hectictar

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 #5
avatar+9460 
+2

Well...I wanted to be consistent with the definitions recommended here  https://en.wikipedia.org/wiki/Gigabyte

 

I just really hope that in the future we can all agree on the same prefixes to mean the same things.

May 21, 2019
 #3
avatar+9460 
+3

I have never heard of Appollonius's Theorem, but does seem like it can be used here.

 

Using the information from here:  https://en.wikipedia.org/wiki/Apollonius%27s_theorem

 

32 + 42  =  2( (2a)2 + a2 )  
25  =  2( 4a2 + a2 )

 

 

25  =  2( 5a2 )  
25  =  10a2

 

 

2.5  =  a2  
a  =  √[ 2.5 ]

 

 

BC  =  2a  =  2√[ 2.5 ]  
May 21, 2019
 #1
avatar+9460 
+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 21, 2019
 #4
avatar+9460 
+2
May 21, 2019