Law of Sines
Topic Index | Algebra2/Trig Index | Regents Exam Prep Center

When you see the triangle below on the left and someone asks you to find the value of x, you immediately know how to proceed.  You call upon your friend the Pythagorean Theorem because you see the right angle.

But what do you do when you see the triangle on the right?  There is no indication of a right angle.

Now, with our knowledge of trigonometry, we are armed to attack any of these perplexing problems!

Let's see how to apply trigonometry to working in triangles which do not contain a right angle.

In this diagram, notice how the triangle is labeled.  The capital letters for the vertices are repeated in small case on the side opposite the corresponding vertex.

side a is opposite <A
side b is opposite <B
side c is opposite <C

working together as partners!

The ratios of each side to the sine of its "partner" are equal to each other.

Law of Sines


These ratios, in pairs, are applied to solving problems.  You never need to use all three ratios at the same time.  Mix and match the ratios to correspond with the letters you need.  Remember when working with proportions, the product of the means equals the product of the extremes (cross multiply).

Example 1:  In , side a = 8, m<A = 30 and m<C = 55.  Find side c to the nearest tenth of an integer.

Since this problem refers to two angles and two sides, use the Law of Sines.

This answer makes sense, since the larger side is opposite the larger angle.


Example 2:        
                                      Find the length of side d.

Again, we are working with two sides and two angles.
Use the Law of Sines:


Example 3:      
Using the Law of Sines:

The ratio of side p to side q is 4 to 3.


If the problem asks to find a missing angle, there is another step required for the solution.

Example 4:  In the diagram, a = 55, c = 20, and m<A = 110.  Find the measure of <C to the nearest degree.

Using the Law of Sines:

Unfortunately, this is NOT the answer!! 
NEW STEP:   Using C = sin-1(.342), we have C = 19.999 = 20

(Since triangle ABC already has an obtuse angle of 110 degrees, we can eliminate the notion that sin is also positioned in Quadrant II, which would give us a second obtuse angle.