In a triangle \[ABC\] \[a,c,A\] are given and \[{b_1},{b_2}\] are two values of the third side \[b\] such that \[{b_2} = 2{b_1}\]. Then what is the value of \[\sin A\]?
A. \[\sqrt {\dfrac{{9{a^2} - {c^2}}}{{8{a^2}}}} \]
B. \[\sqrt {\dfrac{{9{a^2} - {c^2}}}{{8{c^2}}}} \]
C. \[\sqrt {\dfrac{{9{a^2} + {c^2}}}{{8{a^2}}}} \]
D. None of these

Hint: First, use the formula of the cosine of the angle \[A\]. Then, simplify the formula and rewrite it as the quadratic equation. After that, substitute the given conditions in the equation of the sum and product of the roots of the equation. In the end, Solve the equation and calculate the required answer.



Formula Used:If \[\alpha ,\beta \] are the roots of a quadratic equation \[a{x^2} + bx + c = 0\] , then
Sum of the roots: \[\alpha + \beta = \dfrac{{ - b}}{a}\]
Product of the roots: \[\alpha \beta = \dfrac{c}{a}\]
\[{\sin ^2}x + {\cos ^2}x = 1\]



Complete step by step solution:Given:
\[a,b,c\] are the length of the sides of the triangle \[ABC\]
\[{b_1},{b_2}\] are two values of the third side \[b\] such that \[{b_2} = 2{b_1}\]

Use the formula of the cosine of the angle \[A\].
\[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]

\[ \Rightarrow 2bc\cos A = {b^2} + {c^2} - {a^2}\]
\[ \Rightarrow {b^2} - 2bc\cos A = - {c^2} + {a^2}\]
\[ \Rightarrow {b^2} - b\left( {2c\cos A} \right) + \left( {{c^2} - {a^2}} \right) = 0\]
Here, it is given that \[{b_1},{b_2}\] are the roots of the above quadratic equation.
By using the formulas of the sum and product of the roots of the quadratic equation, we get
Sum of the roots: \[{b_1} + {b_2} = 2c\cos A\] \[.....\left( 1 \right)\]
Product of the roots: \[{b_1}{b_2} = {c^2} - {a^2}\] \[.....\left( 2 \right)\]

It is also given that, \[{b_2} = 2{b_1}\]
So, substitute \[{b_2} = 2{b_1}\] in the equation \[\left( 1 \right)\].
\[{b_1} + 2{b_1} = 2c\cos A\]
\[ \Rightarrow 3{b_1} = 2c\cos A\]
\[ \Rightarrow {b_1} = \dfrac{2}{3}c\cos A\] \[.....\left( 3 \right)\]

Again, substitute \[{b_2} = 2{b_1}\] in the equation \[\left( 2 \right)\].
\[{b_1}\left( {2{b_1}} \right) = {c^2} - {a^2}\]
\[ \Rightarrow 2{b_1}^2 = {c^2} - {a^2}\] \[.....\left( 4 \right)\]

Substitute equation \[\left( 3 \right)\] in the equation \[\left( 4 \right)\].
\[2{\left( {\dfrac{2}{3}c\cos A} \right)^2} = {c^2} - {a^2}\]
\[ \Rightarrow 2\left( {\dfrac{4}{9}{c^2}{{\cos }^2}A} \right) = {c^2} - {a^2}\]
\[ \Rightarrow \dfrac{8}{9}{c^2}{\cos ^2}A = {c^2} - {a^2}\]
\[ \Rightarrow {\cos ^2}A = \dfrac{{9{c^2} - 9{a^2}}}{{8{c^2}}}\]
Apply the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] on the left-hand side.
 \[1 - {\sin ^2}A = \dfrac{{9{c^2} - 9{a^2}}}{{8{c^2}}}\]
\[ \Rightarrow {\sin ^2}A = 1 - \dfrac{{9{c^2} - 9{a^2}}}{{8{c^2}}}\]
\[ \Rightarrow {\sin ^2}A = \dfrac{{8{c^2} - 9{c^2} + 9{a^2}}}{{8{c^2}}}\]
\[ \Rightarrow {\sin ^2}A = \dfrac{{9{a^2} - {c^2}}}{{8{c^2}}}\]
Take the square root on both sides.
\[\sin A = \sqrt {\dfrac{{9{a^2} - {c^2}}}{{8{c^2}}}} \]



Option ‘B’ is correct



Note: Students often mistake to apply the sum of roots and product of roots. They take the sum of roots is \[\dfrac{b}{a}\] and product of roots is \[\dfrac{a}{c}\] which are incorrect formulas. The correct formulas are the sum of roots is \[ - \dfrac{b}{a}\] and product of roots is \[\dfrac{c}{a}\].