Examples on Sine Rule Set 2

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Example – 37

Let O be a point inside  \(\Delta ABC\) such that  \(\angle OAB = \angle OBC = \angle OCA = \omega \)  .

Prove the following.

(a)  \(\cot \omega  = \cot A + \cot B + \cot C\)

(b)  \({\text{cose}}{{\text{c}}^2}\omega  = {\text{cose}}{{\text{c}}^2}A + {\text{cose}}{{\text{c}}^2}B + {\text{cose}}{{\text{c}}^2}C\)

Solution: (a) In  \(\Delta \,OAB\) ,

\[\frac{{OB}}{{\sin \omega }} = \frac{c}{{\sin \angle AOB}}\]

But  \(\angle AOB = \pi  - (\omega  + \angle ABO) = \pi  - (\omega  + \angle B - \omega ) = \pi  - \angle B\)

\[ \Rightarrow   \frac{{OB}}{{\sin \omega }} = \frac{c}{{\sin B}} \Rightarrow OB = \frac{{c\sin \omega }}{{\sin B}}\]

Similarly, from

\[\begin{align}&\qquad\quad\Delta OBC,\;\;\;\;OB = \frac{{a\sin (C - \omega )}}{{\sin C}}\qquad\;({\text{Verify}}\,{\text{!}})\\ & \Rightarrow\quad  \frac{{c\sin \omega }}{{\sin B}} = \frac{{a\sin (C - \omega )}}{{\sin C}} \qquad\qquad\qquad ...(1)  \end{align} \]

Further, from the sine rule,  \(c = \lambda \sin C,\;a = \lambda \sin A\)  . Substituting this in (1) gives

\[\begin{align} &\qquad\;\frac{{\sin C\sin \omega }}{{\sin B}} = \frac{{\sin A\sin (C - \omega )}}{{\sin C}}\\   \Rightarrow\quad   &{\sin ^2}C\sin \omega  = \sin A\sin B\sin (C - \omega ) \\   \Rightarrow\quad   &\sin C\sin (A + B)\sin \omega  = \sin A\sin B\sin (C - \omega ) \\ \end{align}\]

Simplifying this yields the desired result.

(b) Recall that in any \(\Delta ABC\)  , 

\[\begin{align}  \tan A + \tan B + \tan C = \tan A\tan B\tan C  \\  \cot B\cot C + \cot C\cot A + \cot A\cot B = 1 \\ \end{align} \]

Now, in part - (a), we proved that

\[\cot \omega  = \cot A + \cot B + \cot C\]

Squaring this yields

\[\begin{align}&\qquad\quad{\cot ^2}\omega = {\cot ^2}A + {\cos ^2}B + {\cot ^2}C + 2 \\ &  \Rightarrow\quad {\text{cose}}{{\text{c}}^2}\omega - 1 = ({\text{cose}}{{\text{c}}^2}A - 1) + ({\text{cose}}{{\text{c}}^2}B - 1) + ({\text{cose}}{{\text{c}}^2}C - 1) + 2 \\&  \Rightarrow\quad {\text{cose}}{{\text{c}}^2}\omega = {\text{cose}}{{\text{c}}^2}A + {\text{cose}}{{\text{c}}^2}B + {\text{cose}}{{\text{c}}^2}C \\ \end{align} \]

Example – 38

ABCD is a trapezium such that AB and CD are parallel and CB is perpendicular to them. If   \(\angle ADB = \theta ,\;BC = p\;{\text{and}}\;CD = q\) , show that

\[AB = \frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{p\cos \theta  + q\sin \theta }}\]

Solution:

Applying the sine rule to  \(\Delta ABD\), we have

\[\begin{align} & \qquad \frac{AB}{\sin \theta }=\frac{BD}{\sin \left( \theta +\phi  \right)} \\  & \Rightarrow \quad AB=\frac{BD\sin \theta }{\sin \theta \cos \phi +\cos \theta \sin \phi }=\frac{\left( \sqrt{{{p}^{2}}+{{q}^{2}}} \right)\sin \theta }{\begin{align}{\sin \theta \left( \frac{q}{\sqrt{{{p}^{2}}+{{q}^{2}}}} \right)+\cos \theta \left( \frac{p}{\sqrt{{{p}^{2}}+{{q}^{2}}}} \right)}\end{align}} \\  & \qquad \qquad =\frac{\left( {{p}^{2}}+{{q}^{2}} \right)\sin \theta }{p\cos \theta +q\sin \theta } \end{align}\]

Example - 39

In a triangle of base a, the ratio of the other two sides is r ( < 1). Prove that the maximum altitude of the triangle can be   \(\begin{align}\frac{{ar}}{{1 - {r^2}}}\end{align}\) . Find the vertical angle of the triangle in case the altitude has this maximum value.

Solution:

Assuming a fixed, what we’ve to do here is express h as a function of a, r and the angles of the triangle, and then find the maximum value that h can take.

Note that

\[\begin{align}&\Delta  = \frac{1}{2}ah = \frac{1}{2}bc\sin A \\   \Rightarrow\quad   &h = \frac{{bc\sin A}}{a} = \frac{{abc\sin A}}{{{a^2}}}  \\\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;\;= \frac{{abc\sin A\left( {{{\sin }^2}B - {{\sin }^2}C} \right)}}{{{a^2}\left( {{{\sin }^2}B - {{\sin }^2}C} \right)}} \\  \,\,\,\,\,\,\,\,\,\,\,&\qquad\qquad\quad\;\; = \frac{{abc\;{{\sin }^2}A\sin (B - C)}}{{({b^2} - {c^2})\;{{\sin }^2}A}}\qquad\quad ({\text{how}}?)  \\    \,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;\;= \frac{{ar\sin (B - C)}}{{1 - {r^2}}}  \\   \Rightarrow\quad   &{h_{\max }} = \frac{{ar}}{{1 - {r^2}}}  \\ \end{align} \]

In this case,

\[\begin{align}&B - C = \frac{\pi }{2} \qquad  \Rightarrow\qquad B = \frac{\pi }{2} + C \\  & \Rightarrow \qquad \frac{c}{b} = \frac{{\sin C}}{{\sin B}} = \frac{{\sin C}}{{\cos C}} = \tan C = r   \quad\Rightarrow\quad C = {\tan ^{ - 1}}r  \\ \end{align} \]

\[\begin{align}&\Rightarrow\qquad  A = \pi  - (B + C) \\\,\,\,\,\, &\qquad\qquad\;= \frac{\pi }{2} - 2C \\\,\,\,\,\, &\qquad\qquad\;= \frac{\pi }{2} - 2{\tan ^{ - 1}}r \\ \end{align} \]

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