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