Examples on Cosine Rule Set 3

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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.

 

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