Properties and Solution of Triangles
8.0 Radius of circumcentre
8.0 Radius of circumcentre
In geometry, the circumscribed circle or circumcircle of a triangle is a circle which passes through all the vertices of a triangle. The center of this circle is called circumcentre and the radius of this circle is called circumradius.
Radius of circumcircle is given as $$\frac{a}{{2\sin A}} = \frac{b}{{2\sin B}} = \frac{c}{{2\sin C}} = \frac{{abc}}{{4\Delta }}$$ where $\Delta$ is the area of triangle $ABC$.
Proof: Consider a triangle $ABC$ circumscribed in a circle with center $O$. Draw a perpendicular from point $O$ on $AB$.
As it is clear from the figure that $BC=a$, $CA=b$ and $AB=c$
In triangle $BDO$, we have $$\sin C = \frac{{\frac{c}{2}}}{R} = \frac{c}{{2R}}$$
$\Delta $ is the area of triangle $ABC$
$$\begin{equation} \begin{aligned} \Delta = \frac{1}{2}ab\sin C \\ \Delta = \frac{1}{2}ab\frac{c}{{2R}} \\ \Delta = \frac{{abc}}{{4R}} \\ R = \frac{{abc}}{{4\Delta }} \\\end{aligned} \end{equation} $$
putting the value of $\Delta $ in above expression we get $$\begin{equation} \begin{aligned} R = \frac{{abc}}{{4 \times \frac{1}{2} \times ab\sin C}} \\ R = \frac{c}{{2\sin C}} \\\end{aligned} \end{equation} $$
Question 15. In a triangle $ABC$, prove that $$\sin A + \sin B + \sin C = \frac{s}{R}$$
Solution: In a triangle$ABC$, we know that $$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}=2R$$
$$\begin{equation} \begin{aligned} \sin A = \frac{a}{{2R}} \\ \sin B = \frac{a}{{2R}} \\ \sin C = \frac{a}{{2R}} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} \sin A + \sin B + \sin C = \frac{{a + b + c}}{{2R}} = \frac{{2s}}{{2R}} \\ \sin A + \sin B + \sin C = \frac{s}{R} \\\end{aligned} \end{equation} $$
Question 16. In a triangle$ABC$, prove that $$\frac{1}{{s - a}} + \frac{1}{{s - b}} + \frac{1}{{s - c}} - \frac{1}{s} = \frac{{4R}}{\Delta }$$
Solution: $$\begin{equation} \begin{aligned} \frac{1}{{s - a}} + \frac{1}{{s - b}} + \frac{1}{{s - c}} - \frac{1}{s} \\ (\frac{1}{{s - a}} + \frac{1}{{s - b}}) + (\frac{1}{{s - c}} - \frac{1}{s}) \\ \frac{{2s - a - b}}{{(s - a)(s - b)}} + \frac{{s - s + c}}{{s(s - c)}} \\\end{aligned} \end{equation} $$ Using $2s=a+b+c$
$$\begin{equation} \begin{aligned} \frac{c}{{(s - a)(s - b)}} + \frac{c}{{s(s - c)}} \\ c[\frac{{(s - a)(s - b) + s(s - c)}}{{(s - a)(s - b)s(s - c)}}] \\ c[\frac{{2{s^2} - s(a + b + c) + ab}}{{{\Delta ^2}}}] \\ c[\frac{{2{s^2} - s(2s) + ab}}{{{\Delta ^2}}}] \\ \frac{{abc}}{{{\Delta ^2}}} \\\end{aligned} \end{equation} $$
Using $$\Delta = \frac{{abc}}{{4R}}$$, we get $$\frac{{4R\Delta }}{{{\Delta ^2}}} = \frac{{4R}}{\Delta }$$
which is equal to RHS.
Question 17. If the circumradius of an isosceles triangle $ABC$ such that $AB=AC$, then find the value of angle $A$.
Solution: We know that $$\begin{equation} \begin{aligned} \sin B = \frac{b}{{2R}} \\ = \frac{{AC}}{{2R}} \\ = \frac{R}{{2R}} \\\end{aligned} \end{equation} $$ Using the given condition that $R=AC$
$$\begin{equation} \begin{aligned} = \frac{1}{2} \\ B = \frac{\pi }{6}or\frac{{5\pi }}{6} \\\end{aligned} \end{equation} $$
When $B = \frac{{5\pi }}{6}$
then $$B = \frac{{5\pi }}{6},C = \frac{{5\pi }}{6}...(AB = AC;\angle B = \angle C)$$
In that case $$B + C > \pi $$ which is not possible so $B = \frac{{5\pi }}{6}$ not possible.
So $$B = C = \frac{\pi }{6}$$
therefore, $$\begin{equation} \begin{aligned} A = \pi - (\frac{\pi }{6} + \frac{\pi }{6}) \\ A = \frac{{2\pi }}{3} \\\end{aligned} \end{equation} $$
Question 18. In a triangle $ABC$, prove that $$s = 4R\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$$
Solution: In a triangle$ABC$
$$\begin{equation} \begin{aligned} \cos \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{bc}}} \\ \cos \frac{B}{2} = \sqrt {\frac{{s(s - b)}}{{ac}}} \\ \cos \frac{C}{2} = \sqrt {\frac{{s(s - c)}}{{ab}}} \\ R = \frac{{abc}}{{4\Delta }} \\\end{aligned} \end{equation} $$
Consider RHS:$$\begin{equation} \begin{aligned} 4R\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} \\ 4 \times \frac{{abc}}{{4\Delta }} \times \sqrt {\frac{{s(s - a)}}{{bc}}} \times \sqrt {\frac{{s(s - b)}}{{ac}}} \times \sqrt {\frac{{s(s - c)}}{{ab}}} \\ \frac{{abc}}{\Delta } \times s \times \sqrt {\frac{{s(s - a)(s - b)(s - c)}}{{{{(abc)}^2}}}} \\ = s \\\end{aligned} \end{equation} $$
which is equal to LHS.
