+0

# AHellelepoeowoe GELP

0
131
2

May 21, 2019

#1
+8756
+3

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) $$A\,=\,\pi-B-C \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\\~\\ A\,=\,\pi-\arcsin(\frac56)-\frac{\pi}{6}\\~\\ A\,=\,\frac{5\pi}{6}-\arcsin(\frac56)\\~\\ \sin(A)\,=\,\sin(\frac{5\pi}{6}-\arcsin(\frac56))\\~\\ \sin(A)\,=\,(\frac12)(\frac{\sqrt{11}}{6})-(-\frac{\sqrt3}{2})(\frac56)\\~\\ \sin(A)\,=\,\frac{\sqrt{11}\,+\,5\sqrt3}{12}$$   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  =  $$10\sqrt3$$

.
May 22, 2019
#2
+103917
+2

Nice, hectictar  !!!

CPhill  May 22, 2019