Right Triangles and Trigonometry
Law of Sines
The Law of Sines can be used to find the angles and side lengths of triangles that aren't right triangles. It has little to do with crop circles, aliens, or Mel Gibson.
The triangle above represents any non-right triangle. The Law of Sines says that for such a triangle,
We can prove, too. All we have to do is cut that triangle in half.
Using the trig ratios we learned, we can find the sine of angles A and B for the right triangles we made.
Isolate for the altitude h and then set the two equations equal to each other.
b × sin A = h
a × sin B = h
b × sin A = a × sin B
Dividing both sides by ab will give us our Law of Sines.
If draw different altitudes and repeat all that business, we'll end up with and . Solid proof, we think. Now that we know it's true, we can start putting it to use.
We'll begin by finding the value of x. The Law of Sines is still fresh in our minds.
We'll set A to 82° and B to 32°. That means a is 7 and b is x.
x ≈ 3.7
The Law of Sines is so useful that with it, we're unstoppable. We can solve triangles, which means finding every side length and every angle measurement.
Solving this triangle should be pretty easy. We'll start with the Law of Sines.
If A is 60°, then a is 4. The only other side we have is 3.2, so that'll be b.
We want B, not sin B. If we take the inverse sine of each side, we get:
B ≈ 43.9°
Wonderful. Now we have two out of three angles and two out of three sides. Knowing that all the angles in a triangle add up to 180° allows us to find the last angle.
A + B + C = 180°
Substitute in the angles we know.
60° + 43.9° + C = 180°
Solve for the remaining angle.
C = 180° – (60° + 43.9°)
C = 76.1°
One side to go. We can use either the ratio between sin(60°) and 4 or sin(43.9°) and 3.2.
Substitute in what we know. We're looking for the last side, c.
c ≈ 4.5
Double-check. Do we know all the angles and all the sides?
A = 60°; B = 43.9°; C = 76.1°
a = 4; b = 3.2; c = 4.5
Yep. We solved that triangle real good.