Question

In a \[\vartriangle ABC\], the values of 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 \[\sin A = \] ?
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

Answer 1

Hint: We use the law of cosines to calculate the cosine of angle A. Form a quadratic equation with variable ‘b’. Since \[{b_1},{b_2}\] are two values of the third side b means that \[{b_1},{b_2}\] are two roots of quadratic equation. Use the relation of roots with the coefficients of the equation to find the value of sine of angle A.
* In triangle ABC, if side a has opposite angle A, side b has opposite angle B and side c has opposite angle C then \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
* If a quadratic equation \[a{x^2} + bx + c = 0\] has roots \[\alpha ,\beta \] then we have relation between coefficients of equation and roots of equation as: \[\alpha + \beta = \dfrac{{ - b}}{a}\] and \[\alpha \beta = \dfrac{c}{a}\]

Complete step by step solution:
We have a triangle ABC, sides opposite to angle A, B and C are a, b and c respectively.

Use law of cosines
\[ \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
Cross multiply the values
\[ \Rightarrow 2bc\cos A = {b^2} + {c^2} - {a^2}\]
Shift all values to one side of the equation
\[ \Rightarrow {b^2} - 2bc\cos A + {c^2} - {a^2} = 0\]
This forms a quadratic equation in ‘b’
\[ \Rightarrow {b^2} - \left( {2c\cos A} \right)b + \left( {{c^2} - {a^2}} \right) = 0\]...............… (1)
Since it is given \[{b_1},{b_2}\]are two values of the third side b
Then \[{b_1},{b_2}\]are the roots of the equation (1)
Then we know \[{b_1} + {b_2} = \dfrac{{ - ( - 2c\cos A)}}{1}\]and\[{b_1}{b_2} = \dfrac{{{c^2} - {a^2}}}{1}\]
I.e. \[{b_1} + {b_2} = 2c\cos A\]and\[{b_1}{b_2} = {c^2} - {a^2}\]....................… (2)
Also, \[{b_2} = 2{b_1}\]
Substitute the value of \[{b_2} = 2{b_1}\]in \[{b_1} + {b_2} = 2c\cos A\]
\[ \Rightarrow {b_1} + 2{b_1} = 2c\cos A\]
\[ \Rightarrow 3{b_1} = 2c\cos A\]
Divide both sides by 3
\[ \Rightarrow {b_1} = \dfrac{{2c\cos A}}{3}\].................… (3)
Substitute the value of \[{b_2} = 2{b_1}\]in \[{b_1}{b_2} = {c^2} - {a^2}\]
\[ \Rightarrow {b_1} \times 2{b_1} = {c^2} - {a^2}\]
\[ \Rightarrow 2{b_1}^2 = {c^2} - {a^2}\]
Divide both sides by 2
\[ \Rightarrow {b_1}^2 = \dfrac{{{c^2} - {a^2}}}{2}\].............… (4)
Square equation (3) and equate it to equation (4)
\[ \Rightarrow {\left( {\dfrac{{2c\cos A}}{3}} \right)^2} = \dfrac{{{c^2} - {a^2}}}{2}\]
Cross multiply the values
\[ \Rightarrow \dfrac{{2 \times 4{c^2}{{\cos }^2}A}}{9} = {c^2} - {a^2}\]
\[ \Rightarrow \dfrac{{8{c^2}{{\cos }^2}A}}{9} = {c^2} - {a^2}\]
Cross multiply the values
\[ \Rightarrow 8{c^2}{\cos ^2}A = 9{c^2} - 9{a^2}\]
Divide both sides by \[8{c^2}\]
\[ \Rightarrow {\cos ^2}A = \dfrac{{9{c^2} - 9{a^2}}}{{8{c^2}}}\]
Since we know \[{\cos ^2}x = 1 - {\sin ^2}x\]
\[ \Rightarrow 1 - {\sin ^2}A = \dfrac{{9{c^2} - 9{a^2}}}{{8{c^2}}}\]
Shift all constant values to one side of the equation
\[ \Rightarrow 1 - \dfrac{{9{c^2} - 9{a^2}}}{{8{c^2}}} = {\sin ^2}A\]
Take LCM in LHS
\[ \Rightarrow \dfrac{{8{c^2} - 9{c^2} + 9{a^2}}}{{8{c^2}}} = {\sin ^2}A\]
\[ \Rightarrow \dfrac{{9{a^2} - {c^2}}}{{8{c^2}}} = {\sin ^2}A\]
Take Square root on both sides of the equation
\[ \Rightarrow \sqrt {\dfrac{{9{a^2} - {c^2}}}{{8{c^2}}}} = \sqrt {{{\sin }^2}A} \]
Cancel square root by square power
\[ \Rightarrow \sin A = \sqrt {\dfrac{{9{a^2} - {c^2}}}{{8{c^2}}}} \]
\[\therefore \]The value of \[\sin A\] is \[\sqrt {\dfrac{{9{a^2} - {c^2}}}{{8{c^2}}}} \].

\[\therefore \]Correct option is B.

Note: Students are likely to make the mistake of assuming the variable in the quadratic equation as ‘c’ which is wrong because we are given the values of a, c and A are given. The question states the values of ‘b’ are \[{b_1},{b_2}\] so the variable has to be ‘b’ in the quadratic equation then only we can find its roots.