Cosine Shaped Pizza

The area under $\cos x$ , $x \in \left(0, \frac \pi 2\right)$ is $A = \displaystyle \int_0^\frac \pi 2 \cos x \ dx = \sin x \bigg|_0^\frac \pi 2 = 1$ . Let the straight line through origin that cuts $A$ into half be $y = mx$ , and let the two lines intersect at $P(a, \cos a)$ . Then, $m=\dfrac {\cos a}a$ and we have:

$\int_0^a (\cos x - mx) dx = \sin x - \frac {mx^2}2 \bigg|_0^a = \sin a - \frac a2 \cos a = \frac 12$

Solving numerically, we get $a \approx 0.894563$ , $\implies m = \dfrac {\cos a}a \approx 0.699626195$ and the angle $y=mx$ makes with the $x$ -axis, $\theta = \tan^{-1} m \approx 0.610475045 \text{ rad} \approx \boxed{35}^\circ$ .