Question

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

• A. $\frac{2}{\sqrt{3}}-\frac{\pi }{6}$
• B. $\sqrt{3}-\frac{\pi }{3}$
• C. $\frac{\pi }{3}-\frac{\sqrt{3}}{6}$
• D. $\sqrt{3}(1-\frac{\pi }{6})$

Answer 1

The correct option is B $\sqrt{3}-\frac{\pi }{3}$

$\angle OAP=\angle OBP={90}^{\circ }$ (Line from center to tangent at point of contact is a perpendicular)

Also, $PA=PB$ (tangents from common external point to circle)

and $OA=OB$ (radii of same circle)

Hence, $\angle OPA=\angle OPB={30}^{\circ }$ (given $\angle APB={60}^{\circ }$)

Hence, $\angle AOP=\angle BOP={60}^{\circ }$

Area of sector $AOB=\frac{\pi }{{360}^{\circ }}\times {120}^{\circ }=\frac{\pi }{3}$

Area of quadrilateral $BPOA=2\times area\left(△POA\right)=2\times \frac{1}{2}\times 1\times \sqrt{3}=\sqrt{3}$

Hence, required area $=area\left(PBOA\right)-area\left(sectorAOB\right)=\sqrt{3}-\frac{\pi }{3}$

This is the required answer.