In week 17 of precalc, we went over Sine law and Cosine law. We use these set of rules to find angles and sides of NON RIGHT triangles (triangles without a 90° angle).
Sine law has the formula SinA/a=SinB/b=SinC/c OR: a/SinB=b/SinB=c/SinC. For reference, the capitals are the angles and lowercase letters are sides. The thing with sine law is that you need to know which version to use for what you’re finding. Basically, when you are finding the angle, put angles (SinA,B,C) on the top and the sides (abc) on the bottom and vice versa for when you are finding sides.
The sine rule is useful when you have two angles and a side. The sine rule helps to find the remaining sides and angles. It’s also helpful when you are given two sides and a non-included angle. The sine rule can help to find the other angles and the third side, though this can sometimes lead to the ambiguous case where two different triangles satisfy the given conditions.
Cosine law has the formula b^2=a^+b^2(2(ac))cosB, or more simplified:
The cosine law, similarly to Pythagoras (A^2+B^2=C^2), you can interchange the variables, like to find B, just replace the values accordingly.
The cosine rule is useful when you have two sides and an included angle. cosine rule helps to find the third side. It also works when you know all three sides of the triangle. the cosine rule helps to find any of the angles.
Now let’s go through example to show the use of the cosine rule
Problem: You have a triangle with sides a=7units, b units, and the included angle C=60 degrees. Find the length of the side c. Since we know two sides and the included angle we use the cosine rule: c^2=a^2+b^2−2abcos(C). Substituting the given values: c^2=7^2+5^2−2⋅7⋅5⋅cos(60). cos 60 is just .5 so c^2=49+25−2⋅7⋅5⋅0.5. Simplifying it turns the equation into c^2=49+25−35 and then c^2=39 so our answer is c=√39.