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

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 $\cos A,\cos B{\text{ and }}\cos C$.

\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}}$$.