Multiple choice

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

  1. $\displaystyle \frac{2}{\sqrt{3}} - \frac{\pi}{6}$
  2. $\displaystyle \sqrt{3} - \frac{\pi}{3}$
  3. $\displaystyle \frac{\pi}{3} - \frac{\sqrt{3}}{6}$
  4. $\displaystyle \sqrt{3} (1 - \frac{\pi}{6})$
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B Correct answer
AI explanation

The external point A and the two points of contact with the circle form a triangle where the angle at A is 60 degrees and the two radii are perpendicular to the tangents. The area of this quadrilateral, formed by the two radii and the two tangents, is given by r^2 cot(theta/2), where theta is 60 degrees and r is 1, giving an area of cot(30 degrees) = sqrt(3). The sector of the circle enclosed by the radii has a central angle of 180 - 60 = 120 degrees, and its area is (120/360) * pi * 1^2 = pi/3. Subtracting the sector area from the quadrilateral area gives the required enclosed area as sqrt(3) - pi/3.

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