Examples on Cosine Rule Set 3
Example - 48
Suppose that the medians of a \(\Delta ABC\) make angles \(\alpha ,\;\beta ,\;\gamma \) with each other. Find the value of \(S = \cot \alpha + \cot \beta + \cot \gamma + \cot A + \cot B + \cot C\)
Solution:
To proceed, we first evaluate the lengths of the medians of the triangle in terms of the sides. Consider the following diagram:
By the pythagoras theorem,
\[{x^2} = {p^2} + D{E^2} = \frac{1}{2}\left\{ {{p^2} + {p^2} + 2D{E^2}} \right\}\]
Note that \(\begin{align}{p^2} = {c^2} - B{E^2} = {b^2} - C{E^2},\;{\text{and}}\;DE = \frac{a}{2} - BE = CE - \frac{a}{2}\end{align}\) .
Thus,
\[\begin{align}&{x^2} = \frac{1}{2}\left\{ {({c^2} - B{E^2}) + ({b^2} - C{E^2}) + 2\left( {\frac{a}{2} - BE} \right)\left( {CE - \frac{a}{2}} \right)} \right\}\\ \,\,\,\,\,\,\, &\quad= \frac{1}{2}\left\{ {{b^2} + {c^2} - (B{E^2} + C{E^2}) - \frac{{{a^2}}}{2} + a(CE + BE) - 2BE \cdot CE} \right\} \\ \,\,\,\,\,\,\, &\quad= \frac{1}{2}\left\{ {{b^2} + {c^2} - \frac{{{a^2}}}{2} + {a^2} - {{(BE + CE)}^2}} \right\} \\ \,\,\,\,\,\,\, &\quad= \frac{1}{2}\left\{ {{b^2} + {c^2} - \frac{{{a^2}}}{2}} \right\} = \frac{{2{b^2} + 2{c^2} - {a^2}}}{4} \qquad\qquad...(1) \end{align} \]
Similar expressions will hold for the other medians. Now, we return to the original problem.
Applying the cosine rule in \(\Delta BGC\) , we have
\[B{C^2} = {a^2} = B{G^2} + C{G^2} - 2BG \cdot CG \cdot \cos \alpha \]
Now, since G is the centroid, we have \(\begin{align}BG = \frac{2}{3}BE,\;CG = \frac{2}{3}CF,\end{align}\) so that
\[{a^2} = \frac{4}{9}B{E^2} + \frac{4}{9}C{F^2} - \frac{8}{9}BE \cdot CF \cdot \cos \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]
Also, if the area of \(\Delta BGC\) is \({\Delta _1}\) , then
\[{\Delta _1} = \frac{1}{2} \cdot BG \cdot CG \cdot \sin \alpha = \frac{2}{9}BE \cdot CF \cdot \sin \alpha \]
But \({\Delta _1}\) is one thirds of the total area \(\Delta \) of \(\Delta ABC\) (why?), so that
\[BE \cdot CF = \frac{{3\Delta }}{{2\sin \alpha }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)\]
Using (3) in (2), we have
\[\begin{align} &\qquad\quad \;\;\,{a^2} = \frac{4}{9}\left( {B{E^2} + C{F^2}} \right) - \frac{{4\Delta }}{3}\cot \alpha \\ & \Rightarrow \quad \cot \alpha = \frac{{4(B{E^2} + C{F^2}) - 9{a^2}}}{{12\Delta }} \\ \end{align} \]
Similar expressions hold for \( \cot \beta \;{\text{and}}\;\cot \gamma \) . Summing the expressions for the three ‘cot’s, and using the result of (1) and simiplfying, we have
\[\begin{align} \cot \alpha + \cot \beta + \cot \gamma = \frac{{8(A{D^2} + B{E^2} + C{F^2}) - 9({a^2} + {b^2} + {c^2})}}{{12\Delta }} \\ \,\,\,\,\,\,\,\,\,\,\, = \frac{{ - ({a^2} + {b^2} + {c^2})}}{{4\Delta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 4 \right) \\ \end{align} \]
Also, if you refer to Example – 46, you’ll note that we’ve already calculated the sum \(\cot A + \cot B + \cot C\) :
\[\cot A + \cot B + \cot C = \frac{{({a^2} + {b^2} + {c^2})}}{{4\Delta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 5 \right)\]
From (4) and (5), the required sum is zero.
Example - 49
Consider a triangle \(\Delta ABC\) . Let \(\Psi \) be the circumcircle of \(\Delta ABC\) . At each of the three points A, B, C, tangents are drawn to \(\Psi \) to form another triangle, \(\Delta PQR\) . Find the sides of \(\Delta PQR\) .
Solution:
Let us evaluate the side PQ.
Note that
\[\angle PCB = \angle PBC = \angle A\qquad\qquad(why?)\]
so that
\[\angle P = \pi - 2\angle A\]
Applying the sine rule in \(\Delta PBC\) , we have
\[\begin{align} & \qquad \frac{{PB}}{{\sin A}} = \frac{{PC}}{{\sin A}} = \frac{{BC}}{{\sin 2A}} \\ &\Rightarrow \quad PC = \frac{{BC\sin A}}{{\sin 2A}} = \frac{a}{{2\cos A}} \\ \end{align} \]
Similarly, \(\begin{align}QC = \frac{b}{{2\cos B}}\end{align}\) . Thus,
\[\begin{align}&PQ = PC + CQ = \frac{a}{{2\cos A}} + \frac{b}{{2\cos B}} = \frac{1}{2}\frac{{(a\cos B + b\cos A)}}{{\cos A\cos B}} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;= \frac{{\cos C}}{{2\cos A\cos B}} \\ \end{align} \]
Similar expressions will hold for QR and RP.