Lesson Law of sines

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<H3>Law of Sines</H3>
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 <TD>
Theorem (the 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/sin(alpha)}}} = {{{b/sin(beta)}}} = {{{c/sin(gamma)}}}.

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

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.
 </TD>
 <TD>
{{{drawing( 200, 200, 0.5, 5.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 4, 4 ),
 line( 5, 1, 4, 4), 
 locate ( 1, 1, A),
 locate ( 4.2, 4, B),
 locate ( 5, 1, C),

locate ( 4.7, 2.8, a),
            locate ( 2.0, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.5, 1.4, alpha),
 locate ( 3.8, 3.6, beta),
 locate ( 4.5, 1.4, gamma)
)}}}
Figure 1a. Acute Triangle 
 </TD>
 <TD>
{{{drawing( 300, 200, 0.5, 7.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 7, 4 ),
 line( 5, 1, 7, 4), 
 locate ( 1, 1, A),
 locate ( 7.2, 4, B),
 locate ( 5, 1, C),

locate ( 6.3, 2.7, a),
            locate ( 3.7, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.8, 1.4, alpha),
 locate ( 6.3, 3.6, beta),
 locate ( 4.7, 1.4, gamma)
)}}}
Figure 1b. Obtuse Triangle
 </TD>
 </TR>
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The Proof of the Theorem

Let us draw an altitude of the triangle from one vertex (shown as B) perpendicularly&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
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 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{h = c*sin(alpha)}}}.

In the triangle CDB, the angle BCD is equal to {{{gamma}}} or {{{pi-gamma}}} (see the Figures 2a) and 2b)).
Since {{{sin(gamma)=sin(pi-gamma)}}}, we have for this triangle &nbsp;&nbsp;{{{h=a*sin(gamma)}}}.
Combining these two expressions for the altitude h, we obtain {{{c*sin(alpha) = a*sin(gamma)}}}}.
Dividing both sides by {{{sin(alpha)*sin(gamma)}}}, we get the required equality

{{{a/sin(alpha)}}} = {{{c/sin(gamma)}}}.

For the ratio {{{b/sin(beta)}}} the proof is similar.

</TD>
 <TD>
{{{drawing( 200, 200, 0.5, 5.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 4, 4 ),
 line( 5, 1, 4, 4), 
 locate ( 1, 1, A),
 locate ( 4.2, 4, B),
 locate ( 5, 1, C),

locate ( 4.7, 2.8, a),
            locate ( 2.0, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.5, 1.4, alpha),
            locate ( 4.5, 1.4, gamma),

red (line (4,4, 4,1)),
       locate (4, 1, D),
            locate ( 3.6, 2.2, h)

)}}}
Figure 2a. Acute Triangle 
 </TD>
 <TD>
{{{drawing( 300, 200, 0.5, 7.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 7, 4 ),
 line( 5, 1, 7, 4), 
 locate ( 1, 1, A),
 locate ( 7.2, 4, B),
 locate ( 5, 1, C),

locate ( 6.3, 2.7, a),
            locate ( 3.7, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.8, 1.4, alpha),
            locate ( 6.3, 3.6, beta),
            locate ( 4.7, 1.4, gamma),

red (line (7,4, 7,1)),
       red (line (5,1, 7,1)),
       locate (7, 1, D),

locate ( 7.2, 2.2, h)
)}}}
Figure 2b. Obtuse Triangle
 </TD>
 </TR>
</TABLE> 
Below are couple examples illustrating how to use the Law of Sines to solve triangles.
Examples in this lesson relate to the simplest cases, when the solution always exists and is unique.
More complicated cases are considered in another lesson, <A HREF = http://www.algebra.com/algebra/homework/Trigonometry-basics/Solve-triangles-using-Law-of-Sines.lesson>Solve triangles using Law of Sines</A> under the topic Trigonometry of the section Algebra-II in this site.
The term "to solve the triangle" means "to calculate unknown elements of the triangle using given data".

