BQ#1:UnitP: Concept 1-4:Non-Right Tiangle Equations

1. Law Of Sines
For non-right triangles we need an equation that will help us find the angles and sides. We can't use the Pythagorean theorem because that only applies to right triangles. So observing a non-right triangle, we can try and find an equation to help us. First of all, let's try and make that non-right triangle into two right triangles by taking angle B and putting a straight line through it.

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

Above, we can see how the line that cuts in between angle B has created two right triangles. Also, the line is now being called h. Now, we use Sine to find and equation with h in it.
In order to do so, we use Sin A=h/c and for the other side we use Sin C=h/a. Then we cross multiply to get cSinA=h and aSinc=h. Now, here's where it gets a little tricky.

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

In the above picture, we can see how we already did the first two steps. Now to put it simply, all we had to do was equal both equations to each other since they both equal h. Then we divide both by the side by the side that it has. So we divide by ac. Then one we do that we get the equation SinA/a=SinC/c. Now this is the equation for sines. Since we have this, we can safely assume that SinB/b also equals the others. And if you are skeptical, we can do the exact same thing but instead just cut a straight line through angle A

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

We did the same thing we did at the beginning but the line cuts through angle A. Now we solve the exact same way but for angle B. We did the same thing by cutting a straight line but only to a different angle. Then let's call it k. It could be h but we already used it.

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

 we solved the exact same way but for angle B and got the same and similar answer. So now we know that not only do they all equal each other but also that this works for all triangles

4. Now let's try and find the area. As we know, for right triangles, the area equation is A=1/2bh. Now of course' this is for right triangles. Now, to find one for non-right triangles, we simply replace h with

aSinC. Now you're probably wondering how that is so. It's simple. Remember when we were finding the origins of the law of Sines each one was equal to h? Well instead of using the equation area for right triangles we simply replace the h with the sin that we got when fixing the Law of Sine.

(http://pcb2s11.blogspot.com/2011/06/chapter-6-summary.html)

References:
http://pcb2s11.blogspot.com/2011/06/chapter-6-summary.htmlhttp://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm