Question

Given \[\dfrac{{b + c}}{{11}} = \dfrac{{c + a}}{{12}} = \dfrac{{a + b}}{{13}}\] for \[\Delta ABC\] with usual notation. If \[\dfrac{{cosA}}{\alpha } = \dfrac{{\cos B}}{\beta } = \dfrac{{\cos C}}{\gamma }\], then what will be values for ordered triad \[(\alpha, \beta, \gamma)\] ?
(a) (3, 4, 5)
(b) (19, 7, 25)
(c) (7, 19, 25)
(d) (5, 12, 13)

Answer 1

Hint: To derive the values of \[\alpha ,\beta ,\gamma \], we have to calculate values of a, b and c. We will get these values by equating given equations with any constant. After that we can use following cosine formulas for Angles A, B and C to get values of \[\cos A,\cos B{\text{ and }}\cos C\].
\[\begin{array}{l}
\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}
\end{array}\]

Complete step-by-step solution:
By comparing given equation of cosine with calculated cosines equations, we will get values of \[\alpha ,\beta {\rm{ and }}\alpha \].
\[\begin{array}{l}
{\text{Let’s consider }}\dfrac{{b + c}}{{11}} = \dfrac{{c + a}}{{12}} = \dfrac{{a + b}}{{13}} = \lambda (say)\\
b + c = 11\lambda {\rm{ }}...{\rm{(i)}}\\
c + a = 12\lambda {\rm{ }}...{\rm{(ii)}}\\
a + b = 13\lambda {\rm{ }}...{\rm{(iii)}}
\end{array}\]
By adding Eqs. (i), (ii) and (iii), we get
\[\begin{array}{l}
b + c + a + c + a + b = 11\lambda + 12\lambda + 13\lambda \\
 \Rightarrow 2a + 2b + 2c = 36\lambda \\
 \Rightarrow 2(a + b + c) = 36\lambda \\
 \Rightarrow a + b + c = 18\lambda {\rm{ }}...{\rm{(i)}}
\end{array}\]
Now, we will find values of a in terms of \[\lambda \]
Subtracting Eq.(i) from Eq.(iv), we get
\[\begin{array}{l}
a + b + c - b - c = 18\lambda - 11\lambda \\
 \Rightarrow a = 7\lambda
\end{array}\]
Now, we will find values of b in terms of \[\lambda \]
Subtracting Eq.(ii) from Eq.(iv), we get
\[\begin{array}{l}
a + b + c - a - c = 18\lambda - 12\lambda \\
 \Rightarrow b = 6\lambda
\end{array}\]
Now, we will find values of c in terms of \[\lambda \]
Subtracting Eq.(iii) from Eq.(iv), we get
\[\begin{array}{l}
a + b + c - a - b = 18\lambda - 13\lambda \\
 \Rightarrow c = 5\lambda
\end{array}\]
Now, we get \[a = 7\lambda ,b = 6\lambda {\text{ and }}c = 5\lambda \]
To derive values of \[\cos A,\cos B{\text{ and }}\cos C\], we can use cosine formulas.
Firstly, we will derive value of \[\cos A\]
\[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
Now, we can put values of a, b and c in this equation.
\[ = \dfrac{{{{(6\lambda )}^2} + {{(5\lambda )}^2} - {{(7\lambda )}^2}}}{{2 \times 6\lambda \times 7\lambda }}\]
This equation can be written as,
\[ = \dfrac{{36{\lambda ^2} + 25{\lambda ^2} - 49{\lambda ^2}}}{{60{\lambda ^2}}}\]
Now, we can solve this equation to get value of \[\cos A\].
\[\begin{array}{l}
 = \dfrac{{61{\lambda ^2} - 49{\lambda ^2}}}{{60{\lambda ^2}}}\\
 = \dfrac{{12{\lambda ^2}}}{{60{\lambda ^2}}}\\
 = \dfrac{1}{5}
\end{array}\]
Now, we will find value of \[\cos B \]
\[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]
Now, we can put values of a, b, and c in this equation.
\[ = \dfrac{{{{(7\lambda )}^2} + {{(5\lambda )}^2} - {{(6\lambda )}^2}}}{{2 \times 7\lambda \times 5\lambda }}\]
This equation can be written as,
\[ = \dfrac{{49{\lambda ^2} + 25{\lambda ^2} - 36{\lambda ^2}}}{{70{\lambda ^2}}}\]
Now, we can solve this equation to get value of \[\cos B\].
\[\begin{array}{l}
 = \dfrac{{74{\lambda ^2} - 36{\lambda ^2}}}{{70{\lambda ^2}}}\\
 = \dfrac{{38{\lambda ^2}}}{{70{\lambda ^2}}}\\
 = \dfrac{{19}}{{35}}
\end{array}\]
Now, we will find value of \[\cos C\]
\[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
Now, we can put values of a, b, and c in this equation.
\[ = \dfrac{{{{(7\lambda )}^2} + {{(6\lambda )}^2} - {{(5\lambda )}^2}}}{{2 \times 7\lambda \times 6\lambda }}\]
This equation can be written as,
\[ = \dfrac{{49{\lambda ^2} + 36{\lambda ^2} - 25{\lambda ^2}}}{{84{\lambda ^2}}}\]
Now, we can solve this equation to get value of \[\cos B\].
\[\begin{array}{l}
 = \dfrac{{85{\lambda ^2} - 25{\lambda ^2}}}{{84{\lambda ^2}}}\\
 = \dfrac{{60{\lambda ^2}}}{{84{\lambda ^2}}}\\
 = \dfrac{5}{7}
\end{array}\]
We can rewrite these equations in ratio form as
\[\begin{array}{l}
\cos A:\cos B:\cos C = \dfrac{1}{5}:\dfrac{{19}}{{35}}:\dfrac{5}{7}\\
\therefore \dfrac{{\cos A}}{7}:\dfrac{{\cos B}}{{19}}:\dfrac{{\cos C}}{{25}}{\rm{ }}...{\rm{(v)}}
\end{array}\]
But, given equation is
 \[\dfrac{{cosA}}{\alpha } = \dfrac{{\cos B}}{\beta } = \dfrac{{\cos C}}{\gamma }{\rm{ }}...{\rm{(vi)}}\]
By comparing Eqs. (v) and (vi), we get
\[ \Rightarrow \alpha = 7,\beta = 19,\gamma = 25\]
Hence, the option (c) is correct.

Note: Calculation plays an important role in these types of trigonometric problems. Students get confused when to use cosine rules between \[\cos A, \cos B\text{ }and\text { }\cos C \]. One can do mistake like \[\cos B = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\] instead of \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\].