Example 1. Use the Law of Sines to solve ASA Triangle.
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Solve the triangle: &nbsp;&nbsp;{{{alpha}}} = 35�, &nbsp;&nbsp;b = 5, &nbsp;&nbsp;{{{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+beta+gamma}}}=180�, we have {{{beta}}} = 180�-{{{alpha+gamma}}}=180�-35�-70�=75�.
Next calculate the side a using the Law of Sines with the known side b and the angles {{{alpha}}} and {{{beta}}}:&nbsp;&nbsp;&nbsp;&nbsp;

{{{a/sin(alpha)}}} = {{{b/sin(beta)}}}, &nbsp;&nbsp;&nbsp;&nbsp;{{{a = b*(sin(alpha)/sin(beta))}}} = b*sin(35�)/sin(75�) = 5*0.574/0.966 = 2.969.

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

{{{c/sin(gamma)}}} = {{{b/sin(beta)}}}, &nbsp;&nbsp;&nbsp;&nbsp;{{{c = b*(sin(gamma)/sin(beta))}}} = b*sin(70�)/sin(75�) = 5*0.940/0.966 = 4.864.
 </TD>
 <TD>
{{{drawing( 200, 200, 0.5, 5.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 4, 3.2 ),
 line( 5, 1, 4, 3.2), 
 locate ( 1, 1, A),
 locate ( 4.2, 3.2, B),
 locate ( 5, 1, C),

locate ( 4.7, 2.3, a),
            locate ( 2.0, 2.3, c),
            locate ( 3.0, 1,   b),

locate ( 1.5, 1.4, alpha),
 locate ( 3.8, 3.0, beta),
 locate ( 4.5, 1.4, gamma)
)}}}
Figure 3. Example 1
 </TD>
 </TR>
</TABLE> 
The solution is unique in this case.

Example 2. Use the Law of Sines to solve SAA Triangle.
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 <TR>
 <TD>
Solve the triangle: &nbsp;&nbsp;{{{alpha}}} = 35�, &nbsp;&nbsp;b = 5, &nbsp;&nbsp;{{{beta}}}=75�.
This is so named SAA case: the side of the triangle, one adjacent and one opposite angle are&nbsp;&nbsp;&nbsp;&nbsp;
given.
Figure 4 illustrates the triangle we are going to solve.

First find the third angle, {{{gamma}}}. Since
{{{alpha+beta+gamma}}}=180�, we have {{{gamma}}} = 180�-{{{alpha+beta}}}=180�-35�-75�=70�.
Next calculate the side a using the Law of Sines with the known side b and the angles {{{alpha}}} and {{{beta}}}:&nbsp;&nbsp;&nbsp;&nbsp;

{{{a/sin(alpha)}}} = {{{b/sin(beta)}}}, &nbsp;&nbsp;&nbsp;&nbsp;{{{a = b*(sin(alpha)/sin(beta))}}} = b*sin(35�)/sin(75�) = 5*0.574/0.966 = 2.969.

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

locate ( 4.7, 2.3, a),
            locate ( 2.0, 2.3, c),
            locate ( 3.0, 1,   b),

locate ( 1.5, 1.4, alpha),
 locate ( 3.8, 3.0, beta),
 locate ( 4.5, 1.4, gamma)
)}}}
Figure 4. Example 2
 </TD>
 </TR>
</TABLE> 
The solution is unique in this case.

As I said above, more examples and more complicated cases are considered in the lesson <A HREF = http://www.algebra.com/algebra/homework/Trigonometry-basics/Solve-triangles-using-Law-of-Sines.lesson>Solve triangles using Law of Sines</A> under the topic Trigonometry of the section Algebra-II in this site.

Another, more geometric proof of the Law of Sines theorem is presented in the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Law-of-sines-the-Geometric-Proof.lesson>Law of sines - the Geometric Proof</A> under the current topic Triangles of the section Geometry in this site.

For navigation over the lessons on Properties of Triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/Compendium-of-properties-of-triangles.lesson>Properties of Trianles</A>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.