A strange value

$$ First, we calculate $\cos 15^\circ$ . $$ $\begin{aligned} \cos 15^\circ&=\cos(60^\circ-45^\circ)\\ &=\cos60^\circ\cos45^\circ+\sin60^\circ\sin45^\circ\\ &=\frac{1}{2}\cdot\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}\cdot\frac{1}{2}\sqrt{2}\\ &=\frac{1}{4}\sqrt{2}(1+\sqrt{3}). \end{aligned}$ $$ Let $\overrightarrow{BC}=x$ , then $\overrightarrow{BC}$ can be calculated by using cosine formula. $$ $\begin{aligned} 9^2&=x^2+6^2-2x(6)\cos 15^\circ\\ 81&=x^2+36-12x\cdot\frac{1}{4}\sqrt{2}(1+\sqrt{3})\\ x^2-3\sqrt{2}(1+\sqrt{3})x-45&=0\\ x_{1,2}&=\frac{3\sqrt{2}(1+\sqrt{3})\pm\sqrt{(3\sqrt{2}(1+\sqrt{3}))^2-4(1)(-45)}}{2}\\ &=\frac{3\sqrt{2}+3\sqrt{6}\pm6\sqrt{7+\sqrt{3}}}{2}\\ &=\frac{3\sqrt{6}+3\sqrt{2}}{2}\pm3\sqrt{7+\sqrt{3}} \end{aligned}$ $$ We obtain $a=3$ , $b=2$ , $c=6$ , and $d=7$ . Therefore, $a+b+c+d=3+2+6+7=\boxed{18}$ .