The doctor and Amy are standing at the seashore ten miles apart. The coastline is a straight line between them. Both can see the same ship in the water. The angle between the coastline and the line between the ship and the doctor is thirty-five degrees. The angle between the coastline and the line between the ship and Amy is forty-five degrees. How far is the ship from the doctor?
+1006 Far enough that it will never be canon.
...
Okay, I guess I'll stop being that one guy that makes everybody awkward at a reference 
"The doctor and Amy are standing at the seashore ten miles apart. The coastline is a straight line between them. Both can see the same ship in the water. The angle between the coastline and the line between the ship and the doctor is thirty-five degrees. The angle between the coastline and the line between the ship and Amy is forty-five degrees. How far is the ship from the doctor?"
Okay, so the distance between them is the only distance known, and that is 10 miles.
Yes, yes, admire my handiwork.
So, we know that one angle of the triangle is 35°, one is 45°, and taking those 2 angles the last angle must be 100°.
Now, we have to do a bunch of work some trigonometry! Fun! I hope you know the Law of Sines.
$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{need, to, put, something, here, just, please, ignore, it}}}}}{{solve}}{\left(\begin{array}{l}{\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}}{{\mathtt{a}}}}={\mathtt{0}}\\
\mathrm{ERROR}={\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{b}}\right)}}{{\mathtt{b}}}}\\
\mathrm{ERROR}={\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{c}}\right)}}{{\mathtt{c}}}}\\
\mathrm{ERROR}={i}\end{array}\right)}$$
PLEASE NOTE THAT THE ABOVE FORMULA IS REALLY REALLY BAD. I JUST THOUGHT IT LOOKED FUNNY.
What I meant to say is:
sin(a)/a = sin(b)/b = sin(c)/c
So if sin(80)/10 were to be taken, in theory wouldn't it be equivalent to sin(35)/x?
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{80}}^\circ\right)} = {\mathtt{0.984\: \!807\: \!753\: \!012}}$$
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{35}}^\circ\right)} = {\mathtt{0.573\: \!576\: \!436\: \!351}}$$
$${\frac{{\mathtt{0.984\: \!807\: \!753\: \!012}}}{{\mathtt{10}}}} = {\frac{{\mathtt{0.573\: \!576\: \!436\: \!351}}}{{\mathtt{x}}}}$$
Cross-multiply.
$${\mathtt{0.984\: \!807\: \!753\: \!012}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{5.735\: \!764\: \!363\: \!51}}$$
Divide 5.73576436351 by 0.984807753012
$${\frac{{\mathtt{5.735\: \!764\: \!363\: \!51}}}{{\mathtt{0.984\: \!807\: \!753\: \!012}}}} = {\mathtt{5.824\: \!247\: \!774\: \!215\: \!186\: \!4}}$$
The Doctor is about 5.824 miles away from the ship, which is still too far for it to ever be canon.
