Lesson Law of sines

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Law of Sines



Theorem (Law of Sines)

In any triangle, the ratio of the length of each side to the sine of the angle
opposite that side is the same for all three sides:

a%2Fsin%28alpha%29 = b%2Fsin%28beta%29 = c%2Fsin%28gamma%29.

Figures 1a) and 1b) illustrate the Theorem, showing an acute and an obtuse triangles.        

The side a is opposite to the vertex A and the angle alpha.
The side b is opposite to the vertex B and the angle beta.
The side c is opposite to the vertex C and the angle gamma.
As it is stated, the Theorem is valid for any triangle.
drawing%28+200%2C+200%2C+0.5%2C+5.5%2C+0.5%2C+4.5%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+5%2C+1+%29%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+4%2C+4+%29%2C%0D%0A++++++++++++line%28+5%2C+1%2C+4%2C+4%29%2C+%0D%0A++++++++++++locate+%28+1%2C+1%2C+A%29%2C%0D%0A++++++++++++locate+%28+4.2%2C+4%2C+B%29%2C%0D%0A++++++++++++locate+%28+5%2C+1%2C+C%29%2C%0D%0A%0D%0A++++++++++++locate+%28+4.7%2C+2.8%2C+a%29%2C%0D%0A++++++++++++locate+%28+2.0%2C+2.8%2C+c%29%2C%0D%0A++++++++++++locate+%28+3.0%2C+1%2C+++b%29%2C%0D%0A%0D%0A++++++++++++locate+%28+1.5%2C+1.4%2C+alpha%29%2C%0D%0A++++++++++++locate+%28+3.8%2C+3.6%2C+beta%29%2C%0D%0A++++++++++++locate+%28+4.5%2C+1.4%2C+gamma%29%0D%0A%29
Figure 1a. Acute Triangle

drawing%28+300%2C+200%2C+0.5%2C+7.5%2C+0.5%2C+4.5%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+5%2C+1+%29%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+7%2C+4+%29%2C%0D%0A++++++++++++line%28+5%2C+1%2C+7%2C+4%29%2C+%0D%0A++++++++++++locate+%28+1%2C+1%2C+A%29%2C%0D%0A++++++++++++locate+%28+7.2%2C+4%2C+B%29%2C%0D%0A++++++++++++locate+%28+5%2C+1%2C+C%29%2C%0D%0A%0D%0A++++++++++++locate+%28+6.3%2C+2.7%2C+a%29%2C%0D%0A++++++++++++locate+%28+3.7%2C+2.8%2C+c%29%2C%0D%0A++++++++++++locate+%28+3.0%2C+1%2C+++b%29%2C%0D%0A%0D%0A++++++++++++locate+%28+1.8%2C+1.4%2C+alpha%29%2C%0D%0A++++++++++++locate+%28+6.3%2C+3.6%2C+beta%29%2C%0D%0A++++++++++++locate+%28+4.7%2C+1.4%2C+gamma%29%0D%0A%29
Figure 1b. Obtuse Triangle


The Proof of the Theorem

Let us draw an altitude of the triangle from one vertex (shown as B) perpendicularly to the opposite side. The altitude is shown in red in Figures 2a) and 2b). For the obtuse triangle, the altitude is located outside the triangle. Triangles ADB and CDB are right triangles.
From the triangle ADB       h+=+c%2Asin%28alpha%29.

In the triangle CDB, the angle BCD is equal to gamma or pi-gamma.
Since sin%28gamma%29=sin%28pi-gamma%29, we have for this triangle   h=a%2Asin%28gamma%29.
Combining these two expressions for the altitude h, we obtain c%2Asin%28alpha%29+=+a%2Asin%28gamma%29}.
Dividing both sides by sin%28alpha%29%2Asin%28gamma%29, we get the required equality

a%2Fsin%28alpha%29 = c%2Fsin%28gamma%29.

