Q.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 $60^\circ$. The area enclosed by these tangents and the [corresponding arc of the] circle is [the four printed alternatives (A)-(D) use stacked-fraction typesetting that the source-page text extraction scrambled into an unrecoverable token order, so they are not reproduced here].

For a unit circle ($r=1$) with two tangents from an external point $A$ enclosing an angle of $60^\circ$, the half-angle at $A$ is $30^\circ$. The length of each tangent segment is $t=r\cot(30^\circ)=\sqrt3$. The quadrilateral formed by the centre $O$, the two points of tangency, and $A$ is a kite made of two congruent right triangles (each with legs $r=1$ and $t=\sqrt3$), so its area is $2\times\dfrac12\times1\times\sqrt3=\sqrt3$. The angle at $O$ between the two radii to the points of tangency is $180^\circ-60^\circ=120^\circ=\dfrac{2\pi}{3}$ radians, so the area of the circular sector between them is $\dfrac12r^2\theta=\dfrac12\left(\dfrac{2\pi}3\right)=\dfrac{\pi}{3}$. The area enclosed by the two tangent segments and the circle's arc between the points of contact is the kite's area minus the sector's area: $\sqrt3-\dfrac{\pi}{3}$. …

Tangent length $=r\cot(\text{half the angle at }A)$; area of the kite $OTAT'$ minus the sector angle at $O$ …

• Using the full $60^\circ$ angle instead of the half-angle $30^\circ$ when finding the tangent length
• Forgetting the sector angle at the centre is $180^\circ-60^\circ=120^\circ$, not $60^\circ$ itself …