Do you want to learn the Sine Rule quickly?
Well, you've come to the right place.
Please keep watching.
When studying triangles that do NOT contain a right angle,
there are three basic trigonometric equations that we need to know ...
the Sine Rule, the Cosine Rule and the Area Rule.
And, interestingly, all three of them start [by] using the same diagram.
So, let's create that.
We'll draw a triangle without a right angle in it.
We'll label it A, B and C.
And, of course, following convention,
opposite the vertex with the capital A we label the side little a,
opposite the B we'll label the side little b,
and opposite the C, the side little c.
And the construction is this:
draw a perpendicular line from C to the base of the triangle,
and we'll call this height h.
That's all we're going to use for the Sine Rule.
Now, in this triangle (I'll shade it) ... in this triangle here ... what is Sine A?
Well, we have a right-angled triangle.
The Sine is opposite over hypotenuse, which is h/b.
In this triangle here (which I'll shade differently) ...
it's also a right-angled triangle (this little one) ...
and what is Sine B?
By the way, I'm looking for the sines of angles
because it is the Sine Rule that I'm trying to prove (or derive).
Sine B is opposite over hypotenuse which, in this triangle, is h/a.
Now I have two equations and I wish to eliminate the h from both.
So, the simple thing is
to get the b away from this fraction and the a away from that one.
To do that I multiply both sides of this equation by b. So I have bSinA = h.
Similarly, here, I'll multiply both sides by a and get aSinB = h.
This is not particularly difficult.
And since we have only the one height ... it is the same height for both triangles ...
if they're equal, then so are these. So I can write bSinA = aSinB.
I'll even write 'therefore.'
And, to finish off, I'll divide both sides by ab.
b will divide out the b; the a will divide out the a.
And I'll have SinA/a = SinB/b.
Now, [if we drew] instead of having the height drawn here,
if we drew it from another vertex, we would be able to prove in the same way
that it equals SinC/c. And there's your Sine Rule!
Of course, you can quote it
with the lengths on top and the sine values on the bottom.
a/SinA is the same as b/SinB or c/SinC.
This derivation is quite simple.
And you'll notice that it uses our Sine Rule for right-angled triangles ...
we've got simultaneous equations that we're practicing ...
all within the space of four lines.
This will take you about a minute-and-a-half to two minutes
each afternoon to practice (just once).
And, within a week or so ... probably within two or three days ...
you'll be able to produce this at great speed.
And, within a week or so, you would understand this very well indeed!
So, how do we learn it?
You set time aside.
This is so that you can actually concentrate on learning.
Don't do it while you're watching TV or you're distracted in any way.
Actually focus on the task.
Derive ... just once ... and use it.
Find some questions and use this to solve those problems ...
the trigonometric problems.
And then, when you're confident that you understand it
and you're doing the work quickly, you get your diary
and you ask yourself the question that you should understand
from my previous videos: 'How long before I think I might forget this?'
And, if you think you can go for two weeks, for example,
then two weeks ahead in your diary,
you make a note to derive this again and to do a few more questions.
If you think you'll survive two or three months,
then you put it ahead two or three months.
And use your diary to continue practising this
at whatever intervals you require.
And, within a matter of months or a year or two,
this will be so ingrained in your memory that you'll probably not forget it for years.
So, there it is: quick derivation and practice ...
makes a great deal of difference.
I hope this has helped, and thank you for watching this video.