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<title>Solving SSS Triangles</title>
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<h1 align="center">Solving SSS Triangles</h1>
<p align="center"><i>&quot;SSS&quot; means &quot;Side, Side, Side&quot;</i></p>
<table border="0" align="center">
<tr>
<td><img src="images/triangle-sss.svg" alt="SSS Triangle" /></td>
<td>
<p>&quot;<i>SSS</i>&quot; is when we know <b>three sides</b> of the triangle, and want to find the <b>missing angles</b>.</p>
</td>
</tr>
</table>
<br />
<table align="center">
<tr>
<td>
<p>To solve an SSS triangle:</p>
<ul>
<li>use <a href="trig-cosine-law.html">The Law of Cosines</a> first to calculate one of the angles </li>
<li>then use The Law of Cosines again to find another angle</li>
<li>and finally use <a href="../proof180deg.html">angles of a triangle add to 180&deg;</a> to find the last angle. </li>
</ul>
</td>
</tr>
</table>
<p>We use the &quot;angle&quot; version of the Law of Cosines:</p>
<p class="center large">cos(C) = <span class="intbl">
<em>a<sup>2</sup> + b<sup>2</sup> &minus; c<sup>2</sup></em>
<strong>2ab</strong>
</span></p>
<p class="center large">cos(A) = <span class="intbl">
<em>b<sup>2</sup> + c<sup>2</sup> &minus; a<sup>2</sup></em>
<strong>2bc</strong>
</span></p>
<p class="center large">cos(B) = <span class="intbl">
<em>c<sup>2</sup> + a<sup>2</sup> &minus; b<sup>2</sup></em>
<strong>2ca</strong>
</span></p>
<p class="center"><i>(they are all the same formula, just different labels)</i></p>
<div class="example">
<h3>Example 1</h3>
<p> <img src="images/trig-sssex1.gif" width="180" height="137" alt="trig SSS example 6,7,8" /></p>
<p>In this triangle we know the three sides:</p>
<ul>
<li>a = 8, </li>
<li>b = 6 and </li>
<li>c = 7. </li>
</ul>
<p>Use The Law of Cosines first to find one of the angles. It doesn't matter which one. Let's find angle <b>A</b> first:</p>
<div class="so">cos A = (b<sup>2</sup> + c<sup>2</sup> &minus; a<sup>2</sup>) / 2bc </div>
<div class="so"> cos A = (6<sup>2</sup> + 7<sup>2</sup> &minus; 8<sup>2</sup>) / (2&times;6&times;7) </div>
<div class="so"> cos A = (36 + 49 &minus; 64) / 84</div>
<div class="so"> cos A = 0.25</div>
<div class="so"> A = cos<sup>&minus;1</sup>(0.25)</div>
<div class="so"> A = 75.5224...&deg;</div>
<div class="so"> A = <b>75.5&deg;</b> to one decimal place.</div>
<p> Next we will find another side. We use The Law of Cosines again, this time for angle B:</p>
<div class="so">cos B = (c<sup>2</sup> + a<sup>2</sup> &minus; b<sup>2</sup>)/2ca </div>
<div class="so"> cos B = (7<sup>2</sup> + 8<sup>2</sup> &minus; 6<sup>2</sup>)/(2&times;7&times;8)</div>
<div class="so"> cos B = (49 + 64 &minus; 36) / 112 </div>
<div class="so"> cos B = 0.6875</div>
<div class="so"> B = cos<sup>&minus;1</sup>(0.6875)</div>
<div class="so"> B = 46.5674...&deg;</div>
<div class="so"> B = <b>46.6&deg;</b> to one decimal place</div>
<p> Finally, we can find angle C by using 'angles of a triangle add to 180&deg;': </p>
<div class="so"> C = 180&deg; &minus; 75.5224...&deg; &minus; 46.5674...&deg;</div>
<div class="so"> C = <b>57.9&deg;</b> to one decimal place</div>
<p> Now we have completely solved the triangle i.e. we have found all its angles.</p>
</div>
<p>The triangle can have letters other than ABC:</p>
<div class="example">
<h3>Example 2</h3>
<p> <img src="images/trig-sssex2.gif" alt="trig SSS example" /></p>
<p>This is also an SSS triangle.</p>
<p>In this triangle we know the three sides x = 5.1, y = 7.9 and z = 3.5. Use The Law of Cosines to find angle X first:</p>
<div class="so">cos X = (y<sup>2</sup> + z<sup>2</sup> &minus; x<sup>2</sup>)/2yz </div>
