+9466 
By the Law of Sines:
\(\frac{\sin B}{10}\,=\,\frac{\sin( \frac{\pi}{6})}{6}\\~\\ \sin B\,=\,\frac{10\sin( \frac{\pi}{6})}{6}\\~\\ \sin B\,=\,\frac56\\~\\ B\,\approx\,56.44°\qquad\text{or}\qquad B\,\approx\,123.56°\)
Both options are valid in this case because neither make the current sum of the angles exceed 180° .
Using the first possible value of B, that is, B = arcsin(5/6)
| Using the second possible value of B, that is, B = π - arcsin(5/6)
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\(A\,=\,\pi-B-C \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\\~\\ A\,=\,\pi-\arcsin(\frac56)-\frac{\pi}{6}\\~\\ A\,=\,\frac{5\pi}{6}-\arcsin(\frac56)\\~\\ \sin(A)\,=\,\sin(\frac{5\pi}{6}-\arcsin(\frac56))\\~\\ \sin(A)\,=\,(\frac12)(\frac{\sqrt{11}}{6})-(-\frac{\sqrt3}{2})(\frac56)\\~\\ \sin(A)\,=\,\frac{\sqrt{11}\,+\,5\sqrt3}{12}\)
By the Law of Sines:
\(\frac{\sin A}{BC}\,=\,\frac{\sin \frac{\pi}{6}}{6}\\~\\ \frac{\sin A}{BC}\,=\,\frac{1}{12}\\~\\ \frac{BC}{\sin A}\,=\,\frac{12}{1}\\~\\ BC=12\sin A\\~\\ BC\,=\,12(\frac{5\sqrt3+\sqrt{11}}{12})\\~\\ BC\,=\,5\sqrt3+\sqrt{11}\) | \(A\,=\,\pi-B-C\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\\~\\ A\,=\,\pi-(\pi-\arcsin(\frac56))-\frac{\pi}{6}\\~\\ A\,=\,\arcsin(\frac56)-\frac{\pi}{6}\\~\\ \sin(A)\,=\,\sin(\arcsin(\frac56)-\frac{\pi}{6}) \\~\\ \sin(A)\,=\, (\frac56)(\frac{\sqrt3}{2})-(\frac{\sqrt{11}}{6})(\frac12) \\~\\ \sin(A)\,=\,\frac{5\sqrt3-\sqrt{11}}{12}\)
By the Law of Sines:
\(\frac{\sin A}{BC}\,=\,\frac{\sin \frac{\pi}{6}}{6}\\~\\ \frac{\sin A}{BC}\,=\,\frac{1}{12}\\~\\ \frac{BC}{\sin A}\,=\,\frac{12}{1}\\~\\ BC=12\sin A\\~\\ BC\,=\,12(\frac{5\sqrt3-\sqrt{11}}{12})\\~\\ BC\,=\,5\sqrt3-\sqrt{11}\) |
the first possible value of BC + the second possible value of BC = \((5\sqrt3+\sqrt{11})+(5\sqrt3-\sqrt{11})\)
the first possible value of BC + the second possible value of BC = \(10\sqrt3\)
please help quickly thank you so much thank you. 
