A pair of tangents are drawn to a unit circle | vedclass

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(B) Let the circle be $x^2 + y^2 = 1$,so the radius $r = 1$. Let the tangents intersect at point $A$. The angle between the tangents is $60^o$,so the

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

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(B) Let the circle be $x^2 + y^2 = 1$,so the radius $r = 1$. Let the tangents intersect at point $A$. The angle between the tangents is $60^o$,so the angle between each tangent and the line joining the origin to $A$ is $30^o$. Let $O$ be the origin and $P$ be the point of tangency. In $	riangle OPA$,$\angle OPA = 90^o$ and $\angle OAP = 30^o$. Thus,$PA = r \cot(30^o) = 1 \cdot \sqrt{3} = \sqrt{3}$. The area of the quadrilateral formed by the origin,the two points of tangency,and $A$ is $2 	imes (	ext{Area of } 	riangle OPA) = 2 	imes (\frac{1}{2} \cdot PA \cdot r) = 2 	imes (\frac{1}{2} \cdot \sqrt{3} \cdot 1) = \sqrt{3}$. The angle subtended by the two points of tangency at the center is $180^o - 60^o = 120^o = \frac{2\pi}{3}$ radians. The area of the circular sector is $\frac{1}{2} r^2 	heta = \frac{1}{2} \cdot 1^2 \cdot \frac{2\pi}{3} = \frac{\pi}{3}$. The area enclosed by the tangents and the arc is the area of the quadrilateral minus the area of the sector: $\sqrt{3} - \frac{\pi}{3}$.

$A$ pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at a point $A$ enclosing an angle of $60^o$. The area enclosed by these tangents and the arc of the circle is

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