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 it, 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 two 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.
Sample Problem
We'll start 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
Simple enough.
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.
Sample Problem
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.
Example 1
Find x if ∆XYZ has a side y = 9.2 and angles of X = 38° and Y = 62°. |
Example 2
What is the perimeter, P, of this triangle?
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Example 3
Solve this triangle.
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Exercise 1
Find the value t of ∆TUV where ∠T = 54°, ∠U = 67°, and u = 14.
Exercise 2
Find the value of x.
Exercise 3
Find the value of x.
Exercise 4
Will, Bill, and Steve are playing baseball. Will, the pitcher, stands 60.5 ft away from Bill at home plate and 50 ft away from first base, where Steve is tying his shoe. If home plate is the vertex of a 44° angle between first base and the pitcher's mound, how far away are Bill and Steve, approximately?
Exercise 5
A spaceship has traveled some distance away from its home planet, Zargon, along a straight path. Alluron, a nearby star, is behind the spaceship at an angle of 32°. Zargon is the vertex of a 73° angle between Alluron and the spaceship, and Zargon and Alluron are known to be 25 million light-years away from each other. How far has the spaceship traveled, in light-years?
Exercise 6
The roof of a house makes an isosceles triangle with a vertex of 50°. If each side of the roof is 18 feet long, how wide is the house?