For the ratio b%2Fsin%28beta%29 the proof is similar.

drawing%28+200%2C+200%2C+0.5%2C+5.5%2C+0.5%2C+4.5%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+5%2C+1+%29%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+4%2C+4+%29%2C%0D%0A++++++++++++line%28+5%2C+1%2C+4%2C+4%29%2C+%0D%0A++++++++++++locate+%28+1%2C+1%2C+A%29%2C%0D%0A++++++++++++locate+%28+4.2%2C+4%2C+B%29%2C%0D%0A++++++++++++locate+%28+5%2C+1%2C+C%29%2C%0D%0A%0D%0A++++++++++++locate+%28+4.7%2C+2.8%2C+a%29%2C%0D%0A++++++++++++locate+%28+2.0%2C+2.8%2C+c%29%2C%0D%0A++++++++++++locate+%28+3.0%2C+1%2C+++b%29%2C%0D%0A%0D%0A++++++++++++locate+%28+1.5%2C+1.4%2C+alpha%29%2C%0D%0A++++++++++++locate+%28+4.5%2C+1.4%2C+gamma%29%2C%0D%0A%0D%0A+++++++red+%28line+%284%2C4%2C+4%2C1%29%29%2C%0D%0A+++++++locate+%284%2C+1%2C+D%29%2C%0D%0A++++++++++++locate+%28+3.6%2C+2.2%2C+h%29%0D%0A%0D%0A%29
Figure 2a. Acute Triangle

drawing%28+300%2C+200%2C+0.5%2C+7.5%2C+0.5%2C+4.5%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+5%2C+1+%29%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+7%2C+4+%29%2C%0D%0A++++++++++++line%28+5%2C+1%2C+7%2C+4%29%2C+%0D%0A++++++++++++locate+%28+1%2C+1%2C+A%29%2C%0D%0A++++++++++++locate+%28+7.2%2C+4%2C+B%29%2C%0D%0A++++++++++++locate+%28+5%2C+1%2C+C%29%2C%0D%0A%0D%0A++++++++++++locate+%28+6.3%2C+2.7%2C+a%29%2C%0D%0A++++++++++++locate+%28+3.7%2C+2.8%2C+c%29%2C%0D%0A++++++++++++locate+%28+3.0%2C+1%2C+++b%29%2C%0D%0A%0D%0A++++++++++++locate+%28+1.8%2C+1.4%2C+alpha%29%2C%0D%0A++++++++++++locate+%28+6.3%2C+3.6%2C+beta%29%2C%0D%0A++++++++++++locate+%28+4.7%2C+1.4%2C+gamma%29%2C%0D%0A%0D%0A+++++++red+%28line+%287%2C4%2C+7%2C1%29%29%2C%0D%0A+++++++red+%28line+%285%2C1%2C+7%2C1%29%29%2C%0D%0A+++++++locate+%287%2C+1%2C+D%29%2C%0D%0A%0D%0A++++++++++++locate+%28+7.2%2C+2.2%2C+h%29%0D%0A%29
Figure 2b. Obtuse Triangle

Below are couple examples to illustrate usage the Law of Sines in solving triangles.
Examples in this lesson relate to the simplest cases, when the solution always exists and is unique.
Consideration of more complicated cases is given in another lesson, Solve triangles using Law of Sines, in the module Trigonometry.
The term "to solve the triangle" means "to calculate unknown elements of the triangle using given data".

Example 1. Using the Law of Sines to Solve ASA Triangle.

Solve the triangle:   alpha = 35°,   b = 5,   gamma=70°.
This is so named ASA case when the side of the triangle and two its adjacent angles are given.
Figure 3 is the sketch that illustrates the triangle we are going to solve.

First, find the third angle, beta. Since
alpha%2Bbeta%2Bgamma=180°, we have beta = 180°-alpha%2Bgamma=180°-35°-70°=75°.
Next calculate side a using Law of Sines with known side b along with angles alpha and beta:

a%2Fsin%28alpha%29 = b%2Fsin%28beta%29,     a+=+b%2A%28sin%28alpha%29%2Fsin%28beta%29%29 = b*sin(35°)/sin(75°) = 5*0.574/0.966 = 2.969.

