'sin'tiplying

From the reference (equation (24)) , we have $\displaystyle \prod_{k=1}^{n-1} \sin \left(\frac {k\pi}n\right) = 2^{1-n} n$ . Then

$\begin{aligned} \prod_{k=1}^{179} \sin \left(\frac {k\pi}{180}\right) & = \frac {180}{2^{179}} \\ \implies \prod_{k=1}^{179} \sin k^\circ & = \frac {180}{2^{179}} \\ \prod_{k=1}^{90} \sin k^\circ \color{#3D99F6} \prod_{k=91}^{179} \sin k^\circ & = \frac {180}{2^{179}} & \small \color{#3D99F6} \text{Since }\sin (180-x)^\circ = \sin x^\circ \\ \prod_{k=1}^{90} \sin k^\circ \color{#3D99F6} \prod_{k=1}^{89} \sin k^\circ & = \frac {180}{2^{179}} & \small \color{#3D99F6} \text{Since }\sin 90^\circ = 1 \\ \prod_{k=1}^{90} \sin k^\circ \color{#3D99F6} \prod_{k=1}^{\color{#D61F06}90} \sin k^\circ & =\frac {180}{2^{179}} \end{aligned}$

$\begin{aligned} \implies \prod_{k=1}^{90} \sin k^\circ & = \sqrt{\frac {180}{2^{179}}} = \frac {3\sqrt{10}}{2^{89}} \end{aligned}$

Therefore, $a+b+c+d = 3+10+2+89 = \boxed{104}$ .

I will add a proof of the formula used by Chew-Seong Cheong.

Let $P=X^{n}-1$ .

Using the roots of unity, we have:

$P=(X-1)\displaystyle\prod_{k=1}^{n-1}(X-e^{\frac{i2k\pi}{n}})$

Therefore,

$\begin{aligned} Q & =\dfrac{X^{n}-1}{X-1}=X^{n-1}+X^{n-2}+\cdots + 1\\ & =\displaystyle\prod_{k=1}^{n-1}(X-e^{\frac{i2k\pi}{n}}) \end{aligned}$

Plugging $1$ into $Q$ :

$\begin{aligned} n & =\displaystyle\prod_{k=1}^{n-1}(1-e^{\frac{i2k\pi}{n}})\\ & = \displaystyle\prod_{k=1}^{n-1} e^{\frac{ik\pi}{n}} ( e^{\frac{-ik\pi}{n}} -e^{\frac{ik\pi}{n}})\\ & = \displaystyle\prod_{k=1}^{n-1} -2i e^{\frac{ik\pi}{n}} \sin(\frac{k\pi}{n}) \\ & = (-1)^{n-1} i^{n-1} 2^{n-1} (e^{\frac{i\pi}{n}})^{\frac{n(n-1)}{2}} \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n})\\ & = (-1)^{n-1} i^{n-1} 2^{n-1} (e^{\frac{i\pi}{2}})^{n-1} \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n})\\ & = (-1)^{n-1} i^{n-1} 2^{n-1} i^{n-1} \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n})\\ & = (-1)^{n-1} (i^2)^{n-1} 2^{n-1} \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n})\\ & = (-1)^{n-1} (-1)^{n-1} 2^{n-1} \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n})\\ & = 2^{n-1} \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n}) \end{aligned}$

So,

$\displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n})=\dfrac{n}{2^{n-1}}$