Properties and Solution of Triangles
1.0 Sine rule
1.0 Sine rule
It is the law relating the angles and sides of a triangle. In a triangle $ABC$, the sine of angles of a triangle is proportional to the opposite sides of a triangle.
$$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}$$
Proof: Consider an acute triangle $ABC$, draw an altitude $h$ from the vertex $A$ of a triangle
Now by the definition of sine function, we can write $$\begin{equation} \begin{aligned} \sin B = \frac{h}{c} \\ \sin C = \frac{h}{b} \\\end{aligned} \end{equation} $$
or, $$\begin{equation} \begin{aligned} h = c\sin B \\ h = b\sin C \\\end{aligned} \end{equation} $$
As both are equal to $h$ so these are equal $$c\sin B = b\sin C$$
Divide throughout by $\sin B\sin C$, we get $$\begin{equation} \begin{aligned} \frac{{c\sin B}}{{\sin B\sin C}} = \frac{{b\sin C}}{{\sin B\sin C}} \\ \frac{c}{{\sin C}} = \frac{b}{{\sin B}} \\\end{aligned} \end{equation} ...(i)$$
Repeat the above procedure by drawing the altitude from the vertex $B$.
Using the same method, we can write as $$\frac{c}{{\sin C}} = \frac{a}{{\sin A}}...(ii)$$
So by using these two relations $(i)$ and $(ii)$, we have
$$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}$$
The above relation holds good for the acute triangle. In the case of the obtuse triangle, we need to draw one interior altitude and one exterior altitude.
For the interior altitude, drawn from the vertex $A$, the procedure remains the same and hence we get $$\frac{c}{{\sin C}} = \frac{b}{{\sin B}}...(iii)$$
Draw an exterior altitude from the vertex $B$. To do so, we need to extend the side $b$ as shown in the figure. The angles $BAC$ and $BAK$ are supplementary so the sine of both is same.
Here angle $A$ is $BAK$, so $$\begin{equation} \begin{aligned} \sin A = \frac{h}{c} \\ h = c\sin A \\\end{aligned} \end{equation} $$
and in the triangle $CBK$, we can write $$\begin{equation} \begin{aligned} \sin C = \frac{h}{a} \\ h = a\sin C \\\end{aligned} \end{equation} $$
Since both are equal to $h$ so $$c\sin A = a\sin C$$
Divide throughout by $\sin A\sin C$, we get
$$\begin{equation} \begin{aligned} \frac{{c\sin A}}{{\sin A\sin C}} = \frac{{a\sin C}}{{\sin A\sin C}} \\ \frac{c}{{\sin C}} = \frac{a}{{\sin A}} \\\end{aligned} \end{equation} ...(iv) $$
By using $(iii)$ and $(iv)$, we get $$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}$$
Question 1. In any triangle $ABC$, prove that $$({b^2} - {c^2})\cot A + ({c^2} - {a^2})\cot B + ({a^2} - {b^2})\cot C = 0$$
Solution: From sine rule, we can write as $$\begin{equation} \begin{aligned} a = k\sin A \\ b = k\sin B \\ c = k\sin C \\\end{aligned} \end{equation} $$
therefore, $$\begin{equation} \begin{aligned} ({b^2} - {c^2})\cot A = {k^2}({\sin ^2}B - {\sin ^2}C)\cot A = {k^2}\sin (B + C)\sin (B - C)\cot A \\ {k^2}\sin (\pi - A)\sin (B - C)\cot A \\ {k^2}\sin A\sin (B - C)\frac{{\cos A}}{{\sin A}} \\ {k^2}\sin (B - C)\cos A \\\end{aligned} \end{equation} $$
By using $$(\cos A = - \cos (B + C))$$
$$\begin{equation} \begin{aligned} = - {k^2}\sin (B - C)\cos (B + C) \\ = - \frac{{{k^2}}}{2}[2\sin (B - C)\cos (B + C)] \\ = - \frac{{{k^2}}}{2}[\sin 2B - \sin 2C]...(i) \\\end{aligned} \end{equation} $$
Similarly, we can write $$\begin{equation} \begin{aligned} ({c^2} - {a^2})\cot B = - \frac{{{k^2}}}{2}[\sin 2C - \sin 2A]...(ii) \\ ({a^2} - {b^2})\cot C = - \frac{{{k^2}}}{2}[\sin 2A - \sin 2B]...(iii) \\\end{aligned} \end{equation} $$
By adding $(i)$,$(ii)$ and $(iii)$, we get
$$\begin{equation} \begin{aligned} - \frac{{{k^2}}}{2}[\sin 2B - \sin 2C] - \frac{{{k^2}}}{2}[\sin 2C - \sin 2A] - \frac{{{k^2}}}{2}[\sin 2A - \sin 2B] \\ - \frac{{{k^2}}}{2}[\sin 2B - \sin 2C + \sin 2C - \sin 2A + \sin 2A - \sin 2B] \\ = 0 \\\end{aligned} \end{equation} $$
