Question

Let $A(0,-1,1),B(0,0,1),C(1,0,1)$ are the vertices of a $\Delta ABC$. If $R$ and $r$ denotes the circumradius and inradius of $\Delta ABC$, then $\frac{r}{R}$ has value equal to

• A. $\tan \frac{3\pi }{8}$
• B. $\cot \frac{3\pi }{8}$
• C. $\tan \frac{\pi }{12}$
• D. $\cot \frac{\pi }{12}$

Answer 1

The correct option is B $\cot \frac{3\pi }{8}$

$\begin{array}{c}\overrightarrow{AB}=\left(0,1,0\right)\\ \overrightarrow{BC}=\left(1,0,0\right)\\ \overrightarrow{CA}=\left(-1,-1,0\right)\\ \overrightarrow{AB}.\overrightarrow{BC}=\left|\overrightarrow{AB}\right|.\left|\overrightarrow{BC}\right|\cos \theta \\ 0=\cos \theta \\ \theta =\frac{\pi }{2}\\ \tan \left(90-\frac{45}{2}\right)=\cot \frac{45}{2}=\cot {22.5}^{\circ }\\ {\sin }^{2}22.5=\frac{1-\cos 45}{2}=\frac{1-1/\sqrt{2}}{2}=\frac{\sqrt{2}-1}{2}\\ {\tan }^{2}22.5=\frac{\sqrt{2}-1}{\sqrt{2}+1}={\left(\sqrt{2-1}\right)}^2\\ \tan 22.5=\sqrt{2}-1\\ Now\\ 2R=AC=\sqrt{1+1}=\sqrt{2}\\ R=\frac{1}{\sqrt{2}}\\ \left|AB\right|=c=1\\ \left|BC\right|=a=1\\ \left|AC\right|=b=\sqrt{2}\\ A=\frac{1}{2}\times \left|AB\right|\times \left|BC\right|=\frac{1}{2}\\ r=\frac{A}{s}=\frac{1}{2s}=\frac{1}{2+\sqrt{2}}\\ \frac{r}{R}=\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2}\pi \left(2-\sqrt{2}\right)=\sqrt{2}-1=\tan 22.5=\tan \frac{45}{2}\times \frac{\pi }{18}=\tan \frac{\pi }{2}\\ and,\cot \left(\frac{\pi }{2}-\frac{\pi }{8}\right)=\cot \frac{3\pi }{8}\end{array}$