Examples on Cosine Rule Set 2

Go back to  'Trigonometry'

Example - 46

Let  $$\alpha ,\;\beta ,\;\gamma$$  be the lengths of the altitudes in  $$\Delta ABC$$ . Prove that

$\frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} + \frac{1}{{{\gamma ^2}}} = \frac{{\cot A + \cot B + \cot C}}{\Delta }$

Solution: Note that

\begin{align}&\qquad\quad\;\Delta = \frac{1}{2}a\alpha = \frac{1}{2}bc\sin A\\\ & \Rightarrow\qquad \frac{1}{\alpha } = \frac{a}{{2\Delta }}\; \;\;\ {\text{and}} \;\; \sin A = \frac{{2\Delta }}{{bc}} & {\text{etc}} \\ & \Rightarrow\qquad {\text{LHS}} = \frac{{{a^2} + {b^2} + {c^2}}}{{4{\Delta ^2}}} \end{align}

Now,

\begin{align}&{\text{RHS}} = \frac{{\cot A + \cot B + \cot C}}{\Delta }\\\\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad=\frac{{\begin{align}\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}\end{align}}}{\Delta } \\\\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad=\frac{{\begin{align}\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{(2bc)\cdot \left( \frac{2\Delta }{bc} \right)}+\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{(2ca)\cdot \left( \frac{2\Delta }{ca} \right)}+\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{(2ab)\cdot \left( \frac{2\Delta }{ab} \right)}\end{align}}}{\Delta }\\ \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{{{a^2} + {b^2} + {c^2}}}{{4{\Delta ^2}}} = {\text{LHS}} \end{align}

Example - 47

Two circles of radii a and b intersect at angle $$\theta$$ . Find the length of the common chord.

Solution:

We have                                                                                          \begin{align}\angle \,{O_1}PQ = \frac{\pi }{2} - \theta ,\;\angle \,{O_2}PR = \frac{\pi }{2} - \theta \end{align}, so that

\begin{align}&\angle \,{O_1}P{O_2} = \left( {\frac{\pi }{2} - \theta } \right) + \theta + \left( {\frac{\pi }{2} - \theta } \right) \\ &\qquad\quad \,\,\,\, = \pi - \theta \\ \end{align}

Applying the cosine rule in  $$\Delta {O_1}P{O_2}$$   yields

${O_1}O_2^2 = {a^2} + {b^2} + 2ab\cos \theta$

Now, if $$\Delta$$ is the area of  $$\Delta {O_1}P{O_2}$$  , then

$\Delta = \frac{1}{2}({O_1}\,{O_2})\;(PT) = \frac{1}{2}ab\sin \theta$

\begin{align} & \Rightarrow\qquad PT = \frac{{ab\sin \theta }}{{{O_1}{O_2}}} \\ & \Rightarrow\qquad PS = 2PT = \frac{{2ab\sin \theta }}{{{O_1}{O_2}}} \\ &\qquad\qquad\qquad\quad\quad\;= \frac{{2ab\sin \theta }}{{\sqrt {{a^2} + {b^2} + 2ab\cos \theta } }} \end{align}