then we have to find the${c_1} - {c_2}$.
First of all we know using the cosine rule that:
$\cos {\text{A}} = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
Now, this equation can be written as,
$ \Rightarrow {c^2} - 2b\left( {\cos {\text{A}}} \right)c + \left( {{b^2} - {a^2}} \right) = 0{\text{ }} \to \left( 1 \right)$
Now, we have that equation in a quadratic equation in terms of $c$ which is equation (1).
Now, compare the given equation with$x{c^2} + yc + z = 0$, then we have
$x = 1,y = - 2b\cos A$and$z = {b^2} - {a^2}$.
Now, we will let ${c_1}$ and ${c_2}$ as the two roots of the equation (1).
 Then the sum of roots will be given as:
${c_1} + {c_2} = \dfrac{{ - y}}{x}$
We will substitute the values of $y$and$x$.
 ${c_1} + {c_2} = 2b\cos A$
Also the product of roots is given as:
\[{c_1}{c_2} = \dfrac{z}{x}\]
Substitute the value of $x$and$z$.
\[{c_1}{c_2} = {b^2} - {a^2}\]
Now, we have the values:
${c_1} + {c_2} = 2b\cos A$ and \[{c_1}{c_2} = {b^2} - {a^2}\]
Now, we will try to find the value of the difference of roots.
We know that the general formula of algebra:
${\left( {a - b} \right)^2} = {a^2} + {b^2} + 2ab$
Then we can conclude that:
${\left( {{c_1} - {c_2}} \right)^2} = {c_1}^2 + {c_2}^2 - 2{c_1}{c_2}$
Now add and subtract $2{c_1}{c_2}$ in the right hand side of the above equation.
$
  {\left( {{c_1} - {c_2}} \right)^2} = {c_1}^2 + {c_2}^2 - 2{c_1}{c_2} + 2{c_1}{c_2} - 2{c_1}{c_2} \\
   = {c_1}^2 + {c_2}^2 + 2{c_1}{c_2} - 4{c_1}{c_2} \\
 $
Now, we can easily see that at right hand side of equation a formula of ${c_1}^2 + {c_2}^2 + 2{c_1}{c_2} = {\left( {{c_1} + {c_2}} \right)^2}$
So we will right ${c_1}^2 + {c_2}^2 + 2{c_1}{c_2}$ as${\left( {{c_1} + {c_2}} \right)^2}$, then we will obtain
${\left( {{c_1} - {c_2}} \right)^2} = {\left( {{c_1} + {c_2}} \right)^2} - 4{c_1}{c_2}$
${c_1} - {c_2} = \sqrt {{{\left( {{c_1} + {c_2}} \right)}^2} - 4{c_1}{c_2}} $
Now, we will substitute the value of sum of roots and product of roots in the above equation.
${c_1} - {c_2} = \sqrt {{{\left( {2b\cos {\text{A}}} \right)}^2} - 4\left( {{b^2} - {a^2}} \right)} $
Now, we will solve this equation further,
\[{c_1} - {c_2} = \sqrt {{{\left( {2b\cos {\text{A}}} \right)}^2} - 4\left( {{b^2} - {a^2}} \right)} \]
\[{c_1} - {c_2} = \sqrt {4{b^2}{{\cos }^2}{\text{A}} - 4{b^2} + 4{a^2}} \]
\[{c_1} - {c_2} = \sqrt {\left( {4{a^2} - 4{b^2}\left( {1 - {{\cos }^2}{\text{A}}} \right)} \right)} \]
Now, as we know about the identity of cosine angle and sine angle, that is
${\sin ^2}\theta + {\cos ^2}\theta = 1$ .
From the above identity, it can be conclude that:
$1 - {\cos ^2}{\text{A}} = {\sin ^2}{\text{A}}$ .
Now, we will get the equation as:
${c_1} - {c_2} = 2\sqrt {\left( {{a^2} - {b^2}{{\sin }^2}{\text{A}}} \right)} $

Hence, option A is the correct answer.

Note: In the formed quadratic equation using the cosine rule, the variable of the equation is $c$ and other algebraic expressions $a$ and $b$ are taken as constant. On comparing with the quadratic equation, don’t take c as the constant term.