2011SP-II-41

Ans: B
In $\mathrm{\Delta }ATC$,

$\begin{array}{rcl}\mathrm{\angle }TAC& =& {180}^{\circ }-{90}^{\circ }-{20}^{\circ }\\ & =& {70}^{\circ }\end{array}$

Join $AB$.

$\mathrm{\angle }BAT=\mathrm{\angle }ACB={20}^{\circ }$.

In $\mathrm{\Delta }ABC$,

$\begin{array}{rcl}\mathrm{\angle }BAC& =& {70}^{\circ }-{20}^{\circ }\\ & =& {50}^{\circ }\end{array}$

$\begin{array}{rcl}\mathrm{\angle }ABC& =& {180}^{\circ }–{50}^{\circ }-{20}^{\circ }\\ & =& {110}^{\circ }\end{array}$

By sine law, we have

$\begin{array}{rcl}{\displaystyle \frac{AC}{\sin {110}^{\circ }}}& =& {\displaystyle \frac{BC}{\sin {50}^{\circ }}}\\ AC& =& {\displaystyle \frac{6\sin {110}^{\circ }}{\sin {50}^{\circ }}}\\ & =& 7.360089581\text{ cm}\end{array}$

Denote $O$ as the centre of the circle. Join $OA$ and $OC$.

$\begin{array}{rcl}\text{reflex }\mathrm{\angle }AOC& =& 2\times \mathrm{\angle }ABC\\ & =& 2\times {110}^{\circ }\\ & =& {220}^{\circ }\end{array}$

In $\mathrm{\Delta }OAC$,

$\begin{array}{rcl}\mathrm{\angle }AOC& =& {360}^{\circ }–{220}^{\circ }\\ & =& {140}^{\circ }\end{array}$

Let $r\text{ cm}$ be the radius of the circle. By cosine law, we have

$\begin{array}{rcl}A{C}^2& =& O{A}^2+O{C}^2-2(OA)(OC)\cos \mathrm{\angle }AOC\\ ({\displaystyle \frac{6\sin {110}^{\circ }}{\sin {50}^{\circ }}}{)}^2& =& r^2+r^2-2r^2\cos {140}^{\circ }\\ ({\displaystyle \frac{6\sin {110}^{\circ }}{\sin {50}^{\circ }}}{)}^2& =& r^2(2-2\cos {140}^{\circ })\\ r^2& =& ({\displaystyle \frac{6\sin {110}^{\circ }}{\sin {50}^{\circ }}}{)}^2\times {\displaystyle \frac{1}{2-2\cos {140}^{\circ }}}\\ r^2& =& 15.33679372\\ r& =& 3.916221868\end{array}$

Therefore, the radius is $3.9\text{ cm}$ correct to the nearest $0.1\text{ cm}$.