# 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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