Answer: (a) \(\tan\phi=2\phi\). (b) \(\tan\phi=2+\phi\).

Answer: (a) \(\tan\phi=2\phi\). (b) \(\tan\phi=2+\phi\).

(a) The area of sector \(OPQ\) is

\(\displaystyle \frac12a^2(2\phi)=a^2\phi\).

Since \(PT\) and \(QT\) are tangents, \(OP\perp PT\) and \(OQ\perp QT\). In right triangle \(OPT\), the angle at \(O\) is \(\phi\), so

\(\displaystyle \tan\phi=\frac{PT}{a}\).

Hence

\(PT=a\tan\phi\).

The area of triangle \(OPT\) is

\(\displaystyle \frac12\times a\times a\tan\phi=\frac12a^2\tan\phi\).

There are two congruent triangles, so the area of kite \(OPTQ\) is

\(a^2\tan\phi\).

The shaded area is the area of the kite minus the sector:

\(a^2\tan\phi-a^2\phi\).

It is given that the shaded area equals the sector area, so

\(a^2\tan\phi-a^2\phi=a^2\phi\).

Dividing by \(a^2\) gives

\(\tan\phi-\phi=\phi\),

and therefore

\(\tan\phi=2\phi\).

(b) The perimeter of the sector is the two radii plus the arc length:

\(2a+2a\phi\).

The perimeter of the shaded region is

\(PT+QT+\text{arc }PQ\).

Since \(PT=QT=a\tan\phi\), this is

\(2a\tan\phi+2a\phi\).

The sector perimeter is half the shaded perimeter, so

\(2a+2a\phi=\frac12(2a\tan\phi+2a\phi)\).

Thus

\(2a+2a\phi=a\tan\phi+a\phi\).

Dividing by \(a\) and rearranging gives

\(\tan\phi=2+\phi\).