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3.Trigonometrical Ratios, Functions and Identities
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Let $S_1,S_2$ and $S_3$ be three circles of unit radius which touch each other externally. The common tangent to each pair of circles are drawn and extended so that they can intersect and form a triangle $ABC$ with circumradius $R,$ then $R$ is equal to
A
$4+2\sqrt 3$
B
$2(1+\frac{1}{\sqrt 3})$
C
$4(1+\sqrt 3)$
D
$\frac{3(1+\sqrt 3)}{2}$
Solution
Ans. $( 2 )$
$\mathrm{BM}=\mathrm{NC}=\cot 30^{\circ}=\sqrt{3}$
and $\mathrm{MN}=2$
$\therefore \mathrm{BC}=2(1+\sqrt{3})$
$\therefore$ By sine Rule
$\mathrm{R}=\frac{2(1+\sqrt{3})}{2 \sin 60^{\circ}}=2\left(1+\frac{1}{\sqrt{3}}\right)$
Standard 11
Mathematics