<div class="so"> cos X = ((7.9)<sup>2</sup> + (3.5)<sup>2</sup> &minus; (5.1)<sup>2</sup>)/(2&times;7.9&times;3.5) </div>
<div class="so"> cos X = (62.41 + 12.25 &minus; 26.01)/55.3 </div>
<div class="so"> cos X = 48.65/55.3 = 0.8797... </div>
<div class="so"> X = cos<sup>&minus;1</sup>(0.8797...)</div>
<div class="so">X = 28.3881...&deg;</div>
<div class="so"> X = <b>28.4&deg;</b> to one decimal place</div>
<p> Next we will use The Law of Cosines again to find angle Y:</p>
<div class="so">cos Y = (z<sup>2</sup> + x<sup>2</sup> &minus; y<sup>2</sup>)/2zx </div>
<div class="so"> cos Y = &minus;24.15/35.7 = &minus;0.6764... </div>
<div class="so"> cos Y = (12.25 + 26.01 &minus; 62.41)/35.7 </div>
<div class="so"> cos Y = &minus;24.15/35.7 = &minus;0.6764... </div>
<div class="so"> Y = cos<sup>&minus;1</sup>(&minus;0.6764...)</div>
<div class="so">Y = 132.5684...&deg;</div>
<div class="so"> Y = <b>132.6&deg;</b> to one decimal place.</div>
<p> Finally, we can find angle Z by using 'angles of a triangle add to 180&deg;':</p>
<div class="so">Z = 180&deg; &minus; 28.3881...&deg; &minus; 132.5684...&deg;</div>
<div class="so"> Z = <b>19.0&deg;</b> to one decimal place </div>
</div>
<h2>Another Method</h2>
<table align="center">
<tr>
<td>
<p>Here is another (slightly faster) way to solve an SSS triangle:</p>
<ul>
<li>use <a href="trig-cosine-law.html">The Law of Cosines</a> first to calculate the <b>largest</b> angle </li>
<li>then use <a href="trig-sine-law.html">The Law of Sines</a> to find another angle</li>
<li>and finally use <a href="../proof180deg.html">angles of a triangle add to 180&deg;</a> to find the last angle. </li>
</ul>
</td>
</tr>
</table>
<h3>Largest Angle?</h3>
<p>Why do we try to find the largest angle first? That way the other two angles must be <b>acute</b> (less than 90&deg;) and the Law of Sines will give correct answers. </p>
<div class="center80">
<p>The <a href="trig-sine-law.html">Law of Sines</a> is difficult to use with<b> angles above 90&deg;</b>. There can be two answers either side of 90&deg; (example: 95&deg; and 85&deg;), but a calculator will only give you the smaller one.</p>
</div>
<p>So by calculating the largest angle first using the Law of Cosines, the other angles are less than 90&deg; and the Law of Sines can be used on either of them without difficulty.</p>
<div class="example">
<h3>Example 3</h3>
<p><img src="images/trig-sssex3.gif" width="276" height="187" alt="trig SSS example" /></p>
<p> B is the largest angle, so find B first using the Law of Cosines:</p>
<div class="so">cos B = (a<sup>2</sup> + c<sup>2</sup> &minus; b<sup>2</sup>) / 2ac</div>
<div class="so">cos B = (11.6<sup>2</sup> + 7.4<sup>2</sup> &minus; 15.2<sup>2</sup>) / (2&times;11.6&times;7.4)</div>
<div class="so">cos B = (134.56 + 54.76 &minus; 231.04) / 171.68</div>
<div class="so">cos B = &minus;41.72 / 171.68</div>
<div class="so">cos B = &minus;0.2430...</div>
<div class="so">B = <b>104.1&deg;</b> to one decimal place</div>
<p>Use the Law of Sines, sinC/c = sinB/b, to find angle A:
<br /> </p>
<div class="so">sin C / 7.4 = sin 104.1&deg; / 15.2</div>
<div class="so">sin C = 7.4 &times; sin 104.1&deg; / 15.2</div>
<div class="so">sin C = 0.4722...</div>
<div class="so">C = <b>28.2&deg;</b> to one decimal place</div>
<p>Find angle A using &quot;angles of a triangle add to 180&quot;:</p>
<div class="so">A = 180&deg; &minus; (104.1&deg; + 28.2&deg;)</div>
<div class="so">A = 180&deg; &minus; 132.3&deg;</div>
<div class="so">A = <b>47.7&deg;</b> to one decimal place</div>
<p>&nbsp;</p>
<p align="center" class="larger">So A = 47.7&deg;, B = 104.1&deg;, and C = 28.2&deg;</p>
</div>
<p>&nbsp;</p>
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