Now calculate side c using Law of Sines with known side b and angles alpha and gamma:

c%2Fsin%28gamma%29 = b%2Fsin%28beta%29,     c+=+b%2A%28sin%28gamma%29%2Fsin%28beta%29%29 = b*sin(70°)/sin(75°) = 5*0.940/0.966 = 4.864.
drawing%28+200%2C+200%2C+0.5%2C+5.5%2C+0.5%2C+4.5%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+5%2C+1+%29%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+4%2C+3.2+%29%2C%0D%0A++++++++++++line%28+5%2C+1%2C+4%2C+3.2%29%2C+%0D%0A++++++++++++locate+%28+1%2C+1%2C+A%29%2C%0D%0A++++++++++++locate+%28+4.2%2C+3.2%2C+B%29%2C%0D%0A++++++++++++locate+%28+5%2C+1%2C+C%29%2C%0D%0A%0D%0A++++++++++++locate+%28+4.7%2C+2.3%2C+a%29%2C%0D%0A++++++++++++locate+%28+2.0%2C+2.3%2C+c%29%2C%0D%0A++++++++++++locate+%28+3.0%2C+1%2C+++b%29%2C%0D%0A%0D%0A++++++++++++locate+%28+1.5%2C+1.4%2C+alpha%29%2C%0D%0A++++++++++++locate+%28+3.8%2C+3.0%2C+beta%29%2C%0D%0A++++++++++++locate+%28+4.5%2C+1.4%2C+gamma%29%0D%0A%29
Figure 3. Example 1

The solution is unique in this case.

Example 2. Using the Law of Sines to Solve SAA Triangle.

Solve the triangle:   alpha = 35°,   b = 5,   beta=75°.
This is so named SAA case: the side of the triangle, one adjacent and one opposite angle are    
given.
Figure 4 illustrates the triangle we are going to solve.

First find the third angle, gamma. Since
alpha%2Bbeta%2Bgamma=180°, we have gamma = 180°-alpha%2Bbeta=180°-35°-75°=70°.
Next calculate side a using Law of Sines with known side b and angles alpha and beta:

a%2Fsin%28alpha%29 = b%2Fsin%28beta%29,     a+=+b%2A%28sin%28alpha%29%2Fsin%28beta%29%29 = b*sin(35°)/sin(75°) = 5*0.574/0.966 = 2.969.

Now calculate side c using Law of Sines with known side b and angles alpha and gamma:

c%2Fsin%28gamma%29 = b%2Fsin%28beta%29,     c+=+b%2A%28sin%28gamma%29%2Fsin%28beta%29%29 = b*sin(70°)/sin(75°) = 5*0.940/0.966 = 4.864.
drawing%28+200%2C+200%2C+0.5%2C+5.5%2C+0.5%2C+4.5%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+5%2C+1+%29%2C+%0D%0A++++++++++++line%28+1%2C+1%2C+4%2C+3.2+%29%2C%0D%0A++++++++++++line%28+5%2C+1%2C+4%2C+3.2%29%2C+%0D%0A++++++++++++locate+%28+1%2C+1%2C+A%29%2C%0D%0A++++++++++++locate+%28+4.2%2C+3.2%2C+B%29%2C%0D%0A++++++++++++locate+%28+5%2C+1%2C+C%29%2C%0D%0A%0D%0A++++++++++++locate+%28+4.7%2C+2.3%2C+a%29%2C%0D%0A++++++++++++locate+%28+2.0%2C+2.3%2C+c%29%2C%0D%0A++++++++++++locate+%28+3.0%2C+1%2C+++b%29%2C%0D%0A%0D%0A++++++++++++locate+%28+1.5%2C+1.4%2C+alpha%29%2C%0D%0A++++++++++++locate+%28+3.8%2C+3.0%2C+beta%29%2C%0D%0A++++++++++++locate+%28+4.5%2C+1.4%2C+gamma%29%0D%0A%29
Figure 4. Example 2

The solution is unique in this case.

As I said above, the usage of the Law of Sines for more complicated cases is described in another lesson, Solve triangles using Law of Sines, in the module Trigonometry.
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