Without using convexity (and without using calculator, of course) we can make the following.
Since, $33+29+28=90$, we need to prove that: $$33(\sin33^{\circ}-\sin30^{\circ})>29(\sin30^{\circ}-\sin29^{\circ})+28(\sin30^{\circ}-\sin28^{\circ})$$ or $$33(3-4\sin^20.5^{\circ})\cos31.5^{\circ}>29\cos29.5^{\circ}+56\cos0.5^{\circ}\cos29^{\circ}$$ and since $$132\sin^20.5^{\circ}\cos31.5^{\circ}=132\sin^2\left(\frac{\pi}{360}\right)^2\cos31.5^{\circ}<\frac{132\cdot10}{360^2}\cdot1<1,$$ $$99\cos31.5^{\circ}>99\cos36^{\circ}=\frac{99(\sqrt5+1)}{4}=\frac{11\sqrt{405}+99}{4}>\frac{220+99}{4}=\frac{319}{4}$$ and $$\cos29.5^{\circ}<\cos29^{\circ},$$ it's enough to prove that $$\frac{319}{4}>85\cos29^{\circ}.$$ Now, $$\cos29^{\circ}=\frac{\sqrt3}{2}\cos1^{\circ}+\frac{1}{2}\sin1^{\circ}<\frac{\sqrt3}{2}+\frac{1}{2}\cdot\frac{\pi}{180}<\frac{\sqrt3}{2}+\frac{1}{100}.$$ Thus, it's enough to prove that $$319>170\sqrt3+\frac{17}{5}$$ or $$31.56>17\sqrt3,$$ which is true because $$17\sqrt3=\sqrt{289\cdot3}<\sqrt{900}=30<31.56.$$
$\sin$ is a concave function on $[0,\pi],$ which says that Jensen can't help.
It remains Karamata, but I don't see, how to use it here.
Thank you!
Let $f(x)=x\sin x.$ Then, $f''(x)=2\cos x-x\sin x>0$ at least for $x\in[0,\pi/4]$ hence $f$ is strictly convex on this interval, which contains the numbers $$a:=33m,\quad b:=29m,\quad c:=28m\quad\text{where}\quad m=\frac\pi{180}.$$ Therefore, $$f(a)+f(b)+f(c)>3f\left(\frac{a+b+c}3\right)=3f(30m)=45m,$$ which is what you wanted.