Examples on Sine Rule Set 3

Example – 40

Find the ratio a : b : c in a  \(\Delta ABC\)  : if

\[\cos A\cos B + \sin A\sin B\sin C = 1\]

Solution: Note that

\[\begin{align}&\sin C = \frac{{1 - \cos A\cos B}}{{\sin A\sin B}} \leqslant 1  \\   \Rightarrow\qquad   &\sin A\sin B + \cos A\cos B = \cos (A - B) \geqslant 1\\ \end{align} \]

This necessarily implies that

\[\begin{align}&\cos (A - B) = 1\\   \Rightarrow\qquad   &\angle A = \angle B  \\ \end{align} \]

\[\begin{align}&\Rightarrow  \quad \sin C = \frac{{1 - {{\cos }^2}A}}{{{{\sin }^2}A}} = 1 \quad  \Rightarrow\quad  C = \frac{\pi }{2}  \\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\Rightarrow  A = B = \frac{\pi }{4}  \\ \end{align} \]

Since \(\begin{align}\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\end{align}\)  from the sine rule, we have

\[\begin{align}&\frac{a}{{1/\sqrt 2 }} = \frac{b}{{1/\sqrt 2 }} = \frac{c}{1}  \\   \Rightarrow\qquad   &a:b:c = 1:1:\sqrt 2   \\ \end{align}\]

Example - 41

Let  \(\alpha ,\;\beta ,\;\gamma \)   be the lengths of the internal angle bisectors of angles A, B, C respectively in a  \(\Delta ABC\) . Find an expression for

\[S = \frac{1}{\alpha }\cos \frac{A}{2} + \frac{1}{\beta }\cos \frac{B}{2} + \frac{1}{\gamma }\cos \frac{C}{2}\]

in terms of a, b, c

Solution: Let us first find an expression for  \(\alpha \) :

Note that

\[\begin{align}&{\text{area}}\;(\Delta ABD) + {\text{area}}\;(\Delta ACD) = {\text{area}}\;(\Delta ABC)  \\   \Rightarrow\qquad  &\frac{1}{2}c\alpha \sin \frac{A}{2} + \frac{1}{2}b\alpha \sin \frac{A}{2} = \frac{1}{2}bc\sin A \\   \Rightarrow\qquad  &  \alpha  = \frac{{bc\sin A}}{{(b + c)\sin \frac{A}{2}}} = \frac{{2bc}}{{b + c}}\cos \frac{A}{2}  \\ \end{align} \]

Similar expressions hold for  \(\beta \) and   \(\gamma \) . We note that

\[\begin{align}&\frac{1}{\alpha }\cos \frac{A}{2} = \frac{{b + c}}{{2bc}}  \\   \Rightarrow\qquad &S = \frac{{b + c}}{{2bc}} + \frac{{c + a}}{{2ca}} + \frac{{a + b}}{{2ab}} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\;\;= \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ \end{align} \]

Example – 42

In a  \(\Delta ABC\) , the sides are in A.P. and the greatest angle exceeds the least by  \(\begin{align}\frac{\pi }{2}\end{align}\) . Prove that the squares of two its sides will be proportional to the roots of the equation \({x^2} - 16x + 36\)  .

Solution: Let  \(\angle \,C\)  be the greatest angle and  \(\angle \,A\)  the least, so that a, b, c is an increasing A.P.

\[\angle \,C = \angle \,A + \frac{\pi }{2},\;\;2b = a + c\]

Using the sine rule on the second expression,

\[2\sin B = \sin A + \sin C = \sin A + \cos A\]

Squaring, and noting that  \(\begin{align}2\angle A = \frac{\pi }{2} - \angle B,\end{align}\) we have

\(\begin{align}\cos B = \frac{3}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(verify!)} \end{align}\)

Since \(\begin{align}2\angle \,A = \frac{\pi }{2} - \angle \,B\end{align}\) ,

\[\begin{align}&\cos 2A = \sin B \\   \Rightarrow\quad   &1 - 2{\sin ^2}A = \sin B = \sqrt {1 - {{\cos }^2}B}  = \frac{{\sqrt 7 }}{4} \\   \Rightarrow\quad   &  {\sin ^2}A = \frac{{8 - 2\sqrt 7 }}{{16}} \\   \Rightarrow\quad   &   {\sin ^2}C = {\cos ^2}A = \frac{{8 + 2\sqrt 7 }}{{16}}  \\   \Rightarrow \quad   &  {c^2}:{a^2} = 8 + 2\sqrt 7 :8 - 2\sqrt 7  \\ \end{align} \]

If we consider the quadratic \({x^2} - 16x + 36 = 0\) , it roots    \(\alpha \,\,and\,\,\beta \) are given by

\[\begin{align} &\qquad\quad  \alpha ,\ \beta =\frac{16\pm 4\sqrt{7}}{2}=8\pm 2\sqrt{7} \\  & \Rightarrow \quad\;\; \alpha :\beta =8+2\sqrt{7}:8-2\sqrt{7} \\  & \Rightarrow \quad  {{c}^{2}}:{{a}^{2}}=\alpha :\beta  \end{align}\]