[Mathematics] The Sine Rule 04-26-2015, 04:55 PM
#1
@"Lux"y and @"roger_smith" requested a thread on trigonometry (specifically the sine rule), so here it is. I'll go over general SOH CAH TOA and the cosine rule in separate threads.
The sine rule is a simple statement:
a/sin(A) = b/sin(B) = c/sin( C)
for any triangle. (Both Right-Angled and Non-Right-Angled triangles.) What the formula is saying, simply, is that any side divided by the opposite angle is equal to any other side divided by their own respected angles.
Let's take the following triangle as a start:
![[Image: X9zhN2o.png]](http://i.imgur.com/X9zhN2o.png)
The letters can be switched around but the general rule is that angle A has to be opposite to side a, angle B has to be opposite side b, and so on.
An important thing to note is that the formula is slightly different depending on whether you're trying to find the side or the angle, just to make things easier when rearranging. Like so:
![[Image: r0zfnmo.png]](http://i.imgur.com/r0zfnmo.png)
Let's do a basic example:
![[Image: xefdWWG.png]](http://i.imgur.com/xefdWWG.png)
We're finding a side so we'll use the first form of the sine rule. You should note that you only need two of the values. So, either a and b, b and c, or a and c.
What you do first is look for two opposite sides and angles with a known value, so in our case, that's 7 and 60. If there aren't, then chances are that Pythagoras's Theorem will give you the third side if you know two.
Putting the known values into the formula:
![[Image: iHzjbRS.png]](http://i.imgur.com/iHzjbRS.png)
Now let's try to find an angle:
![[Image: DeCay2l.png]](http://i.imgur.com/DeCay2l.png)
Putting it into the formula:
![[Image: yQiyC5y.png]](http://i.imgur.com/yQiyC5y.png)
You see what we did differently? We got sin(x) on it's own so we has to do the inverse sine of sin(x) to get x.
Now let's use this on a trigonometry question:
![[Image: SZoII0G.png]](http://i.imgur.com/SZoII0G.png)
Now, they'd expect you to use gay-ass SOH-CAH-TOA on this question, but you can just use the sine rule: (Learn SOH CAH TOA too, just in case. The more you know...
)
![[Image: 9E8W4TQ.png]](http://i.imgur.com/9E8W4TQ.png)
Hope you learned something. As always, if you know something about maths that someone else doesn't, make a thread!
The sine rule is a simple statement:
a/sin(A) = b/sin(B) = c/sin( C)
for any triangle. (Both Right-Angled and Non-Right-Angled triangles.) What the formula is saying, simply, is that any side divided by the opposite angle is equal to any other side divided by their own respected angles.
Let's take the following triangle as a start:
The letters can be switched around but the general rule is that angle A has to be opposite to side a, angle B has to be opposite side b, and so on.
An important thing to note is that the formula is slightly different depending on whether you're trying to find the side or the angle, just to make things easier when rearranging. Like so:
Let's do a basic example:
We're finding a side so we'll use the first form of the sine rule. You should note that you only need two of the values. So, either a and b, b and c, or a and c.
What you do first is look for two opposite sides and angles with a known value, so in our case, that's 7 and 60. If there aren't, then chances are that Pythagoras's Theorem will give you the third side if you know two.
Putting the known values into the formula:
Now let's try to find an angle:
Putting it into the formula:
You see what we did differently? We got sin(x) on it's own so we has to do the inverse sine of sin(x) to get x.
Now let's use this on a trigonometry question:
Now, they'd expect you to use gay-ass SOH-CAH-TOA on this question, but you can just use the sine rule: (Learn SOH CAH TOA too, just in case. The more you know...
)Hope you learned something. As always, if you know something about maths that someone else doesn't, make a thread!
![[+]](https://sinister.ly/images/modern/collapse_collapsed.png)
![[Image: tumblr_noac9s6rgw1tvnnaxo1_500.gif]](https://33.media.tumblr.com/95c52b25dceea2bedb89888221404c11/tumblr_noac9s6rgw1tvnnaxo1_500.gif)
