In the given figure,AB is tangent to the circ | vedclass

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(D) In $\Delta OAB$,$\angle OAB = 90^{\circ}$ (since $AB$ is tangent to the circle at $A$).Given $\angle AOB = 60^{\circ}$ and radius $OA = 2$.In $\De

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$\frac{3\sqrt{3}}{\pi} - 1$

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(D) In $\Delta OAB$,$\angle OAB = 90^{\circ}$ (since $AB$ is tangent to the circle at $A$).Given $\angle AOB = 60^{\circ}$ and radius $OA = 2$.In $\Delta OAB$,$	an 60^{\circ} = \frac{AB}{OA}$ $\Rightarrow \sqrt{3} = \frac{AB}{2}$ $\Rightarrow AB = 2\sqrt{3}$.Area of $\Delta OAB = \frac{1}{2} 	imes OA 	imes AB = \frac{1}{2} 	imes 2 	imes 2\sqrt{3} = 2\sqrt{3}$.Area of sector $OAC = \frac{	heta}{360^{\circ}} 	imes \pi r^2 = \frac{60^{\circ}}{360^{\circ}} 	imes \pi (2)^2 = \frac{1}{6} 	imes 4\pi = \frac{2\pi}{3}$.The shaded region is the area of $\Delta OAB$ minus the area of sector $OAC$,which is $2\sqrt{3} - \frac{2\pi}{3}$.The unshaded region is the area of sector $OAC$,which is $\frac{2\pi}{3}$.Ratio = $\frac{	ext{Shaded}}{	ext{Unshaded}} = \frac{2\sqrt{3} - \frac{2\pi}{3}}{\frac{2\pi}{3}} = \frac{2\sqrt{3}}{\frac{2\pi}{3}} - 1 = \frac{6\sqrt{3}}{2\pi} - 1 = \frac{3\sqrt{3}}{\pi} - 1$.

$List-I$	$List-II$
$(P) \alpha \text{ equals}$	$(1) (-2,4)$
$(Q) r \text{ equals}$	$(2) \sqrt{5}$
$(R) A_1 \text{ equals}$	$(3) (-2,6)$
$(S) B_1 \text{ equals}$	$(4) 5$
	$(5) (2,4)$

In the given figure,$AB$ is tangent to the circle with centre $O$. The ratio of the shaded region to the unshaded region of the triangle $OAB$ is

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