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Ok... Basically I can get an awnser, but there must be an easier way.
You have a triangle with one angle, you have the lengths of the sides erm... jabbing into the angle. Nothing else, sides are not equal. Try to find out the length of the third side. Confused? Not so easy to explain. Diagram 55 is the angle.
/\
x / \ 15
/ \
------ 55*
14
Only way I could think of doing it was by cutting it in half. But! I know there must be an easier way because it is a fixed relationship. Any clues?
Well, there's the cosine rule: a²=b²+c²-2bcCosA, meaning that x²=14²+15²-2*14*15*cos55, so x=√(14²+15²-2*14*15*cos55)=13.4, to one decimal place.
You could also cut it in half, but that involves calculating three more lengths before you're allowed to find x, so I think that this way's actually easier.
Why did the vector cross the road?
It wanted to be normal.
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Nice, thanks, I'll use that rule
But, er... What does that A represent? The capital A. I realise it represents the angle, but why a capital A? Won't be confused with the lowercase a?
If you want to use it for other triangles, remember to label them like this:
A
/\
/ \
b / \ c
/ \
-------------
C a B
Big letters are angles, small letters are sides and matching lettered sides and angles are opposite each other.
Also, remember the cosine rule's sister equation, a/SinA=b/SinB=c/SinC.
Why did the vector cross the road?
It wanted to be normal.
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I see, thanks again.
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