blog dismissed,bring in the dancin lobstahs.

So this was a four day week and we took a test on Thursday and question day was Wednesday and we had a quiz Tuesday and class work Monday so I do not recall learning anything new. That being said, I’ve decided that this week my blog will be on Law of Sines. It’s a rather simple concept. You use this for non-right triangles. This should be your first choice in solving the triangle because it is shorter and much less complicated than Law of Cosines.

Formula:

SinA/A=sinB/B=sinC/C

Solve triangle ABC if BC=3, A=62 degrees, and B=43 degrees

-First you draw your triangle. (As always you should write your answers in as you get them.)

To find C, simply subtract 180-62-43.

C=75 degrees

Next we’ll solve for AC.

Sin62degrees/3=sin3degrees/B

You’re gonna cross multiply. So Bsin62degrees=3sin3degrees

Solve for b.

B=3sin3degrees/sin62degrees

B=.178

Use the same process to solve for AB.

Sin62degrees/3=sin75degrees/c

Csin62degrees=3sin75degrees

C=3sin75degrees/sin62degrees

C=3.381

Your final triangle should look like this J

That’s that.

--Sarah