$\noindent Please keep in mind I have never derived trig functions before as I'm not up to that chapter, so I just used symbolab.$

$\noindent 14. a) \angle QOM = \frac{\pi}{2} (A line from the centre of a circle that bisects a chord is perpendicular to the chord)$

$\noindent It is clear that \Delta QOM is a right-angled triangle and thus sin\theta = QM/1 = QM (radius is 1)$

$\noindent Similarly cos\theta = OM$

$\noindent b) OP = 1 (radius), therefore from a) PM = cos\theta + 1$

$\noindent From a), QR = 2sin\theta$

$A = \frac{1}{2}2sin\theta(cos\theta + 1) = sin\theta(cos\theta + 1)$

$\noindent c) \frac{dA}{d\theta} = cos\theta(cos\theta + 1) - sin^{2}\theta (again I have no idea how this is done I just plugged into symbolab, I'm guessing you know how to because you are studying this chapter)$

$Stationary points occur when \frac{dA}{d\theta} = 0$

$cos\theta(cos\theta + 1) - sin^{2}\theta = 0$

$cos\theta(cos\theta + 1) - (1 - cos^{2}\theta) = 0$

$2cos^2\theta + cos\theta - 1 = 0$

$Solving this quadratic we get cos\theta = \frac{1}{2} and cos\theta = -1$

$Obviously cos\theta \neq -1 as cos\theta is a length! (OM)$

$\noindent So cos\theta = \frac{1}{2} or \theta = \frac{\pi}{3} is the only stationary point which we should consider (It's not 420 or 780 or any other multiple of \pi as \theta is acute as stated). Using a table of values we can determine that there is a maximum turning point at \theta = \frac{\pi}{3} , and thus the area is maximum when \theta = \frac{\pi}{3}$

$\noindent Now \Delta PQR = \theta (angle at centre is twice angle at circumference), therefore the area of \Delta PQR is max when \theta = \frac{\pi}{3} , and since \Delta PQR is isosceles, \angle PQR = \angle PRQ = \frac{\pi}{3} (which is an equilateral triangle!)$

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